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4 years ago

7

An atomic nucleus has. nuclear potential energy. chemical potential energy. electrical potential energy. static potential energy

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1 answer:

kolezko [41]4 years ago

7 0

It would have nuclear potential energy if I remember correctly

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What is the molar mass for Agno3

Aloiza [94]

Answer:

169.87 g/mol

Explanation:

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3 years ago

A reaction that occurs in the internal combustion engine is n2(g) + o2(g) ⇌ 2 no(g) (a) calculate δh o and δs o for the reaction

jekas [21]

1) ΔrH = 2mol·ΔfH(NO) - (ΔfH(O₂) + ΔfH(N₂)).
ΔrH = 2 mol · 90.3 kJ/mol - (0 kJ/mol + 0 kJ/mol).
ΔrH = 180.6 kJ.
2) ΔS = 2mol·ΔS(NO) - (ΔS(O₂) + ΔS(N₂)).
ΔS = 2mol · 210.65 J/mol·K - (1mol · 205 J/mol·K + 1 mol · 191.5 J/K·mol).
ΔS = 24.8 J/K.
3) ΔG = ΔH - TΔS.
55°C: ΔG = 180.6 kJ - 328.15 K · 24.8 J/K = 172.46 kJ.
2570°C: ΔG = 180.6 kJ - 2843.15 K · 24.8 J/K = 110.09 kJ.
3610°C: ΔG = 180.6 kJ - 3883.15 K · 24.8 J/K = 84.29 kJ.

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3 years ago

Which is an example of a catalyst?<br><br> heat<br><br> stirring<br><br> decrease in temperature

Lorico [155]

I think the correct answer is decrease in temperature
I feel like that’s the right one

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4 years ago

Look at the position of Location 1, Location 2, and Location 3 in this picture. Simple diagram of a lake represented by a wavy c

alexgriva [62]

Answer: Location 3 is warmer than location 2 and 1

Explanation:

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3 years ago

Read 2 more answers

Calculate the OH− concentration after 53 mL of the 0.100 M KOH has been added to 25.0 mL of 0.200 M HBr. Assume additive vol- um

Andre45 [30]

Answer:

$\large \boxed{\text{0.0038 mol/L}}$

Explanation:

1. Calculate the initial moles of acid and base

$\text{moles of acid} = \text{0.0250 L} \times \dfrac{\text{0.200 mol}}{\text{1 L}} = \text{0.005 00 mol}\\\\\text{moles of base} = \text{0.053 L} \times \dfrac{\text{0.100 mol}}{\text{1 L}} = \text{0.0053 mol}$

2. Calculate the moles remaining after the reaction

OH⁻ + H₃O⁺ ⟶ 2H₂O

I/mol: 0.0053 0.005 00

C/mol: -0.00500 -0.005 00

E/mol: 0.0003 0

We have an excess of 0.0003 mol of base.

3. Calculate the concentration of OH⁻

Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L

$\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}$

8 0

3 years ago