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3 years ago

34.

Chemistry

1 answer:

Ganezh [65]3 years ago

5 0

Answer:

a. homogeneous

b. heterogeneous

c. homogeneous

d. homogeneous

Hope this helps!!! Good luck!!! ;)

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i really need help with the rest of this assignment. theres 10 pages of multiple choice questions, i can mark brainliest and pay

Zina [86]

Answer:

What's the assignment?

Explanation:

3 0

3 years ago

True or false

svlad2 [7]

Answer:

1. true

2. false

3. true

4. true

Explanation:

I believe these are the answers, I'm just not sure for #3

4 0

3 years ago

If a little old lady is doing 98.3 kilometers/hour, will she get a speeding ticket if the speed limit is 55 miles / hour? [USE:

shutvik [7]

Answer:

It is possible she could get one.

Explanation:

To solve this problem we need to convert 98.3 kilometers/hour to miles/hour.

In other words, we convert km to mi, to do so we multiply 98.3 km by a conversion factor, putting the unit we want to have in the numerator, and the unit we want to convert in the denominator:

98.3 km * $\frac{1000m}{1km} *\frac{100cm}{1m}*\frac{1in}{2.54cm} *\frac{1ft}{12in} *\frac{1mi}{5280ft}$ = 61 mi

Given that the little old lady is doing 61 miles/hour, she could get a speeding ticket.

7 0

3 years ago

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective

katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

$PCl_3,p_1'=6.798 Torr$

$Cl_2,p_2'=26.398 Torr$

$PCl_5,p_3'=223.402 Torr$

Explanation:

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At equilibrium before adding chlorine gas:

Partial pressure of the $PCl_3=p_1=13.2 Torr$

Partial pressure of the $Cl_2=p_2=13.2 Torr$

Partial pressure of the $PCl_5=p_3=217.0 Torr$

The expression of an equilibrium constant is given by :

$K_p=\frac{p_1}{p_1\times p_2}$

$=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245$

At equilibrium after adding chlorine gas:

Partial pressure of the $PCl_3=p_1'=13.2 Torr$

Partial pressure of the $Cl_2=p_2'=?$

Partial pressure of the $PCl_5=p_3'=217.0 Torr$

Total pressure of the system = P = 263.0 Torr

$P=p_1'+p_2'+p_3'$

$263.0Torr=13.2 Torr+p_2'+217.0 Torr$

$p_2'=32.8 Torr$

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At initail

(13.2) Torr (32.8) Torr (13.2) Torr

At equilbriumm

(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr

$K_p=\frac{p_3'}{p_1'\times p_2'}$

$1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}$

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

$p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr$

$p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr$

$p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr$

6 0

3 years ago

calculate the volume occupied by 10g of propane gas, under normal conditions of temperature and pressure

andriy [413]

Answer:

5.5 L

Explanation:

First we convert 10 g of propane gas (C₃H₈) to moles, using its molar mass:

10 g ÷ 44 g/mol = 0.23 mol

Then we use the PV=nRT formula, where:

P = 1 atm & T = 293 K (This are normal conditions of T and P)

n = 0.23 mol

R = 0.082 atm·L·mol⁻¹·K⁻¹

V = ?

1 atm * V = 0.23 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 K

V = 5.5 L

3 0

3 years ago