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Nezavi [6.7K]
3 years ago
9

You have an unknown quantity of oxygen at a pressure of 2.2 atam, a volume of 21 liters and a temperature of 87 Celsius. How man

y miles of gas are left for you to breath
Chemistry
1 answer:
laila [671]3 years ago
6 0

<span>Let's assume that the oxygen gas has ideal gas behavior. 
Then we can use ideal gas formula,
      PV = nRT</span>


Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.

<span>
P = 2.2 atm = 222915 Pa
V = 21 L = 21 x 10</span>⁻³ m³

n = ?

R = 8.314 J mol⁻¹ K⁻¹

<span> T = 87 °C = 360 K

By substitution,
</span>222915 Pa x 21 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻<span>¹ x 360 K
                                       n = 1.56</span><span> mol</span>

<span>
Hence, 1.56 moles of the oxygen gas are </span><span>left for you to breath.</span><span>
</span>
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To what volume should you dilute 55 mL of 12 M stock HNO3 solution to obtain a 0.145 HNO3 solution?
velikii [3]

Answer:

4552 mL

Explanation:

From the question given above, the following data were obtained:

Volume of stock solution (V₁) = 55 mL

Molarity of stock solution (M₁) = 12 M

Molarity of diluted solution (M₂) = 0.145 M

Volume of diluted solution (V₂) =?

The volume of the diluted solution can be obtained by using the dilution formula as illustrated below:

M₁V₁ = M₂V₂

12 × 55 = 0.145 × V₂

660 = 0.145 × V₂

Divide both side by 0.145

V₂ = 660 / 0.145

V₂ ≈ 4552 mL

Thus, the volume of the diluted solution is 4552 mL

7 0
3 years ago
The pH of a solution is measured eight times by one operator using the same instrument. She obtains the following data: 7.15, 7.
sergiy2304 [10]

Answer:

7.186

Explanation:

The mean is the average of some given data from a sample point.

To calculate the mean, we use the formula below:

             Mean = ∑fx/∑f

Where f = frequency

           x = sample data

   

From the given pH of the solutions, we can form a table:

      x                                     f                                          fx

    7.15                                  1                                         7.15

    7.16                                  1                                         7.16

    7.18                                  1                                         7.18

    7.19                                  1                                         7.19

    7.20                                 3                                        21.6

    7.21                                  1                                         7.21

Now ∑fx = 7.15 +7.16 +7.18 + 7.19 + 7.20 + 7.21 = 57.49

         ∑f = 1 + 1 + 1 + 1 + 3 + 1 = 8

The mean = \frac{57.49}{8} = 7.18625 = 7.186

3 0
3 years ago
Compare the properties of protons neutrons and electrons
butalik [34]
Protons are positively charged. Neutrons have no charge. Electrons have a negative charge. Protons and neutrons are in the nucleus. Electrons revolve around the nucleus. 
6 0
4 years ago
Read 2 more answers
If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under
WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\  then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0
3 years ago
Omg pls help i dunno what the frick frack this is
snow_lady [41]

Answer:

1. Mass of KCl produced = 774.8 g of KCl

2. Mass of KNO₃ produced = 13.837g

3. Moles of NaOH made = 0.846 moles

4. Moles of LiCl produced = 0.846 moles

5. Moles of CO₂ produced = 207.6 moles

Explanation:

1. From the equation of reaction, 1 mole of ZnCl₂ produces, 2 moles of KCl.

5.02 moles of ZnCl₂ will produce, 2 × 5.02 moles of KCl = 10.4 moles of KCl

Molar mass of KCl = (39 + 35.5) g/mol = 74.5 g/mol

10.4 moles of KCl = 10.4 × 74.5 g

Mass of KCl produced = 774.8 g of KCl

2. Mole ratio of KNO₃ and KOH = 1:1

O.137 moles of KOH will produce 0.137 moles of KNO₃

Molar mass of KNO₃ = 101 g/mol

Mass of KNO₃ produced = 0.137 × 101 g = 13.837g

3. Molar mas of Ca(OH)₂ = 74.0 g

Moles of Ca(OH)₂ in 31.3 g = 31.3/74.0 = 0.423 moles of Ca(OH)₂

Mole ratio of NaOH and Ca(OH)₂ in the reaction = 2 : 1

Moles of NaOH made = 2 × 0.423 = 0.846 moles

4. Molar mass of MgCl₂ = 95.0 g

Moles of MgCl₂ in 40.2 g = 40.2/95.0 = 0.423 moles

From the reaction equation, mole ratio of MgCl₂ and LiCl = 1:2

Moles of LiCl produced = 2 × 0.423 = 0.846 moles

5. From the equation of reaction, 1 mole of C₆H₁₀O₅ produces 6 moles of cO₂

34.6 moles of C₆H₁₀O₅ will produce 34.6 × 6 moles of CO₂

Moles of CO₂ produced = 207.6 moles

4 0
3 years ago
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