The **atomic radius **of boron atom is 0.09nm

**What is an atomic radius?**

**Atomic radius**, also known as** atomic radii**, is the total distance between an atom's nucleus and its electron's outermost orbital.

**Atomic radius **is thus defined as: The distance from the nucleus's center to the edge of its surrounding electron shells.

The radius of an atom is comparable to a circle's radius. The electron's most distant orbital is likened to the circle's outside edge, and the nucleus to the circle's center. The position of the outermost electron is unclear, making it challenging to calculate the **atomic radii**.

According to the question,

1m = 10-9nm

So, 9*10-11m = 0.09nm

The **atomic radius **of boron atom is 0.09nm

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Consider the following processes: 1. A ─→ B delta H = ─100 kJ 2. B ─→ C + D delta H = ─50 kJ 3. E ─→ D delta H = ─25 kJ What is the delta H for the process: A ─→ C + E?

The **enthalpy change, **ΔH of the process: A → C + E is -125 kJ.

The **enthalpy change** of a reaction is the sum of the heat changes that occur as reactants combine to form products.

The **enthalpy change** of the reaction is given by the following formula:

The delta H for the process A → C + E is given as follows;

1. A ─→ B ΔH = ─100 kJ

2. B ─→ C + D ΔH = ─50 kJ

3. E ─→ D delta H = ─25 kJ

-3. D → E = +25 kJ

A → C + D = 1 + 2 + (- 3)

A → C + E = (- 100 - 50 + 25) kJ

A → C + E = -125 kJ

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A marine biologist is preparing a deep-sea submersible for a dive. The sub stores breathing air under high pressure in aspherical air tank that measures 73.0 cm wide.The biologist estimates she will need 8200. L of air for the dive. Calculate the pressure to which this volume of air must becompressed in order to fit into the air tank. Write your answer in atmospheres. Round your answer to 3 significant digits.0atm0.0XS ?EoloPEBH

To solve the problem we will assume the following:

1. Air behaves as an ideal gas during all the process.

2. The initial air equivalent to 8200L is at atmospheric pressure. It means 1 atm.

3. The temperature remains constant.

Taking into account the above, we can apply the Boyle-Marriote Law that relates the change in pressure and volume at constant temperature. The equation that we will use will be:

[tex]P_1V_1=P_2V_2[/tex]Where,

P1 is the atmospheric pressure. 1atm

V1 is the initial volume of air required, 8200L

P2 is the final pressure we want to find

V2 is the final volume, it means the volume of the spherical air tank. We will calculate this volume using the volume equation for a sphere:

[tex]V_2=\frac{4}{3}\pi r^3[/tex]r is the radius of the sphere, r=73cm/2=36.5cm

So, the volume of the spherical air tank will be:

[tex]\begin{gathered} V_2=\frac{4}{3}\pi\times(36.5cm)^3=20.4\times10^4cm^3 \\ V_2=20.4\times10^4cm^3\times\frac{1L}{1000cm^3}=204L \end{gathered}[/tex]No, we clear P2 from the first equation and replace known data:

[tex]\begin{gathered} P_2=\frac{P_1V_1}{V_2} \\ P_2=\frac{1atm\times8200L}{204L} \\ P_2=40.3atm \end{gathered}[/tex]**The pressure of the gas must be ****40.3 atm**

**Answer: ****40.3 **

The cost of electricity to deposit 10gm of Mg is Rs. 60. How much would it cost to deposit 100gm of copper from CuSO

, ? (At. wt. of Cu = 63.5)

The cost of **e****l****e****c****t****r****i****c****i****t****y**** **to deposit 100g of **c****o****p****p****e****r**** **from CuSO₄ will be **R****s****2****2****7****.****1**

We know, **m**** **= **z****i****t**

1unit current= 1kwh

Energy=**V****i****t**

**m**= (E. weight/F)×it

m proportional to (E. weight)×**c****o****s****t**

m₁/m₂ = E₁/E₂ × C₁/C₂

**E****.****w****e****i****g****h****t** = (Molecular weight/n-factor)

According to question,

Cu² + 2e- **-****-****-**** **** **Cu

Mg² + 2e- **-****-****-**** **Mg { n-factor of both=**2**}

**E****₁**** **= (molecular weight of Mg/2) = (24÷2) = 12

**E****₂**** **= (molecular weight of Cu/2) = (63.5) = 31.7

So, M₁/M₂ **=** E₁×C₁/E₂×C₂ (**C****₁**=cost given)

10/100 **=** 12×60/31.7×C₂

C₂**=** **R****s****2****2****7****.****1**

Therefore the cost of electricity to deposit 100g of copper from CuSO₄ will be **R****s****2****2****7****.****1**

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I cant figure this one out. How do I know ΔH∘ for the preceding reaction without knowing the standard enthalpie of F2?

The **enthalpy of formation **(ΔH°) of the given **chemical reaction** is equal to -905.3 KJ/mol.

The **standard enthalpy of formation **is defined as the change in enthalpy when one mole of a substance is formed from its pure elements under the **standard state **(1 atm of pressure and 298.15 K).

The** standard enthalpy of formation **is the energy released or absorbed when one mole of a substance is formed from its constituents under **standard conditions**. The symbol of the standard enthalpy of formation is represented by [tex]\triangle {H^o_f}[/tex].

The equation for the **standard enthalpy change **of formation for a balanced chemical reaction is shown below,

ΔH° (reaction) = ∑ [ΔH° of (products)− ∑ ΔH° of (Reactants)]

The **standard enthalpy of the formation **of fluorine gas = 0

The standard enthalpy of the formation of **iodine** gas = 62.23 KJ/mol

The standard enthalpy of the formation of** IF₅ **= -843 KJ/mol

ΔH° (reaction) = - 843 - 62.23 = - 905 .3 KJ/mol

Therefore, the **enthalpy of formation (ΔH°)** of the given chemical reaction is -905.3 KJ/mol.

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yes or no ? experimentation with corn and other crops led to the development of new fuels called biofuels

Yes, the **experimentation** with **corn** and other crops led to the development of new fuels called **biofuels**.

**Biofuel** is a **biodegradable**, inexhaustible, fuel that is produced from Biomass. Biofuel is considered the easiest available and pure fuel on the earth. **Biofuels** are manufactured from biomass such as **wood** and **straw**, which are liberated by **direct combustion **of dry matter and converted into a gaseous and liquid fuel.

The** organic matter **such as sludge, sewage, and vegetable oils, can be changed into biofuels by a wet process such as **fermentation** and **digestion**. Biofuel is available in all regions of the world, and mainly includes fuels such as **Biodiesel, Bioethanol, **and** Bio methanol**.

The two types of biofuels are **bioethanol** and **biodiesel** most commonly used these days. Both of these biofuels are the first generation of **biofuel technology.**

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How much work (in J ) is involved in a chemical reaction if the volume decreases from 4.25 to 1.97 L against a constant pressure of 0.838 atm ?

The amount of **work** (in J ) is involved in a chemical reaction if the **volume** decreases from 4.25 to 1.97 L against a constant pressure of 0.838 atm is: 8.78 Joules.

The **work done** by an object can be calculated by multiplying the force acting on it by the perpendicular distance covered by the physical object. Also, the **work done** during a chemical reaction can be determined by multiplying the change in **volume** by the accompanying level of pressure.

Mathematically, the work done by a system is given by the formula:

Work done = PΔV = P(V₂ - V₁)

Where:

P represents the pressure.V represents theSubstituting the given parameters into the formula, we have;

Work done = 0.038(4.25 - 1.97)

Work done = 0.038(2.28)

**Work done** = 0.08664 L.atm

Next, we would convert L.atm to Joules:

Work done = 101.325 × 0.08664

**Work done** = 8.78 Joules.

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An atom with 3 protons in the nucleus and 3 electrons in the orbitals would have what overall charge?

Ans6

wer:

b

Explanation:

Show the conversions required to solve this problem and calculate the grams of Al2O3 .

This problem is an example of a gram-to-gram stoichiometry problem. You are given the mass of

Al

Al

in grams and you are asked to find the mass of

Al

2

O

3

Al

2

O

3

in grams. For questions such as this, the strategy is to convert from grams of

Al

Al

to moles of

Al

Al

, then to moles of

Al

2

O

3

Al

2

O

3

, and finally to grams of

Al

2

O

3

Al

2

O

3

.

grams Al⟶moles Al⟶moles

Al

2

O

3

⟶grams

Al

2

O

3

grams

Al

⟶

moles

Al

⟶

moles

Al

2

O

3

⟶

grams

Al

2

O

3

To convert grams of

Al

Al

to moles of

Al

Al

, you use the molar mass of

Al

Al

,

26.98 g/mol

26.98

g/mol

. The balanced chemical equation is used to relate the moles of

Al

Al

to the moles of

Al

2

O

3

Al

2

O

3

. There are 2 moles of

Al

2

O

3

Al

2

O

3

formed for every 4 moles of

Al

Al

that react. To convert moles of

Al

2

O

3

Al

2

O

3

to grams of

Al

2

O

3

Al

2

O

3

, you use the molar mass of

Al

2

O

3

Al

2

O

3

,

101.96 g/mol

101.96

g/mol

. This can be done one conversion at a time, or you can string the conversions together.

30.6 g Al×

1 mole Al

26.98 g Al

×

2 moles

Al

2

O

3

4 moles Al

×

101.96 g

Al

2

O

3

1 mole

Al

2

O

3

=57.8 g

Al

2

O

3

Hopefully the picture will show up this time.

Al

Al

in grams and you are asked to find the mass of

Al

2

O

3

Al

2

O

3

in grams. For questions such as this, the strategy is to convert from grams of

Al

Al

to moles of

Al

Al

, then to moles of

Al

2

O

3

Al

2

O

3

, and finally to grams of

Al

2

O

3

Al

2

O

3

.

grams Al⟶moles Al⟶moles

Al

2

O

3

⟶grams

Al

2

O

3

grams

Al

⟶

moles

Al

⟶

moles

Al

2

O

3

⟶

grams

Al

2

O

3

To convert grams of

Al

Al

to moles of

Al

Al

, you use the molar mass of

Al

Al

,

26.98 g/mol

26.98

g/mol

. The balanced chemical equation is used to relate the moles of

Al

Al

to the moles of

Al

2

O

3

Al

2

O

3

. There are 2 moles of

Al

2

O

3

Al

2

O

3

formed for every 4 moles of

Al

Al

that react. To convert moles of

Al

2

O

3

Al

2

O

3

to grams of

Al

2

O

3

Al

2

O

3

, you use the molar mass of

Al

2

O

3

Al

2

O

3

,

101.96 g/mol

101.96

g/mol

. This can be done one conversion at a time, or you can string the conversions together.

30.6 g Al×

1 mole Al

26.98 g Al

×

2 moles

Al

2

O

3

4 moles Al

×

101.96 g

Al

2

O

3

1 mole

Al

2

O

3

=57.8 g

Al

2

O

3

Hopefully the picture will show up this time.

how many atoms are in 6 moles of oxygen

Each oxygen molecule has two atoms

6 * 2 = 12

12

6 * 2 = 12

12

Please help me it’s due today!!!!

The following are the examples of a** chemical property** of a **substance;**

1) A banana reacts with orange and turns brown

2) Iron will rust when exposed to oxygen and water

3) Iron will react with acid to form iron chloride.

What is a chemical reaction?The term **chemical reaction** has to do with the change in the properties of substances that are combined together in order to yield a **product**. Recall that the properties of the reactant change as the product is formed.

When a chemical reaction occurs

1: A new color may appear

2) A new gas may be seen

3) The temperature may be changed

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Can someone help me with this? I’m not understanding what it’s asking of me

**Answer: You need to research what ever you are working on and find 2 examples.**

**Explanation:**

34.8 g of Na₂O are used to form a solution with a volume of 450.0 mL L. What is the

molarity?

34.89

**Answer:**

1.25 M

**Explanation:**

**(Step 1)**

**Convert grams to moles using the molar mass of Na₂O.**

Atomic Mass (Na): 22.990 g/mol

Atomic Mass (O): 15.999 g/mol

Molar Mass (Na₂O): 2(22.990 g/mol) + 15.999 g/mol

**Molar Mass (Na₂O):** 61.979 g/mol

34.8 g Na₂O 1 mole

---------------------- x --------------------- = 0.561 moles Na₂O

61.979 g

**(Step 2)**

**Convert milliliters to liters.**

1,000 mL = 1 L

450.0 mL 1 L

------------------ x ------------------- = 0.4500 L

1,000 mL

**(Step 3)**

**Calculate the molarity using the molarity ratio.**

Molarity = moles / volume (L)

Molairty = 0.561 moles / 0.4500 L

Molarity = 1.25 M

Question 1 (3 points)

Listed in the Item Bank are some important labels for sections of the image below.

Select the label for each corresponding area it identifies in

-noble gas

-alkaline earth metals

-transition metals

-alkali metals

-nonmetal

-metalloids

-halogens

The first column represent the **alkali metal**, the second column represent **alkaline earth metals**, the 3rd horizontal middle block is d block, the right 6 columns of the periodic table represent p block and the rightest column is **noble gas**.

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The volume of a gas is 480 mL at 45.0 oC the temperature is increased to 60.0 oC at a constant pressure what is the new volume

**Answer**

The new volume = **0.503 L or 503 mL**

**Explanation**

Given:

Initial volume, V₁ = 480 mL = (480/1000) = 0.480 L

Initial temperature, T₁ = 45.0°C = (45 + 273 K) = 318 K

Final temperature, T₂ = 60.0°C = (60 + 273 K) = 333 K

What to find:

The new volume, V₂ when the temperature increases to 60.0°C.

Step-by-step solution:

Since the pressure is constant, the new volume, V₂ can be calculated using **Charle's law formula.**

Plugging the values of the given parameters into the formula, we have

[tex]V_2=\frac{0.480L\times333K}{318K}=0.503\text{ }L[/tex]The new volume, V₂ when the temperature increased to 60.0°C = **0.503 L or **(0.503 x 1000 mL) = **503 mL**

For the reaction A (g) → 3 B (g), Kp = 67900 at 298 K. When ∆G = -14.2 kJ/mol, what is the partial pressure of A when the partial pressure of B is 2.00 atm for this reaction at 298 K.

The partial pressure is the individual pressure of a **gas**. The partial pressure of gas A is 1.11 × 10⁻⁴atm.

What is **partial pressure**?

In a mixture of gases, the partial pressure of a gas is the individual pressure of that gas in the **mixture** of gases.

Given the reaction; **A (g) → 3 B (g)**

**Kp = P(B)³/PA**

Where Kp = 67900

PB = 2.00 atm

67900 = (2)³/PA

PA = (2)³/67900

PA = 1.11 × 10⁻⁴atm

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explain Maxwell equations and maxwell thermodynamics relations; it's significance and application to ideal gases?

**Answer:**

In order to understand the **Maxwell equations** and relations derived from Maxwell relations, you will have to know Euler's Reciprocity.

It states that for any state thermodynamic quantity (let x and y) having a state function phi, it must satisfy the following condition.

[tex](\frac{ {∂}^{2}\phi }{∂x \cdot∂y}) = (\frac{ {∂}^{2}\phi }{∂y \cdot∂x})[/tex]

**Maxwell equations:** These are a set of equations derived from the application of Euler's Reciprocity. The four Maxwell equations are as follows:

[tex]dH = TdS + VdP[/tex]

[tex]dG = VdP - SdT[/tex]

[tex]dA = -PdV - SdT[/tex]

[tex]dU = TdS - PdV[/tex]

Let's derive each Maxwell relations step by step,

**1) dH = TdS + VdP**

Enthalpy is a function of Entropy & Pressure

[tex] \sf \qquad \qquad H = f(S,P)[/tex]

[tex]dH = TdS + VdP[/tex]

[tex]dH = (\frac{∂H}{∂S})_{_P} dS + (\frac{∂H}{∂P})_{_S}dP[/tex]

Comparing both the above equation,

[tex]T = (\frac{∂H}{∂S})_{_P}[/tex]

[tex]V =(\frac{∂H}{∂P})_{_S}[/tex]

now we know that Enthalpy is a state function hence applying cross reciprocality,

[tex](\frac{∂V}{∂S})_{_P}= (\frac{∂T}{∂P})_{_S}[/tex]

This is called the **first** Maxwell relation,

Similarly,

**2) dG = -SdT + VdP**

Free energy is a function of Temperature & Pressure,

[tex] \sf \qquad \qquad G = f(T,P)[/tex]

[tex]dG = -SdT + VdP[/tex]

[tex]dG = (\frac{∂G}{∂P})_{_T} dP + (\frac{∂G}{∂T})_{_P}dT[/tex]

Comparing both the above equation,

[tex](\frac{∂G}{∂P})_{_T} = V [/tex]

[tex] (\frac{∂G}{∂T})_{_P}= -S[/tex]

now we know that Free energy is a state function hence applying cross reciprocality,

[tex]-(\frac{∂S}{∂T})_{_P}= (\frac{∂V}{∂P})_{_T}[/tex]

This is called the **second** Maxwell relation,

**3) dA = -PdV - SdT **

helmholtz free energy is a function of Temperature & Volume,

[tex] \sf \qquad \qquad A = f(T,V)[/tex]

[tex]dA = -PdV - SdT[/tex]

[tex]dA = (\frac{∂A}{∂V})_{_T} dV + (\frac{∂A}{∂T})_{_V}dT[/tex]

Comparing both the above equation,

[tex](\frac{∂A}{∂V})_{_T} = -P [/tex]

[tex] (\frac{∂A}{∂T})_{_V}= -S[/tex]

now we know that helmholtz free energy is a state function hence applying cross reciprocality,

[tex]-(\frac{∂S}{∂V})_{_T}= -(\frac{∂P}{∂T})_{_V}[/tex]

[tex](\frac{∂S}{∂V})_{_T}= (\frac{∂P}{∂T})_{_V}[/tex]

This is called the **third** Maxwell relation,

**4) dU = TdS - PdV **

Internal energy is a function of Entropy & Volume,

[tex] \sf \qquad \qquad A = f(S,V)[/tex]

[tex]dU = TdS - PdV [/tex]

[tex]dU = (\frac{∂U}{∂S})_{_V} dS + (\frac{∂U}{∂V})_{_S}dV[/tex]

Comparing both the above equation,

[tex](\frac{∂U}{∂V})_{_S} = -P [/tex]

[tex] (\frac{∂U}{∂S})_{_V}= T[/tex]

now we know that Internal energy is a state function hence applying cross reciprocality,

[tex](\frac{∂T}{∂V})_{_S}= -(\frac{∂P}{∂S})_{_V}[/tex]

This is called the **fourth** Maxwell relation,

The **main significance** of the Maxwell relation is that those thermodynamic quantities which are **unmeasurable** can be **replaced with measurable** quantities with the help of the Maxwell relation.

The **derivative** of the extensive asset in relation to the extensive asset gives the **intensive asset**; with respect to the intensive asset, the derivative of the extensive asset gives the **extensive asset.** This is the **result of the overall** **Maxwell relations**.

The coefficient of expansion and compression of a gas in thermodynamics is the application of the Maxwell relations.

There are 3 coefficients introduced,

*Coefficient of thermal expansion (expansivity) α*

[tex]α= \frac{1}{V} (\frac{∂V}{∂T} )_{_P}[/tex]

**Coefficient of isothermal compressibility β**

[tex]β = -\frac{1}{V} (\frac{∂V}{∂P} )_{_T}[/tex]

**Isochoric thermal expansion coefficient ****γ**

**[tex] γ= \frac{1}{P} (\frac{∂P}{∂T} )_{_V}[/tex]**

For ideal gases,

PV = nRT

For one mole ideal gas (n=1),

**PV = RT**

Taking derivative with respect to T at constant pressure,

[tex]V(\frac{∂P}{∂T} )_{_P}+ P(\frac{∂V}{∂T} )_{_P}= R(\frac{∂T}{∂T} )_{_P}[/tex]

At constant pressure ∂P = 0, & R.H.S = 1, hence

[tex](\frac{∂V}{∂T} )_{_P}= \frac{R}{P}[/tex]

[tex]α= \frac{1}{V} (\frac{∂V}{∂T} )_{_P}[/tex]

[tex]α= \frac{1}{V} \cdot \frac{R}{P}[/tex]

[tex]\sf Also, PV=RT\\ \frac{R}{PV} = \frac{1}{T}[/tex]

[tex]\boxed{α= \frac{1}{T}}[/tex]

Following the same procedure, by taking derivating w.r.t. pressure at constant temperature.

[tex]V(\frac{∂P}{∂P} )_{_T}+ P(\frac{∂V}{∂P} )_{_T}= R(\frac{∂T}{∂P} )_{_T}[/tex]

[tex]V+ P(\frac{∂V}{∂P} )_{_T}= 0[/tex]

[tex](\frac{∂V}{∂P} )_{_T}= \frac{-V}{P}[/tex]

**Substituting the above value in,**

**[tex]β = -\frac{1}{V} (\frac{∂V}{∂P} )_{_T}[/tex]**

**[tex]β = -\frac{1}{V} \cdot \frac{-V}{P}[/tex]**

**[tex] \boxed{β = \frac{1}{P}}[/tex]**

Repeating the same procedure again, i.e. derivative w.r.t. T at constant volume

[tex]V(\frac{∂P}{∂T} )_{_V}+ P(\frac{∂V}{∂T} )_{_V}= R(\frac{∂T}{∂T} )_{_V}[/tex]

At constant Volume ∂V = 0, and R.H.S = 1, hence overall equation becomes,

[tex]V(\frac{∂P}{∂T} )_{_V} = R \\ (\frac{∂P}{∂T} )_{_V} = \frac{R}{V}[/tex]

Substituting above value in,

[tex]γ= \frac{1}{P} (\frac{∂P}{∂T} )_{_V}[/tex]

[tex]γ= \frac{1}{P} \cdot \frac{R}{V}[/tex]

[tex] \boxed{γ= \frac{1}{T}}[/tex]

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The **Maxwell **equation in **thermodynamics **is very important and useful because the set of relations allows the scientists to change specific unknown quantities.

The** Maxwell equation** in **thermodynamics **is very useful because this is the set of relations that allows physicists to change certain unknown quantities that are hard to measure in the real world. These quantities need to be replaced by many easily measured quantities. **Maxwell's relations** are a set of equations in **thermodynamics **that are derivable from the second derivatives and from the definitions of the **thermodynamic **potentials. These relations are named for the nineteenth-century physicist James Clerk **Maxwell**. So entropy and pressure are the natural variables of enthalpy. **Maxwell relations** are **thermodynamic **equations that establish the relations between various **thermodynamic **quantities in equilibrium and other fundamental quantities known as **thermodynamical **potentials

So we can conclude that **Maxwell's thermodynamics **are the set of relations allows the scientists to change unknown quantities.

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a gaseous product of a chemical reaction is collected at 285k and 1.3atm. what was the molar mass of the gas in grams per mole if 6.2 g of gas occupies 5.4l

The **molar mass **of the **gas **in grams per mole if 6.2 g of gas occupies 5.4L is 336.97 grams.

**Molar mass **is defined as the **mass **of a **sample **of a certain chemical divided by the quantity of the material, expressed as the number of moles in the sample.

It can also be defined as the **product **of the mass of a **specific substance **and the amount of that substance in the sample.

Given Pressure = 1.3 atm

Temperature = 285 K

Volume = 5.4 L

Gas content = 8.3

So, PV = nRT

n = RT / PV

n = 8.3 x 285 / 1.3 x 5.4

n = 336.97 grams

Thus, the **molar mass **of the **gas **in grams per mole if 6.2 g of gas occupies 5.4L is 336.97 grams.

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At STP, exactly 1 mole of any gas occupies 22.4 L. What size container (in Liters) do you need to hold 2.1mol O₂ gas at STP?(Show all work)A) 44.2B) 47.0C) 40.0D) 55.5

**Answer**

**B) 47.0**

**Explanation**

If at STP, exactly **1 mole of any gas** occupies **22.4 L**

Then, **2.1 moles of O₂ gas **at STP will occupy **x L container**

To get x L, cross multiply and divide both sides by 1 mole.

[tex]x\text{ }L\text{ }container=\frac{2.1\text{ }mol}{1\text{ }mol}\times22.4\text{ }L=47.0\text{ }L[/tex]Therefore, the size of the container (in Liters) needed to hold 2.1 mol O₂ gas at STP is **47.0 L**

Silver Sulfate reacts with Potassium Chloride according to the following reaction:Ag2SO4 + 2KCl -> 2AgC1 + K2SO4a. If 30.0 grams of Ag2SO4 reacts with 10.0 grams of KCl, what mass of AgCl is produced by the reaction b. the limiting reactant is ______c. how many grams ok K2SO4 can be produced.d. how many grams of excess reactant remain after the reaction e. what is the percent yield if there is 10.0g of AgCl

Answer:

Explanations:

Given the chemical reaction

[tex]Ag_2SO_4+2KCl\rightarrow2AgCl+K_2SO_4[/tex]**Given the following**

Mass of Ag2SO4 = 30grams

Mass of KCl = 10grams

Determine the moles of the reactants

[tex]\begin{gathered} mole\text{ of Ag}_2SO_4=\frac{mass}{molar\text{ mass}}molar\text{ mass} \\ mole\text{ of Ag}_2SO_4=\frac{30g}{311.799} \\ mole\text{ of Ag}_2SO_4=0.0962moles \end{gathered}[/tex][tex]\begin{gathered} mole\text{ of KCl}=\frac{10g}{74.5513} \\ moleof\text{ KCl}=0.1341moles \\ 1mole\text{ of KCl}=\frac{0.1341}{2}=0.06707moles \end{gathered}[/tex]B) B) Since the 1 moles of KCl is lower than the moles of Ag2SO4, hence **KCl willl be the limiting reactant.**

A) A) According to stoichiometry, 2 moles of KCl produces 2 moles of AgCl, the **mass of AgCl **produced will be given as;

C) According to stoichiometry, 2 moles of KCl produces 1 moles of K2SO4, the **mole of K2SO4 **produced is;

6. In order to prepare 50.0 mL of 0.100 M NaCl you will add _____ grams of NaCl to _____ mL of wate

The first step to solve this problem is to find the number of moles of NaCl in 50.0mL of 0.100M NaCl. To do it, convert the volume of solution to liters and multiply it by the concentration of the solution (M=mol/L).

[tex]50mL\cdot\frac{1L}{1000mL}\cdot\frac{0.100mol}{L}=0.005mol[/tex]Now, use the molar mass of NaCl to find the mass of 0.005moles of it (MM=58.44g/mol):

[tex]0.005mol\cdot\frac{58.44g}{mol}=0.2922g[/tex]It means that you have to add 0.2922g to 50mL of water.

1. What is the difference between a homogenous and a heterogeneous mixture? 2. Give 3 examples of each mixture and explain its uses in our daily lives.

**Answer:By combining two or more substances, a mixture is produced. A homogeneous solution tends to be identical, no matter how you sample it. Homogeneous mixtures are sources of water, saline solution, some alloys, and bitumen. Sand, oil and water, and chicken noodle soup are examples of heterogeneous mixtures.**

**Explanation:**

The partial pressure of oxygen was observed To be 156 torr

When the **partial pressure** of oxygen is 156 torr and the **atmospheric pressure** is 743 torr, the **mole fraction** of oxygen is 0.210.

Partial pressure is the pressure exerted by an individual gas in a mixture of gases. The partial pressure of a gas depends on its mole fraction.

The relationship between the partial pressure of a gas and the total pressure is given by **Dalton's law**, which states that the sum of the partial pressures is equal to the total pressure.

We can calculate the partial pressure of oxygen using the mathematical expression of Dalton's law:

pO₂ = P × X(O₂)

X(O₂) = pO₂ / P

X(O₂) = 156 torr / 743 torr = 0.210

where,

The **mole fraction** of oxygen is 0.210 when its **partial pressure** is 156 torr and the **atmospheric pressure** is 743 torr.

The complete question is:

The partial pressure of oxygen was observed To be 156 torr in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of oxygen present.

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How many molecules are in 59.73 grams of the theoretical acid borofuric acid, H2B2O2?

remember units and sig figs.

If your answer is a very large number, use scientific notation with an 'E' in this format:

120,000,000 = 1.2E8

The number of **molecules **in 59.73 grams of the theoretical acid, **borofuric acid**, would be 6.44 x [tex]10^2^3[/tex] molecules.

According to **Avogadro**, 1 mole of any substance contains 6.022 x [tex]10^2^3[/tex] molecules.

Recall that: number of **moles **in a substance = mass of the substance/**molar mass** of the substance.

Molar mass of **borofuric **acid, [tex]H_2B_2O_2[/tex]:

H = 1 g/mol

B = 10.8 g/mol

O = 16 g/mol

(1x2) + (10.8x2) + (16x2) = 55.6 g/mol

Number of moles of 59.73 grams [tex]H_2B_2O_2[/tex] = 59.73/55.6

= 1.07 moles

Since 1 mole = 6.022 x [tex]10^2^3[/tex] molecules

1.07 moles of borofuric acid = 1.07 x 6.022 x [tex]10^2^3[/tex] molecules.

= 6.44 x [tex]10^2^3[/tex] molecules

Thus, 59.73 grams of **borofuric acid **contains 6.44 x [tex]10^2^3[/tex] molecules.

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When a bright light is shined on glow in the dark powder what is happening to the electrons.

a.The electrons are absorbing some of the light and are become excited.

b.The atoms become negatively charged ions.

c.Nothing, the electrons are not affected by light.

d.The electrons move to a lower energy state.

A they become excited

A. The electrons are absorbing some of the light and are become excited,

Explain how the following reaction demonstrates that matter is neither created or destroyed in a chemical reaction: Ca(OH)2 + 2HCI-> CaCl2 + 2H20

**Answer:**

In this reaction, Ca(OH)2 is a reducing agent. It reacts with hydrogen chloride to form calcium chloride and water. Therefore, the following reaction shows that matter is neither created nor destroyed in a chemical reaction: Ca(OH)2 + 2HCI -> CaCl2 + 2H20. The formation of calcium chloride and water from the hydrolysis of calcium hydroxide is not an example of matter being created or destroyed in a chemical reaction because it does not involve the breaking down of any bonds between atoms.

**Explanation:**

what is combustion? explain to me pls

**Answer:**

rapid chemical combination of a substance with oxygen, involving the production of heat and light.

**Explanation:**

Combustion is a chemical process in which a substance reacts rapidly with oxygen and gives off heat. The original substance is called the fuel, and the source of oxygen is called the oxidizer.

** if you want to learn more:**

**https://www.britannica.com/science/combustion**

**https://youtu.be/xd1alir07q4**

Imagine the movement of gas particles in a closed container. According to the kinetic molecular theory, which statements below are true of the gas particles? Check all that apply.

Based on the **kinetic molecular theory**, the statements that are true of the gas particles are as follows:

The Kinetic Molecular Theory is the theory that molecules of a **gas** are in constant random motion colliding with one another and with the wall of their container.

These collisions between the gas molecules are perfectly elastic and are responsible for the pressure of gas molecules.

The higher the kinetic energy of the gas molecules, the higher the temperature of the gas.

The **kinetic molecular theory** explains the physical state of gases since gas molecules have negligible intermolecular forces.

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Complete question:

Imagine the movement of gas particles in a closed container. According to the kinetic molecular theory, which statements below are true of the gas particles? Check all that apply. Gas particles act like tiny, solid spheres. Gas particles are in constant, random motion. Gas particles at lower temperatures move faster. Collisions are elastic, there is no energy lost as the particle hits the sides of the container. Slower moving particles collide more often and with more force with the container.

The Ka of a monoprotic weak acid is 0.00469. What is the percent ionization of a 0.141 M solution of this acid?

Use quadratic equation.

The** percent ionization **of an acidic solution can be calculated from the H+ concentration. the percent ionization of the **monoprotic acid** of 0.141 M is 18.23 %.

**Percent ionization** of an acidic solution is the percent of H+ ions in the solution. Thus, mathematically, it is the ratio of H+ ion concentration to the **concentration** of solution multiplied by 100.

Let **HA** be the monoprotic acid when it ionizes, forming equal concentration of H+ and A- let it be x. Thus ionization constant can be written as follows:

Ka = [x]² /[HA]

0.00469 =[x]²/[0.141 M]

[X] = 0.025. = [H+]

**Percentage ionization** = (0.025 M / 0.141 M)× 100

= 18.23 %

Therefore** percentage ionization** of the acid is 18.23%.

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PLEASE HELP ASAP, I will mark the "brainiest"

In the generic reaction: 5A + 2B --> 6C + 3D The molar mass of A is 146.70g/mol while the molar mass of C is 21.31g/mol. If 4.253 g of A reacts with an excess of B, how many grams of C can form? Remember significant figures and units.

The **amount**, in **grams**, of C that can form from the **reaction**, is 0.7414 g.

From the **equation **of the reaction, the mole ratio of A to that of C is 5:6.

In other words, for every 5 moles of A that reacts, 6 moles of C is produced.

Molar mass of A = 146.70 g/mol

Molar mass of C = 21.31 g/mol

Recall that: **mole = mass/molar mass**

**Mole **of 4.253 g of A = 4.253/146.70

= 0.029 mol

From the mole ratio, the equivalent mol of C that will be produced is:

0.029 x 6/5 = 0.03479 mol

**mass = mole x molar mass**

**Mass **of 0.03479 mol of C = 0.03479 x 21.31

= 0.7414 g (to the appropriate sig. fig.)

In other words, the amount of C that will be produced from 4.253 g of A will be 0.7414 g.

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5. During factory inspection, Just Lemons, Inc. discovered that a water valve to the

lemonade mixing station was not functioning. Once they repair it, more water will

enter the mixing station. From what you know about the limiting and excess

ingredients for current lemonade production, what advice would you give engineers

about the upcoming increase in water?

The advise which would be given to the **engineers** is for them to determine the limiting **reactant** so water will not be in excess.

This is referred to as the **reactant** which is completely used up in a chemical reaction and it helps to determine when the reaction stops. In this scenario, either water, lemon juice or sugar is a limiting **reactant**.

The best option in this scenario is to tell the **engineers** to determine the limiting **reactant** so as to ensure that the quality of lemonade made is maximized and is therefore the right choice.

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