Answer:
decreased by a factor of 10
Explanation:
pH is defined in such a way that;
pH= −log10(H)
Where H represents the concentration of Hydronium or Hydrogen ions
Given that pH is changed from 1 to 2,
By rearranging the above formula , we get 10−pH = H
- if pH=1,H=10−1=0.1M
- if pH=2,H=10−2=0.01M
Therefore, 0.1/0.01 = 10 and 0.1 > 0.01
Hence, the concentration of hydronium ions in the solution is decreased by a factor of 10
13.6
a) yes Pb is more reactive that Ag, Pb before Ag
b) no, Cu after H
c) yes, Cl2 is more active than I2
4) yes, Mg is more active
13.7 (as I think)
Al ³⁺ more active than Zn²⁺, Mn can react with Zn²⁺, but not with Al ³⁺ , because Mn after Al but before Zn
Answer:
False
Explanation:
Atoms only achieve complete outer electron shells if they contain an outer shell with 7 electrons before gaining another electron or an outer shell with 1 electron before losing an electron. This is assuming that the octet-rule can be applied to said atom. In addition, the number of valence electrons varies from atom to atom which is why not ALL atoms achieve complete outer electron shells after gaining or losing just ONE electron.
Answer : The products are Silver sulfide, 
and Sodium iodide, 
.
Explanation :
The given balanced chemical reaction is,
From the given balanced reaction, we conclude that the 2 moles of silver iodide react with the 1 mole of sodium sulfide to give product as 1 mole of silver sulfide and 2 moles of sodium iodide.
In a chemical reaction, reactants are represent on the left side of the right-arrow and products are represent on the right side of the right-arrow.
Therefore, in a chemical reaction the products are Silver sulfide and Sodium iodide.
Answer:
Explanation:
<u>1) Rate law, at a given temperature:</u>
- Since all the data are obtained at the same temperature, the equilibrium constant is the same.
- Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:
r = K [A]ᵃ [B]ᵇ
<u>2) Use the data from the table</u>
- Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s
Divide r₂ by r₁: [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0
- Use the first and second set of data to find the exponent a:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s
Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]
2ᵃ = 2 ⇒ a = 1
<u>3) Write the rate law</u>
This means, that the rate is independent of reactant B and is of first order respect reactant A.
<u>4) Use any set of data to find K</u>
With the first set of data
- r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹
Result: the rate constant is K = 0.167 s⁻¹
