It would be 35.8 Calories or calories. Not sure about that part. Hope this helps though.
Answer:
i think it's C
Hope It Helps!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! :D
Given what we know, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.
<h3>Why does it take this much energy to boil the water?</h3>
We arrive at this number by taking into account the energy needed to boil 1g of water to its vaporization point. This results in the use of 2260 J of heat energy. We then take this number and multiply it by the total grams of water being heated, in this case, 5.05g, which gives us our answer of 11.4 kJ of energy required.
Therefore, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.
To learn more about the behavior of water visit:
brainly.com/question/1416592?referrer=searchResults
H₂CO₃ ⇔ HCO₃⁻ + H⁺
I 0.160 0 0
C -x +x +x
E 0.160-x +x +x
Ka1 = [HCO₃⁻][H⁺] / [H₂CO₃]
4.3 x 10⁻⁷ = x² / (0.160-x) (x is neglected in 0.160-x = 0.160)
x² = 6.88 x 10⁻⁸
x = 2.62 x 10⁻⁴
HCO₃⁻ ⇔ CO₃⁻² + H⁺
I 2.62 x 10⁻⁴ 0 2.62 x 10⁻⁴
C -x +x +x
E 2.62 x 10⁻⁴ - x +x 2.62 x 10⁻⁴ + x
Ka2 = [CO₃⁻²][H⁺] / [HCO₃⁻]
5.6 x 10⁻¹¹ = x(2.62 x 10⁻⁴ + x) / (2.62 x 10⁻⁴ - x)
x = 5.6 x 10⁻¹¹
Thus,
[H₂CO₃] = 0.160 - (2.62 x 10⁻⁴) = 0.16 M
[HCO₃⁻] = 2.62 x 10⁻⁴ - ( 5.6 x 10⁻¹¹) = 2.6 x 10⁻⁴ M
[CO₃⁻²] = 5.6 x 10⁻¹¹ M
[H₃O⁺] = 2.62 x 10⁻⁴ + 5.6 x 10⁻¹¹ = 2.6 x 10⁻⁴ M
[OH⁻] = 3.8 x 10⁻¹¹
