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deff fn [24]
3 years ago
10

NEED HELP ASAP PLS

Chemistry
1 answer:
densk [106]3 years ago
5 0

Answer:

31.60% phosphorus

Explanation:

To find the percent by mass, you first calculate the total mass and then the mass of phosphorus, and finally divide phosphorus mass by total.

Phosphorus mass: the molar mass of P is 30.97 g/mol, and since there's 1 mole of P, we just have 30.97 g.

Total mass: we add all the molar masses of the components together.

We have 3 moles of H, so we multiply 3 by 1.008 g/mol = 3.024 g H.

We already calculated the mass of phosphorus: 30.97 g P.

We have 4 moles of O, so we multiply 4 by 16.00 g/mol = 64.00 g O.

The total is then the sum: 3.024 + 30.97 + 64.00 = 97.994 g ≈ 97.99 g

Now, to find the percentage, we take 30.97 g P and divide by 97.99:

30.97/97.99 ≈ 0.3160 ⇒ 31.60% P

Thus, the answer is 31.60% phosphorus.

Hope this helps!

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A 8.00 L tank at 26.9 C is filled with 5.53 g of dinitrogen difluoride gas and 17.3 g of sulfur hexafluoride gas. You can assume
pshichka [43]

Answer:

mole fraction of N_2 O = 0.330

mole of fraction SF_4 = 0.669

PRESSURE OF N_2 O = 39127.053 Pa

pressure of SF_4 = 792126.36

Total pressure   = 118253.413 Pa

Explanation:

Given data:

volume of tank 8 L

Weight of dinitrogen difluoride gas 5.53 g

weight of sulphur hexafluoride gas 17.3 g

Amount of N_2 O = \frac{5.53}{14*2 + 16} = 0.1256 mol

amount of SF_4 = \frac{17.3}{32.1 + 19*4} = 0.254 mol

mole fraction of N_2 O = \frac{0.1256}{0.1256 + 0.254} = 0.330

mole of fractionSF_4 = \frac{0.254}{0.1256 + 0.254} = 0.669

PV = nRT

P of N_2 O = \frac{0.1256 *8.31 (273 + 26.9}{0.008} = 39127.053 Pa

mole of SF_4=\frac{0.254 *8.31*(273+26.9)}{.008} = 79126.36 Pa

Total pressure  = 39127.053 + 79126.36 = 118253.413 Pa

6 0
3 years ago
he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
Leya [2.2K]

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

8 0
3 years ago
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Ann [662]

It is not equal because it dose not obey the conservation of mass. 60+25= 85 not 75.

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3 years ago
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Lana71 [14]

Answer:

The reason why atomic mass is usually not a whole number is because it is a weighted average of the mass numbers of isotopes

Explanation:

8 0
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4 0
3 years ago
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