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Mazyrski [523]
3 years ago
5

Phosphorus trichloride gas is formed by the reaction of solid phosphorus and chlorine gas . Write a balanced chemical equation f

or this reaction.
Chemistry
1 answer:
NARA [144]3 years ago
4 0


P + 3Cl ----> P(Cl)3
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Argon (Ar) and helium (He) are initially in separate compartments of a container at 25°C. The
love history [14]

Answer:

(a) V_B=11.68L

(b) x_{He}=0.533

Explanation:

Hello,

In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

n_A=\frac{0.082\frac{atm*L}{mol*K}*298K}{1.974 atm*6.00L}=2.063mol

Thus, since the final pressure is 3.60 bar, we can write:

P=x_{Ar}P_A+x_{He}P_B\\\\P=\frac{n_{Ar}}{n_{Ar}+n_{He}} P_A+\frac{n_{He}}{n_{Ar}+n_{He}} P_B\\\\3.60bar=\frac{2.063mol}{2.063mol+n_{He}} *2.00bar+\frac{n_{He}}{2.063mol+n_{He}} *5.00bar

The moles of helium could be computed via solver as:

n_{He}=2.358mol

Or algebraically:

3.60bar=\frac{1}{2.063mol+n_{He}} *(4.0126+5.00*n_{He})\\\\7.314+3.60n_{He}=4.013+5.00*n_{He}\\\\7.314-4.013=5.00*n_{He}-3.60n_{He}\\\\n_{He}=\frac{3.3}{1.4}=2.358mol

In such a way, the volume of the compartment B is:

V_B=\frac{n_{He}RT}{P_B}=\frac{2.358mol*0.082\frac{atm*L}{mol*K}*298.15K}{4.935atm}\\  \\V_B=11.68L

Finally, he mole fraction of He is:

x_{He}=\frac{2.358}{2.358+2.063}\\ \\x_{He}=0.533

Regards.

8 0
2 years ago
What would be the total volume of the new solution when it is changed from 0.2 M to 0.04 M?
34kurt
The question is incomplete.

You need two additional data:

1) the original volume
2) what solution you added to change the volume.

This is a molarity problem, so remember molarity definition and formula:

M = n / V in liters: number of moles per liter of solution

To give you the key to answer this kind of questions, supppose the original volumen was 1 ml and that you added only water (solvent).

The original solution was:

V= 1 ml
M = 0.2 M

Using the formula for molarity, M = n / V

n = M×V = 0.2 M × (1 / 10000)l = 0.0002 moles

For the final solution:

n = 0.0002 moles
M = 0.04

From M = n / V ⇒ V = n / M = 0.002 moles / 0.04 M = 0.05 l

Change to ml ⇒ 0.05 l × 1000 ml / l = 50 ml.  This would be the answer for the hypothetical problem that I assumed for you.

I hope this gives you all the cues you need to answer similar problems about molarity.
6 0
3 years ago
Read 2 more answers
If your front yard is 16.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1250 new snowflakes every minu
Free_Kalibri [48]
I think the answer is 3x-x+2=4
3 0
3 years ago
Which of these describes the role of gelada baboons in their ecosystem?
katen-ka-za [31]

Answer:

Gelada baboons plays a significant role.

Explanation:

The role of gelada baboons in their ecosystem is very important because they aerate the soil for plants which is necessary for good plant growth. These gelada baboons also helps in controlling the population of predator in their ecosystem which is very essential for the stability and equilibrium of the ecosystem so gelada baboons has a good effect on both plants and animals in their ecosystem.

6 0
3 years ago
Read 2 more answers
A sample of a compound is decomposed in the laboratory and produces 330 g carbon, 69.5 g hydrogen, and 220.2 g oxygen. Calculate
Zarrin [17]

Answer: Empirical formula is C_2H_5O

Explanation: We are given the masses of elements present in a sample of compound. To evaluate empirical formula, we will be following some steps.

<u>Step 1 :</u> Converting each of the given masses into their moles by dividing them by Molar masses.

Moles=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of Carbon = 12.0 g/mol

Molar mass of Hydrogen = 1.0 g/mol

Molar mass of Oxygen = 16.0 g/mol

Moles of Carbon = \frac{330g}{12g/mol}=27.5moles

Moles of Hydrogen = \frac{69.5g}{1g/mol}=69.5moles

Moles of Oxygen = \frac{220.2g}{16g/mol}=13.76moles

<u>Step 2: </u>Dividing each mole value by the smallest number of moles calculated above and rounding it off to the nearest whole number value

Smallest number of moles = 13.76 moles

\text{Mole ratio of Carbon}=\frac{27.5moles}{13.76moles}=1.99\approx 2

\text{Mole ratio of Hydrogen}=\frac{69.5moles}{13.76moles}=5.05\approx 5

\text{Mole ratio of Oxygen}=\frac{13.76moles}{13.76moles}=1

<u>Step 3:</u> Now, the moles ratio of the elements are represented by the subscripts in the empirical formula

Empirical formula becomes = C_2H_5O

7 0
3 years ago
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