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amid [387]
2 years ago
11

Write a balanced equation depicting the formation of one mole of NaBr(s) from its elements in their standard states.

Chemistry
1 answer:
Lilit [14]2 years ago
6 0

Answer:

Check Explanation.

Explanation:

Formation reactions are chemical reactions where one mole of a compound is produced from its constituent elements in their standard states.

NaBr (s)

The Standard formation reaction is

Na (s) + (1/2)Br₂ (g) → NaBr (s)

Using appendix C, the standard heat of formation of NaBr(s) is

ΔH∘f = -359.8 kJ/mol.

SO₃ (g)

The Standard formation reaction is

S (s) + (3/2) O₂ (g) → SO₃ (g)

Using appendix C, the standard heat of formation of SO₃(g) is

ΔH∘f = -395.2 kJ/mol.

Pb(NO₃)₂ (s)

The Standard formation reaction is

Pb (s) + N₂ (g) + 3O₂ (g) → Pb(NO₃)₂ (s)

Using appendix C, the standard heat of formation of Pb(NO₃)₂(s) is

ΔH∘f = -451.9 kJ/mol.

Hope this Helps!!!

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<span>n = PV / RT hop it helps u ok</span>
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a 9.84 ounces ingot of unknown metal is heated from 73.2 degrees fahrenheit - 191.2 degrees fahrenheit this requires 3.912 calor
Gekata [30.6K]

The SI unit of specific heat is J per gram per degree Celsius. Thus it follows that specific heat could be calculated in this way:

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2 years ago
Can 750 mL of water dissolve 0.60 mol of gold(III) chloride (AuCl3)?
nydimaria [60]

Answer:

Yes  

Explanation:

1. Mass of 0.60 mol of AuCl₃  

\text{Mass} = \text{0.60 mol} \times \dfrac{\text{303.33 g}}{\text{1 mol}} = \text{184 g}

2. Mass of AuCl₃ in 750 mL

The solubility of AuCl₃ is 68 g/100 mL.

In 750 mL of water, you can dissolve

\text{Mass of AuCl}_{3} = \text{750 mL} \times \dfrac{\text{68 g}}{\text{100 mL}} = \text{510 g AlCl}_{3}

∴ Yes, 750 mL of water can dissolve 0.60 mol of AuCl₃.

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Aleksandr-060686 [28]
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754.3 mmHg / 760 mmHg * 1atm = 0.99 atm
760 mmHg = 101.3 KPa
754.3 mmHg/ 760mmHg *101.3 KPa = 100.54 KPa

Hope this helps!
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