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3 years ago

6

Sally adds 3.13 moles of argon to a 5.29 liter balloon that already contained 4.21 moles of argonWhat is the volume of the ballo

on after the addition of the extra gas?

1 answer:

lara31 [8.8K]3 years ago

8 0

11.89, I hope this helped:)

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Under certain conditions, neon (Ne) gas diffuses at a rate of 7.0 centimeters per second. Under the same conditions, an unknown

Nata [24]

Answer : The molar mass of unknown gas is, 41 g/mole

Solution : Given,

Diffusion rate of neon gas = 7 cm/s

Diffusion rate of unknown gas = 4.9 cm/s

Molar mass of neon gas = 20 g/mole

According to the Graham's law, the rate of effusion of a gas is inversely proportional to the square root of the molar mass of the gas.

Formula used :

$\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_1}}$

where,

$R_1$ = diffusion rate of neon gas

$R_2$ = diffusion rate of unknown gas

$M_1$ = molar mass of neon gas

$M_2$ = molar mass of unknown gas

Now put all the given values in the above formula, we get the molar mass of unknown gas.

$\frac{7cm/s}{4.9cm/s}=\sqrt{\frac{M_2}{20g/mole}}$

$M_2=40.81g/mole=41g/mole$

Therefore, the molar mass of unknown gas is, 41 g/mole

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3 years ago

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notka56 [123]

Answer:

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Explanation:

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3 years ago

What are the three important things needed for combustion to occur ?

marissa [1.9K]

Explanation:

Here is the answer hope you do well

4 0

3 years ago

A quantity of 0.225 g of a metal M (molar mass = 27.0 g/mol) liberated 0.303 L of molecular hydrogen (measured at 17°C and 741 m

Lana71 [14]

Answer:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$

Explanation:

0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.

Let moles of hydrogen gas be n.

Temperature of the gas ,T= 17°C =290 K

Pressure of the gas ,P= 741 mmHg= 0.9633 atm

Volume occupied by gas , V = 0.303 L

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9633 atm\times 0.303 L}{0.0821 atm L/mol K\times 290 K}=0.01225 mol$

Moles of hydrogen gas produced = 0.01225 mol

$2M+2xHCl\rightarrow 2MCl_x+xH_2$

Moles of metal = $\frac{0.225 g}{27.0 g/mol}=8.3333 mol$

So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

$\frac{8.3333}{0.01225 mol}=\frac{2}{x}$

x = 2.9 ≈ 3

$2M+6HCl\rightarrow 2MCl_3+3H_2$

$MCl_3\rightarrow M^{3+}+Cl^-$

Formulas for the oxide and sulfate of M will be:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$ .

3 0

4 years ago

What is the mass of 61.9 L of oxygen gas collected at STP?

Tcecarenko [31]

Answer:

D. 44.2 g O₂

General Formulas and Concepts:
<u>Gas Laws</u>

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at <em>1 atm, 273 K</em>

<u>Stoichiometry</u>

Dimensional Analysis
Mole Ratio

Explanation:

<u>Step 1: Define</u>

<em>Identify given.</em>

61.9 L O₂ at STP

<u>Step 2: Convert</u>

We know that the oxygen gas is at STP. Therefore, we can set up and solve for how many <em>moles</em> of O₂ is present:

$\displaystyle 61.9 \ \text{L} \ \text{O}_2 \bigg( \frac{1 \ \text{mol} \ \text{O}_2}{22.4 \ \text{L} \ \text{O}_2} \bigg) = 2.76339 \ \text{mol} \ \text{O}_2$

Recall the Periodic Table (Refer to attachments). Oxygen's atomic mass is roughly 16.00 grams per mole (g/mol). We can use a mole ratio to convert from <em>moles</em> to <em>grams</em>:

$\displaystyle 2.76339 \ \text{mol} \ \text{O}_2 \bigg( \frac{16.00 \ \text{g} \ \text{O}_2}{1 \ \text{mol} \ \text{O}_2} \bigg) = 44.2143 \ \text{g} \ \text{O}_2$

Now we deal with sig figs. From the original problem, we are given 3 significant figures. Round your answer to the <u>exact</u> same number of sig figs:

$\displaystyle 44.2143 \ \text{g} \ \text{O} \approx \boxed{ 44.2 \ \text{g} \ \text{O}_2 }$

∴ our answer is letter choice D.

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Topic: AP Chemistry

Unit: Stoichiometry

6 0

2 years ago