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Finger [1]
2 years ago
9

What is the net ionic equation for this reaction? FeCl3 + 3NaOH → Fe(OH)3 + 3NaCl

Chemistry
1 answer:
jok3333 [9.3K]2 years ago
8 0

Answer:

2.6299

its easy

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PLEASE ANSWER
PolarNik [594]
I think its A.
if one force cannot overcome the other, the object remains stationary.
4 0
3 years ago
Given the following equation, N2O(g) + NO2(g) → 3 NO(g) ΔG°rxn = -23.0 kJ
Lelechka [254]

Answer:

ΔG°rxn = -69.0 kJ

Explanation:

Let's consider the following thermochemical equation.

N₂O(g) + NO₂(g) → 3 NO(g) ΔG°rxn = -23.0 kJ

Since ΔG°rxn < 0, this reaction is exergonic, that is, 23.0 kJ of energy are released. The Gibbs free energy is an extensive property, meaning that it depends on the amount of matter. Then, if we multiply the amount of matter by 3 (by multiplying the stoichiometric coefficients by 3), the ΔG°rxn will also be tripled.

3 N₂O(g) + 3 NO₂(g) → 9 NO(g) ΔG°rxn = -69.0 kJ

8 0
3 years ago
Find the concentration of H+ ions at a pH = 11 and
Black_prince [1.1K]

Explanation:

the pH of the solution defined as negatuve logarithm of H^+ ion concentration.

pH=-\log[H^+]

1. Hydrogen ion concentration when pH of the solution is 11.

11=-\log[H^+]

[H^+]=1\times 10^{-11} mol/L..(1)

At pH = 11, the concentration of H^+ ions is 1\times 10^{-11} mol/L.

2. Hydrogen ion concentration when the pH of the solution is 6.

6=-\log[H^+]'

[H^+]'=1\times 10^{-6} mol/L..(2)

At pH = 6, the concentration of H^+ ions is 1\times 10^{-6} mol/L.

3. On dividing (1) by (2).

\frac{[H^+]}{[H^+]'}=\frac{1\times 10^{-11} mol/L}{1\times 10^{-6} mol/L}=1\times 10^{-5}

The ratio of hydrogen ions in solution of pH equal to 11 to the solution of pH equal to 6 is 1\times 10^{-5}.

4. Difference between the H^+ ions at both pH:

1\times 10^{-6} mol/L-1\times 10^{-11} mol/L=9.99\time 10^{-7} mol/L

This means that Hydrogen ions in a solution at pH = 7 has 9.99\time 10^{-7} mol/L ions fewer than in a solution at a pH = 6

6 0
3 years ago
Read 2 more answers
or a particular isomer of C 8 H 18 , the combustion reaction produces 5104.1 kJ of heat per mole of C 8 H 18 ( g ) consumed, und
beks73 [17]

Answer:

The standard enthalpy of formation of this isomer of C_{8}H_{18} is -220.1 kJ/mol.

Explanation:

The given chemical reaction is as follows.

C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)

\Delta H^{o}_{rxn}= -5104.1kJ/mol

The expression for the entropy change for the reaction is as follows.

\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]

\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol

\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol

\Delta H^{o}_{f}(O_{2})= 0kJ/mol

Substitute the all values in the entropy change expression.

-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]

-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})

\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol

=-220.1kJ/mol

Therefore, The standard enthalpy of formation of this isomer of C_{8}H_{18} is -220.1 kJ/mol.

4 0
3 years ago
Help me solve. Djdjdjfjdkdjdjdkdkckckdkck
ivanzaharov [21]
I would believe it to be 55g. A -> B YIELDS AB. So, 10g + 45g = 55g.
4 0
3 years ago
Read 2 more answers
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