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4 years ago

PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

Chemistry

2 answers:

malfutka [58]4 years ago

8 0

Answer: I think G.

Explanation:

Thepotemich [5.8K]4 years ago

8 0

Answer:

I believe its student 2 as well. it's most definitely g

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Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re

borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

A + B ⇄ C

where;

numbers of moles = 0.386 mol C (g)

Volume = 7.29 L

Molar concentration of C = $\frac{0.386}{7.29}$

= 0.053 M

A + B ⇄ C

Initial 0 0 0.530

Change +x +x - x

Equilibrium x x (0.0530 - x)

$K = \frac{[C]}{[A][B]}$

where

K is given as ; 78.2 atm-1.

So, we have:

$78.2=\frac{[0.0530-x]}{[x][x]}$

$78.2= \frac{(0.0530-x)}{(x^2)}$

$78.2x^2= 0.0530-x$

$78.2x^2+x-0.0530=0$

Using quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where; a = 78.2 ; b = 1 ; c= - 0.0530

= $\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

= $\frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$ or $\frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$

= 0.0204 or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

= 0.0530 - 0.0204

= 0.0326

Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

$V = \frac{nRT}{P}$

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K (fixed constant temperature )

V (volume) = ???

$V=\frac{(0.0734*0.0821*273.15)}{(1.00)}$

V = 1.64604

V ≅ 1.65 L

3 0

3 years ago

CH4 + 2O2 —> 2H2O + CO2

Dovator [93]

Answer: 20 mol CH4 x 1CO2 / 1CH4 = 20mol CO2

Answer: 20mol CO2

Explanation:

4 0

3 years ago

A sample of gas contains 0.1900 mol of CO(g) and 0.1900 mol of NO(g) and occupies a volume of 22.0 L. The following reaction tak

worty [1.4K]

Answer:

V₂ = 16.5 L

Explanation:

To solve this problem we use <em>Avogadro's law, </em>which applies when temperature and pressure remain constant:

V₁/n₁ = V₂/n₂

In this case, V₁ is 22.0 L, n₁ is [mol CO + mol NO], V₂ is our unknown, and n₂ is [mol CO₂ + mol N₂].

n₁ = mol CO + mol NO = 0.1900 + 0.1900 = 0.3800 mol

<em>We use the reaction to calculate n₂</em>:

2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g)

mol CO₂:

0.1900 mol CO * $\frac{2molCO_{2}}{2molCO} =$ 0.1900 mol CO₂

mol N₂:

0.1900 mol NO * $\frac{1molN_{2}}{2molNO} =$ 0.095 mol N₂

n₂ = mol CO₂ + mol N₂ = 0.1900 + 0.095 = 0.2850 mol

Calculating V₂:

22.0 L / 0.3800 mol = V₂ / 0.2850 mol

V₂ = 16.5 L

3 0

4 years ago

How many elements are found in the third period. A.2 B.18 C.8 D.32

mamaluj [8]

Answer:

Explanation:

u can find it on ur chemistry book

3 0

3 years ago

Read 2 more answers

Lead metal is produced by heating solid lead (II) sulfide with solid lead (II) sulfate, resulting in liquid lead and sulfur diox

olya-2409 [2.1K]

Answer:

$PbS_{(s)}+PbSO_4_{(s)}\overset{\Delta}{\rightarrow} 2Pb_{(l)}+2SO_2_{(g)}$

Explanation:

The balanced reaction of heating solid lead (II) sulfide with the solid lead (II) sulfate to produce liquid lead and sulfur dioxide gas is shown below:

$PbS_{(s)}+PbSO_4_{(s)}\overset{\Delta}{\rightarrow} 2Pb_{(l)}+2SO_2_{(g)}$

In the balance reaction above, all the phases are indicated. This reaction is used for the production of lead metal.

4 0

3 years ago