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3 years ago

What would you multiply "moles of oxygen" by to get the units "grams of oxygen"?

Chemistry

1 answer:

photoshop1234 [79]3 years ago

6 0

Answer:

Explanation:

to go from moles to grams you multiply by the Atomic Mass or Molar Mass (Atomic Mass for an element and Molar Mass for a compound).

The formula is:

Mass = moles * MM

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The reaction ch4 (g) + 2 o2 (g) ? co2 (g) + 2 h2o (g) is: the reaction ch4 (g) + 2 o2 (g) co2 (g) + 2 h2o (g) is: an exothermic

barxatty [35]

All of the above
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5 0

3 years ago

A 0.539 g sample of a metal, M, reacts completely with sulfuric acid according to the reaction M(s)+H2SO4(aq)⟶MSO4(aq)+H2(g) A v

Charra [1.4K]

Answer:

The molar mass of the metal is 54.9 g/mol.

Explanation:

When we work with gases collected over water, the total pressure (atmospheric pressure) is equal to the sum of the vapor pressure of water and the pressure of the gas.

Patm = Pwater + PH₂

PH₂ = Patm - Pwater = 1.0079 bar - 0.03167 bar = 0.9762 bar

The pressure of H₂ is:

$0.9762bar.\frac{1atm}{1.013bar} =0.9637atm$

The absolute temperature is:

K = °C + 273 = 25°C + 273 = 298 K

We can calculate the moles of H₂ using the ideal gas equation.

$P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{0.9637atm \times 0.249L }{(0.08206atm.L/mol.K)\times298K} =9.81 \times 10^{-3} mol$

Let's consider the following balanced equation.

M(s) + H₂SO₄(aq) ⟶ MSO₄(aq) + H₂(g)

The molar ratio of M:H₂ is 1:1. So, 9.81 × 10⁻³ moles of M reacted. The molar mass of the metal is:

$\frac{0.539g}{9.81 \times 10^{-3} mol} =54.9g/mol$

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

3 years ago

How many moles of methane are produced when 4.53 mol co2 reacts

ArbitrLikvidat [17]

The reaction formula CH4 + 2O2 → CO2 + 2H2O shows the oxidation of 1 mole of CH4 (Methane) will yield 1 mole of CO2 (Carbon Dioxide). Since 1 mole of CH4 will weigh 12g (for the Carbon) + 4g (1g for each Hydrogen) = 16g, then 32g of CH4 will correspond to 32g / 16g/mole = 2 moles. Therefore the oxidation of 2 moles of CH4 will yield 2 moles of CO2.

6 0

2 years ago

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The variable in an experiment that is changed on purpose

DiKsa [7]

The answer is Independent Variable

4 0

3 years ago