We are given with the initial volume of the substance and the molarity. The first thing that needs to be done is to multiply the equation in order to obtain the number of moles such as shown below.
number of moles = (40 mL) x (1 L / 1000 mL) x (0.3433 moles / L)
number of moles = 0.013732 moles
To get the value of the molarity of the diluted solution, we divide the number of moles by the total volume.
molarity = (0.013732 moles) / (750 mL / 1000 mL/L) = 0.0183 M
Similarly, we can solve for the molarity by using the equation,
M₁V₁ = M₂V₂
Substituting the known values in the equation,
(0.3433 M)(40 mL) = M₂(750 mL)
M₂ = 0.0183 M
The density of the gold is 19.3 grams/cc so each cc weighs 19.3grams. Now we can obtain the volume of gold from the given dimensions ie 4.72x8.21x3.98= 154.23 cc. So for the answer, just multiply the volume or 154.23 x 19.3= 2976.6 grams is the answer.
<u>Answer:</u> The 
for the reaction is 54.6 kJ/mol
<u>Explanation:</u>
For the given balanced chemical equation:
We are given:
- To calculate
for the reaction, we use the equation:
![\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G_f%28reactant%29%5D)
For the given equation:
![\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28COCl_2%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28CO_2%29%7D%29%2B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28CCl_4%29%7D%29%5D)
Putting values in above equation, we get:
![\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%28-204.9%29%29-%28%281%5Ctimes%20%28-394.4%29%29%2B%281%5Ctimes%20%28-62.3%29%29%29%5D%5C%5C%5CDelta%20G%5Eo_%7Brxn%7D%3D46.9kJ%3D46900J)
Conversion factor used = 1 kJ = 1000 J
- The expression of
for the given reaction:
We are given:
Putting values in above equation, we get:
- To calculate the Gibbs free energy of the reaction, we use the equation:
where,
= Gibbs' free energy of the reaction = ?
= Standard gibbs' free energy change of the reaction = 46900 J
R = Gas constant = 
T = Temperature = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
= equilibrium constant in terms of partial pressure = 22.92
Putting values in above equation, we get:
Hence, the for the reaction is 54.6 kJ/mol
Answer:
=342g
Explanation:
atomic mass of C = 12g
atomic mass of H = 1g
atomic mass of O = 16g
Solution;
C12 H22 O11
= 12 (12) + 22 (1) + 11(16)
= 144+ 22 + 176
= 342g
According to that Kc is an equilibrium constant in terms of molar concentrations.
and Kc = [C]^c *[D]^d / [A]^a * [B]^b >>>> (1)
in the general reaction:
aA + bB ↔ cC + dD
and, from our balanced equation:
CH4 + H2O ⇔ Co + 3H2 >>> (2)
So, we need to calculate the concentrations (molarity) of the products and reactants:
the Molarity of CH4 = no. of moles/volume (L)
and no. of moles = weigh / Molecular weight = 42.3 / 16 = 2.643 moles
so the molarity of CH4 = 2.643 / 5 = 0.528 molar
the molarity of H20 = (49.2 / 18) / 5 = 0.546 molar
the molarity of CO = (8.32/28) / 5 = 0.059 molar
the molarity of H2 = (2.63 / 2) / 5 = 0.263 molar
By substitution in (1) according to (2);
∴ Kc = [0.059]*[0.263]^3 / ( [0.528]*[0.546]) = 3.7 * 10 ^-3 >>>> (3)
Kp = Kc (RT)^(Δn) >>> (4)
where R is the gas constant = 0.0821,
and Δn is the change in moles in gas= (3(H2) + 1 (CO) - (1 H2O + 1 CH4) = 2
by substition in (4):
∴ Kp = 3.7*10^-3 (0.0821* 1000)^2= 24.939
