Answer:
3.269 kg
Explanation:
1000g = 1 kg. Since kg is the larger unit, you need to divide by 1000 to get kilograms. You can do the problem as a unit conversion:
3269 g × <u>1 kg</u> = 3.269 kg
1000g
Hope this is helpful!
For the equation to balance out, you would need 2 H2O on the product. That will give a total of 1 Carbon, 4 Hydrogens and 4 Oxygens on each side.
Answer:
I know you have been waiting awhile for this question to be answered :)
Stoichiometry is used in industry quite often to determine the amount of materials required to produce the desired amount of products in a given useful equation. Each one of these products requires stoichiometry. There would be no products from these industries without chemical stoichiometry.
Explanation:
Hopefully this helps :D
Sorry you had to wait so long :(
Answer:
The equation to show the the correct form to show the standard molar enthalpy of formation:

Explanation:
The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.
Given, that 1 mole of
gas and 1 mole of
liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.

Divide the equation by 2.

The equation to show the the correct form to show the standard molar enthalpy of formation:

<u>Answer:</u> The equilibrium concentration of
is 0.332 M
<u>Explanation:</u>
We are given:
Initial concentration of
= 2.00 M
The given chemical equation follows:

<u>Initial:</u> 2.00
<u>At eqllm:</u> 2.00-2x x x
The expression of
for above equation follows:
![K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BCF_4%5D%7D%7B%5BCOF_2%5D%5E2%7D)
We are given:

Putting values in above expression, we get:

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible
So, equilibrium concentration of ![COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M](https://tex.z-dn.net/?f=COF_2%3D%282.00-2x%29%3D%5B2.00-%282%5Ctimes%200.834%29%5D%3D0.332M)
Hence, the equilibrium concentration of
is 0.332 M