Answer:
The mass of the object would be 491.55g/mL. Hope this helped! :)
Explanation:
D= 7.25g/mL M=D•V
V= 67.8mL
M=7.25g/mL•67.8mL
M= 491.55g
Answer:
b. The amount of SO3(g) decreases and the value for K increases.
Explanation:
Hello,
In this case, for the given reaction:
The change in the stoichiometric coefficient is:
In such a way, since the reagents have more moles than the products, based on Le Chatelier's principle, if the volume is increased, the side with more moles is favored. In addition, since the formation of reagent is favored, K is diminished based on the law of mass action shown below:
![K=\frac{[SO_3]_{eq}^2}{[SO_2]_{eq}^2[O_2]_{eq}}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BSO_3%5D_%7Beq%7D%5E2%7D%7B%5BSO_2%5D_%7Beq%7D%5E2%5BO_2%5D_%7Beq%7D%7D)
Therefore the answer is:
b. The amount of SO3(g) decreases and the value for K increases.
Best regards.
Ksp - solubility product constant is equivalent to equilibrium constant, except this constant is used to determine the solubility of ions of a solid in a solution.
ksp is the product of the soluble ions in the compound. Higher the ksp value, higher the degree of solubility.
ZnCO₃ (s) ---> Zn²⁺ (aq) + CO₃²⁻ (aq)
n n
ksp = [Zn²⁺][CO₃²⁻]
In the equation equal amounts of ions Zn²⁺ and CO₃²⁻ ions are soluble.
amount of ions soluble = n
ksp is therefore equal to;
ksp = n x n
ksp = n²
ksp = 1 * 10⁻¹⁰ M
therefore
1 * 10⁻¹⁰ M = n²
n = 1 x 10⁻⁵ M
therefore concentration of CO₃²⁻ = 1 x 10⁻⁵ M
Answer:
<h2>The answer is 3.7 L</h2>
Explanation:
The new volume can be found by using the formula for Boyle's law which is
Since we are finding the new volume
From the question we have
We have the final answer as
<h3>3.7 L</h3>
Hope this helps you
Answer:
1. Cu = +2
2. O = – 2
3. C = +2
4. N = 0
Explanation:
Note:
A. The oxidation state of oxygen is always – 2 except in peroxides where it is – 1.
B. The oxidation state of Hydrogen is always +1 except in metallic hydrides where it is – 1.
Now let us solve the question given above.
1. CuO = 0 (ground state)
Cu + O = 0
O = – 2
Cu + (– 2) = 0
Cu – 2 = 0
Collect like terms
Cu = 0 + 2
Cu = +2
2. H2O = 0 (ground state)
2H + O = 0
H = +1
2(1) + O = 0
2 + O = 0
Collect like terms
O = 0 – 2
O = – 2
3. CO = 0 (ground state)
C + O = 0
O = – 2
C + (– 2) = 0
C – 2 = 0
Collect like terms
C = 0 + 2
C = +2
4. N2 = 0 (ground state)
2N = 0
Divide both side by 2
N = 0/2
N = 0
