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3 years ago

Calculate the mass of 45.0 L of NH3 at 57.0° C and 900. mm Hg.

Chemistry

1 answer:

trasher [3.6K]3 years ago

8 0

Answer:

Mass = 33.515 g

Explanation:

Given data:

Volume of ammonia = 45.0 L

Temperature = 57.0°C (57 +273 = 330 K)

Pressure = 900 mmHg

Mass of ammonia = ?

Solution:

Formula:

PV = nRT

We will use R = 62.364 (L * mmHg)/(mol * K) because pressure is given in mmHg.

900 mmHg × 45.0 L = n × 62.364L * mmHg/mol * K × 330 K

40500 mmHg.L = n × 20580.12L * mmHg/mol

n= 40500 mmHg.L/ 20580.12L * mmHg/mol

n = 1.9679 mol

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 1.9679 mol × 17.031 g/mol

Mass = 33.515 g

When we multiply or divide the values the number of significant figures must be equal to the less number of significant figures in given value.

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1 loop in the primary coil and 8 loops in the secondary. If the secondary voltage is 120 V, what must be the primary voltage

Jet001 [13]

Answer:

$V_{p} = 15 V$

Explanation:

<u>Given the following data;</u>

Number of loops in primary coil, Np = 1 loop.

Number of loops in secondary coil, Ns = 8 loops

Voltage in secondary coil, Vs = 120V

To find the voltage in the primary coil, Vp;

Transformer ratio is given by the formula;

$\frac{V_{p}}{V_{s}} = \frac{N_{p}}{N_{s}}$

Making Vp the subject of formula;

$V_{p} = \frac{N_{p}}{N_{s}} * V_{s}$

Substituting into the equation, we have;

$V_{p} = \frac{1}{8} * 120$

$V_{p} = \frac{120}{8}$

$V_{p} = 15 V$

Therefore, the voltage in the primary coil, Vp is 15 Volts.

4 0

3 years ago

What's is the mass of 0.500 mol NH3

Setler [38]

Answer:

8.5155g NH3

Explanation:

the molar mass of NH3 is 17.031 g/mol

0.5 mol NH3 x 17.031 gNH3/1 mol NH3 = 8.5155g NH3

4 0

3 years ago

What volume (in liters) of a solution contains 0.14 mol of KCl?

oksano4ka [1.4K]

Answer:

$\boxed {\boxed {\sf 0.078 \ L }}$

Explanation:

We are asked to find the volume of a solution given the moles of solute and molarity.

Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:

$molarity= \frac{moles \ of \ solute}{liters \ of \ solution}$

We know there are 0.14 moles of potassium chloride (KCl), which is the solute. The molarity of the solution is 1.8 molar or 1.8 moles of potassium chloride per liter.

moles of solute = 0.14 mol KCl
molarity= 1.8 mol KCl/ L
liters of solution=x

Substitute these values/variables into the formula.

$1.8 \ mol \ KCl/ L = \frac { 0.14 \ mol \ KCl}{x}$

We are solving for x, so we must isolate the variable. First, cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

$\frac {1.8 \ mol \ KCl/L}{1} = \frac{0.14 \ mol \ KCl}{x}$

$1.8 \ mol \ KCl/ L *x = 1*0.14 \ mol \ KCl$

$1.8 \ mol \ KCl/ L *x = 0.14 \ mol \ KCl$

Now x is being multiplied by 1.8 moles of potassium chloride per liter. The inverse operation of multiplication is division, so we divide both sides by 1.8 mol KCl/L.

$\frac {1.8 \ mol \ KCl/ L *x}{1.8 \ mol \ KCl/L} = \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}$

$x= \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}$

The units of moles of potassium chloride cancel.

$x= \frac{0.14 }{1.8 L}$

$x=0.07777777778 \ L$

The original measurements of moles and molarity have 2 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in the ten-thousandth place tells us to round the 7 up to a 8.

$x \approx 0.078 \ L$

There are approximately <u>0.078 liters of solution.</u>

5 0

3 years ago

Calculate the molecules in 9.78 moles of oxygen gas

Hoochie [10]

Answer:

i am quite unsure what you want me to answer but if you explain it to me w ould live to help

Explanation:

7 0

3 years ago

What is the value for (delta)G at 1000 K if (delta)H = -220 kJ/mol and (delta)S = -0.05 kJ/(molK)?

tankabanditka [31]

The system is isothermal, so we use the formula:
(delta)G = (delta)H - T (delta) S

Plugging in the given values:
(delta)G = -220 kJ/ mol - (1000K) (-0.05 kJ/mol K)
(delta)G = -170 kJ/mol

If we take a basis of 1 mol, the answer is
D. -170 kJ

6 0

3 years ago

Read 2 more answers