Answer:
a) 3.98 x 10^-10
Explanation:
Hello,
In this case, for the given pH, we can compute the concentration of hydronium by using the following formula:
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
Hence, solving for the concentration of hydronium:
![[H^+]=10^{-pH}=10^{-9.40}\\](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-9.40%7D%5C%5C)
![[H^+]=3.98x10^{-10}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D3.98x10%5E%7B-10%7DM)
Therefore, answer is a) 3.98 x 10^-10
Best regards.
Answer:
a) Li2CO3
b) NaCLO4
c) Ba(OH)2
d) (NH4)2CO3
e) H2SO4
f) Ca(CH3COO)2
g) Mg3(PO4)2
f) Na2SO3
Explanation:
a) 2Li + CO3 ↔ Li2CO3
b) NaOH * HCLO4 ↔ NaCLO4 + H2O
c) Ba + 2H2O ↔ Ba(OH)2 +
d) 2NH4 + H2CO3 ↔ (NH4)2CO3 + H2O
c) SO2 + NO2 +H2O ↔ H2SO4 + NOx
f) 2CH3COOH + CaO ↔ Ca(CH3COOH)2 + H2O
g) 3MgO + 2H3PO4 ↔ Mg3(PO4)2 + H2O
h) NaOH + H2SO3 ↔ Na2SO3 + H2O
Could you attach a picture because I can tell you didn't post the entire question.
ΔG⁰ = ΔH⁰ - TΔS
ΔH⁰ = Hf,(CH₃OH) - Hf,(CO) = -238.7 + 110.5 = -128.2 kJ/mol
ΔS = S(CH₃OH) - S(CO) - 2S(H₂) = 126.8 - 197.7 - 2 x 130.6 = -332.1 J/mol.K
So
ΔG⁰ = - 128200 + 332.1 T
For the reaction to be spontaneous:
ΔG⁰ < 0
So: -128200 + 332.1 T < 0
332.1 T < 128200
T < 386.028 K
<span>U-236 spontaneously decays to Br-87, X and three neutrons. The element X is 4. La, also known as Lanthanum, number 57 in the periodic system of elements.</span>
