Answer:
C
Explanation:
it's always colder at night and close to night especially near water and that's when wind tends to blow the most
Answer:
c = 0.898 J/g.°C
Explanation:
1) Given data:
Mass of water = 23.0 g
Initial temperature = 25.4°C
Final temperature = 42.8° C
Heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Specific heat capacity of water is 4.18 J/g°C
ΔT = 42.8°C - 25.4°C
ΔT = 17.4°C
Q = 23.0 g × × 4.18 J/g°C × 17.4°C
Q = 1672.84 j
2) Given data:
Mass of metal = 120.7 g
Initial temperature = 90.5°C
Final temperature = 25.7 ° C
Heat released = 7020 J
Specific heat capacity of metal = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 25.7°C - 90.5°C
ΔT = -64.8°C
7020 J = 120.7 g × c × -64.8°C
7020 J = -7821.36 g.°C × c
c = 7020 J / -7821.36 g.°C
c = 0.898 J/g.°C
Negative sign shows heat is released.
Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,
pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
Now put all the given values in this expression, we get:
![6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=6.0%3D8.0%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
![\frac{[Deprotonated]}{[Protonated]}=0.01](https://tex.z-dn.net/?f=%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D%3D0.01)
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
![\frac{[Protonated]}{[Deprotonated]}=100](https://tex.z-dn.net/?f=%5Cfrac%7B%5BProtonated%5D%7D%7B%5BDeprotonated%5D%7D%3D100)
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100
Mg(OH)₂ ⇄ Mg²⁺ + 2 OH⁻
Ksp = [Mg²⁺] [OH⁻]²
6.0 x 10⁻¹⁰ = 0.10 x [OH⁻]²
[OH⁻] = 7.746 x 10⁻⁵ M
when Mg(OH)₂ 1st precipitates, [OH⁻] = 7.746 * 10⁻⁵ M
Fe(OH)₂ <—> Fe²⁺ + 2OH⁻
Ksp = [Fe²⁺] [OH⁻]²
7.9 x 10⁻¹⁶ = [Fe²⁺] x (7.746 x 10⁻⁵)²
[Fe²⁺] = 1.32 x 10⁻⁷ M
Answer: 1.32 x 10⁻⁷ M
