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3 years ago

6

The mass of .10 moles of iron is:

2 answers:

Stels [109]3 years ago

8 0

Answer:

<em>so mass in gram=560grams</em>

Explanation:

number of moles=10moles

molar mass=56grams/moles

mass in gram of Fe=?

as we know that

$number of moles=\frac{mass in gram}{molar mass}$

<em>evaluating the formula</em>

<em>number of moles×molar mass=mass in gram</em>

<em>mass in gram=10moles×56grams/moles</em>

<em>mass in gram=560grams</em>

<em>i hope this will help you :)</em>

Elenna [48]3 years ago

5 0

The answer is going to be 560grams, i hope this helps and have a good day

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Which of the following compounds is usually

goblinko [34]

The answer is C which is PbSo

8 0

2 years ago

A containing vessel holds a gaseous mixture of nitrogen and butane. Thepressure in the vessel at 126.9 Cis 3.0 atm. At 0 C, the

Viefleur [7K]

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.

We can calculate the total number of moles using the ideal gas equation.

$P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{3.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.091 mol/L \times V$

At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.

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

$P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.030 mol/L \times V$

The mole fraction of nitrogen in the mixture is:

$X(N_2) = \frac{0.030 mol/L \times V}{0.091 mol/L \times V} = 0.33$

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

Learn more: brainly.com/question/2060778

5 0

3 years ago

Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 L of carbon dioxide at 20.0 °C and

swat32

Answer:

$\%\ mass\ of\ CaCO_3=93.37\ \%$

Explanation:

Given that:

Pressure = 791 mmHg

Temperature = 20.0°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (20 + 273.15) K = 293.15 K

T = 293.15 K

Volume = 100 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 62.3637 L.mmHg/K.mol

Applying the equation as:

791 mmHg × 1.14 L = n × 62.3637 L.mmHg/K.mol × 293.15 K

⇒n of $CO_2$ produced = 0.0493 moles

According to the reaction:-

$CaCO_3 + 2 HCl\rightarrow CaCl_2 + H_2O + CO_2$

1 mole of carbon dioxide is produced 1 mole of calcium carbonate reacts

0.0493 mole of carbon dioxide is produced 0.0493 mole of calcium carbonate reacts

Moles of calcium carbonate reacted = 0.0493 moles

Molar mass of $CaCO_3$ = 100.0869 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0493\ mol= \frac{Mass}{100.0869\ g/mol}$

$Mass_{CaCO_3}=4.93\ g$

Impure sample mass = 5.28 g

Percent mass is percentage by the mass of the compound present in the sample.

$\%\ mass\ of\ CaCO_3=\frac{Mass_{CaCO_3}}{Total\ mass}\times 100$

$\%\ mass\ of\ CaCO_3=\frac{4.93}{5.28}\times 100$

$\%\ mass\ of\ CaCO_3=93.37\ \%$

3 0

2 years ago

Why would an atom become an ion?

crimeas [40]

Answer:

c

Explanation: correct me if im wrong

5 0

3 years ago

What is the pH of a solution which is 0.600 M in dimethylamine ((CH3)2NH) and 0.400 M in dimethylamine hydrochloride ((CH3)2NH2C

Viefleur [7K]

Answer:

pH = 11.05

Explanation:

It is possible to answer this question using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [A] / [HA⁺]

Where A in this case is weak base (dimethylamine) and conjugate acid (HA⁺) is dimethylamine hydrochloride.

As Ka= Kw / Kb = 1x10⁻¹⁴ / 7.4x10⁻⁴ = 1.35x10⁻¹¹ And pKa is -log Ka = <em>10.87 </em> pH of the solution is:

pH = 10.87 + log₁₀ [0.600] / [0.400]

<em>pH = 11.05</em>

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I hope it helps!

3 0

3 years ago