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nadya68 [22]
3 years ago
6

Molar mass (NH4)2SO4

Chemistry
1 answer:
Hoochie [10]3 years ago
6 0

Answer:

<u>132.15</u>

Explanation:

Molar mass N = 14.00

Molar mass H = 1.01

Molar mass H4 = 1.01 x 4 = 4.04

Molar mass NH4 = 14.00 + 4.04 = 18.04

Molar mass (NH4)2 = 18.04 x 2 = 36.08

Molar mass S = 32.07

Molar mass O = 16.00

Molar mass O4 = 16.00 x 4 = 64.00

Molar mass SO4 = 32.07 + 64.00 = 96.07

Molar mass (NH4)2SO4 = 36.08 + 96.07 = <u>132.14</u>

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When 3.51 g of phosphorus was burned in chlorine, the product was a phosphorus chloride. Its vapor took 1.77 times as long to ef
kumpel [21]

Answer:

<em>molar mass of the phosphorus chloride = 138.06 g/mol</em>

<em></em>

Explanation:

<em>mass of phosphorus will be the same as mass of CO2, since it is stated that they are of equal amount.</em>

mass = 3.51 g

<em>lets assume that it took the CO2 1 sec to effuse, then the time taken by the phosphorus chloride will be 1.77 sec</em>

From this we can say that

rate of effusion of CO2 = 3.51/1 = 3.51 g/s

rate of effusion of the phosphorus chloride = 3.51/1.77 = 1.98 g/s

<em>From graham's equation of effusion</em>,

\frac{Rc}{Rp} = \sqrt{\frac{Mp\\}{Mc} }

Rc = rate of effusion of CO2 = 3.51 g/s

Rp = rate of effusion of phosphorus chloride = 1.98 g/s

Mc = molar mass of CO2 = 44.01 g/mol

Mp = molar mass of the phosphorus chloride = ?

Imputing values into the equation, we have

\frac{3.51}{1.98} = \sqrt{\frac{Mp\\}{44.01} }

1.77 = \frac{\sqrt{Mp} }{6.64}

11.75 = \sqrt{Mp}

Mp = 11.75^{2}

Mp = <em>molar mass of the phosphorus chloride = 138.06 g/mol</em>

5 0
3 years ago
Which states of matter only appear in the hydrosphere
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I would say the answer is liquids
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The brightness of a star depends on its _____ a. color b. composition of atmosphere c. distance from Earth, and stars that are c
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Answer:

Either B or C. Composition or the Distance from the Earth.

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Which type of covalent bond is the least stable?
sergejj [24]

Hello ^^

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Explanation:

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Hope this helps!

Have an awesome day!

If you have any questions please ask.

8 0
2 years ago
Read 2 more answers
An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?
umka2103 [35]

<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

K_b = molal boiling point elevation constant = 0.51°C/m

m_{solute} = Given mass of solute (CsCl) = 8.00 g

M_{solute} = Molar mass of solute (CsCl) = 168.4  g/mol

W_{solvent} = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC

Hence, the boiling point of solution is 100.53

6 0
3 years ago
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