Fill in the coefficients that will balance the following reaction:

PLEASE HELP 50 POINTS

stepladder [879]

It’s c add the side

4 0

2 years ago

Read 2 more answers

(I00p)A chemical equation example of a replacement of 1 ion for another in a compound! I need an example

snow_tiger [21]

A single replacement reaction could look like this:

2FeCl3 + 3Ba ➡️ 3BaCl2 + 2Fe

In this reaction, the barium is replacing the iron bound to the chlorine.

8 0

3 years ago

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago

Which of the following events could increase Earth's temperature?

Tcecarenko [31]

I believe the answer Is B

5 0

3 years ago

Read 2 more answers

How many moles of propane<br> react when 294 g of CO2 form?<br><br> C3H8 +502 → 3CO₂ + 4H₂O

tatiyna

2.23 moles of propane react when 294 g of CO₂ is formed .

<h3>What is moles ?</h3>

Moles is a unit which is equal to the molar mass of an element.

A reaction is given

C₃H₈ +50₂ → 3CO₂ + 4H₂O

Grams of CO₂ formed = 294 gm

In moles = 294 /44 = 6.68 moles.

Let x be the moles of C₃H₈ is x

Mole ratio of CO₂ to C₃H₈ = 3 : 1

so

6.68 /x = 3/1

x = 6.68 /3 = 2.23 moles

Therefore 2.23 moles of propane react when 294 g of CO₂ is formed .

To know more about Moles

brainly.com/question/26416088

#SPJ1

7 0

2 years ago