Answer:
x = 4.17
y = 1.86
Explanation:
0.62 = log(x)
x = 10^0.62 = 4.17 ( to the nearest hundredth)
0.62 = ln(y)
y = e^0.62 = 1.86 (to the nearest hundredth)
<u>Answer:</u> The mass of nitrogen gas reacted to produce given amount of energy is 5.99 grams.
<u>Explanation:</u>
The given chemical reaction follows:

We know that:
Molar mass of nitrogen gas = 28 g/mol
We are given:
Enthalpy change of the reaction = 14.2 kJ
To calculate the mass of nitrogen gas reacted, we use unitary method:
When enthalpy change of the reaction is 66.4 kJ, the mass of nitrogen gas reacted is 28 grams.
So, when enthalpy change of the reaction is 14.2 kJ, the mass of nitrogen gas reacted will be = 
Hence, the mass of nitrogen gas reacted to produce given amount of energy is 5.99 grams.
Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
Answer:
C po
Explanation:
baka po pero pwd din ang B pero ok ang C
Answer:
B. Velocity
Explanation:
The answer is Velocity because the def of velocity is the rate of constant speed in a given direction. I hope this helps, have a good night.