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Zarrin [17]
2 years ago
5

Calculate the volume of 38.0 g of carbon dioxide at STP. Enter your answer in the box provided. L

Chemistry
1 answer:
Free_Kalibri [48]2 years ago
7 0

Answer:

19.3 L

Explanation:

V= n × 22.4

where V is volume and n is moles

First, to find the moles of CO2, divide 38.0 by the molecular weight of CO2 which is 44.01

n= m/ MM

n= 38/ 44.01

n= 0.86344012724

V= 0.86344012724 × 22.4

V= 19.3410588502 L

V= 19.3 L

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A chemical reaction occurs between Calcium, Oxygen, and Chlorine. The reaction is demonstrated:
ad-work [718]

Answer:

D.

Double replacement, CaO + Cl2O

Explanation:

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3 years ago
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I need help with 1,2,3, and 4
Schach [20]

Answer:

  • Problem 1: 1.85atm
  • Problem 2: 110mL
  • Problem 3: 290 mL
  • Problem 4: 1.14 atm

Explanation:

Problem 1

<u>1. Data</u>

<u />

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

<u>2. Formula</u>

Since the temeperature is constant you can use Boyle's law for idial gases:

          PV=constant\\\\P_1V_1=P_2V_2

<u>3. Solution</u>

Solve, substitute and compute:

         P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2

        P_2=3.25atm\times755mL/1325mL=1.85atm

Problem 2

<u>1. Data</u>

<u />

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

<u>2. Formula</u>

You assume that the temperature does not change, and then can use Boyl'es law again.

          P_1V_1=P_2V_2

<u>3. Solution</u>

This time, solve for V₂:

           P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2

Substitute and compute:

        V_2=548mmHg\times 125mL/625mmHg=109.6mL

You must round to 3 significant figures:

        V_2=110mL

Problem 3

<u>1. Data</u>

<u />

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

<u>2. Formula</u>

At constant pressure, Charle's law states that volume and temperature are inversely related:

         V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

The temperatures must be in absolute scale.

<u />

<u>3. Solution</u>

a) Convert the temperatures to kelvins:

  • T₁ = 25 + 273.15K = 298.15K

  • T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

        \dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml

You must round to two significant figures: 290 ml

Problem 4

<u>1. Data</u>

<u />

a) P = 865mmHg

b) Convert to atm

<u>2. Formula</u>

You must use a conversion factor.

  • 1 atm = 760 mmHg

Divide both sides by 760 mmHg

       \dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}

<u />

<u>3. Solution</u>

Multiply 865 mmHg by the conversion factor:

    865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer

3 0
3 years ago
Combustion of hydrocarbons such as dodecane (C12H26) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth'
miss Akunina [59]

Answer:

A. 2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)

B. 761.42 L

Explanation:

A. Step 1:

The equation for the reaction.

C12H26(l) + O2(g) —> CO2(g) + H2O(g)

A. Step 2:

Balancing the equation.

The equation can be balance as follow:

C12H26(l) + O2(g) —> CO2(g) + H2O(g)

There are 12 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 12 in front of CO2 as illustrated below:

C12H26(l) + O2(g) —> 12CO2(g) + H2O(g)

There are 26 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 13 in front of H2O as illustrated below:

C12H26(l) + O2(g) —> 12CO2(g) + 13H2O(g)

Now, there are a total of 37 atoms of O2 on the right side and 2 atoms on the left. It can be balance by putting 37/2 in front of O2 as illustrated below:

C12H26(l) + 37/2O2(g) —> 12CO2(g) + 13H2O(g)

Multiply through by 2 to clear the fraction from the equation.

2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)

Now the equation is balanced

B. Step 1:

We'll by obtaining the number of mole of C12H26 in 0.450 kg of C12H26. This is illustrated below:

Molar Mass of C12H26 = (12x12) + (26x1) = 144 + 26 = 170g/mol

Mass of C12H26 = 0.450 kg = 0.450x1000 = 450g

Number of mole of C12H26 =?

Number of mole = Mass/Molar Mass

Number of mole of C12H26 = 450/170

Number of mole of C12H26 = 2.65 moles

B. Step 2:

Determination of the number of mole of CO2 produced by the reaction. This is illustrated below:

2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)

From the balanced equation above,

2 moles of C12H26 produced 24 moles of CO2.

Therefore, 2.65 moles of C12H26 will produce = (2.65x24)/2 = 31.8 moles of CO2.

B. Step 3:

Determination of the volume of CO2 produced by the reaction.

Pressure (P) = 1 atm

Temperature (T) = 19°C = 19°C + 273 = 292K

Gas constant (R) = 0.082atm.L/Kmol

Number of mole (n) = 31.8 moles

Volume (V) =?

The volume of CO2 produced by the reaction can b obtained by applying the ideal gas equation as follow:

PV = nRT

1 x V = 31.8 x 0.082 x 292

V = 761.42 L

Therefore, the volume of CO2 produced is 761.42 L

5 0
3 years ago
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Does the physical form of the material matter for mass-mole<br> and mole-mass calculations?
natulia [17]
Tin metal reacts with hydrogen fluoride to produce tin(II) fluoride and hydrogen gas according to the following balanced equation.

Sn(s)+2HF(g)→SnF2(s)+H2(g)
Sn(s)+2HF(g)→
SnF
2
(s)+
H
2
(g)

How many moles of hydrogen fluoride are required to react completely with 75.0 g of tin?

Step 1: List the known quantities and plan the problem.

Known

given: 75.0 g Sn
molar mass of Sn = 118.69 g/mol
1 mol Sn = 2 mol HF (mole ratio)
Unknown

mol HF
Use the molar mass of Sn to convert the grams of Sn to moles. Then use the mole ratio to convert from mol Sn to mol HF. This will be done in a single two-step calculation.

g Sn → mol Sn → mol HF

Step 2: Solve.

75.0 g Sn×1 mol Sn118.69 g Sn×2 mol HF1 mol Sn=1.26 mol HF
75.0 g Sn×
1
mol Sn
118.69
g Sn
×
2
mol HF
1
mol Sn
=1.26 mol HF

Step 3: Think about your result.

The mass of tin is less than one mole, but the 1:2 ratio means that more than one mole of HF is required for the reaction. The answer has three significant figures because the given mass has three significant figures.
4 0
2 years ago
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**Help!! **<br> How do I do this question
swat32
The answer is D. Okay l hope this helps
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