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Shkiper50 [21]
3 years ago
13

!!!PLEASE HELP!!!!

Chemistry
1 answer:
Lynna [10]3 years ago
4 0

Answer: B

Explanation: molarity = concentration c= n/V = 0.5 mol/ 0.05 l = 10 mol/l

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Calculate AH for the reaction:<br> 2N2 (g) + 6H20 (g) → 302 (g) + 4NH3(g)
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Answer:

Explanation:

4NH₃ (g) + 3O₂ (g) ⇒ 2N₂ (g) + 6H₂ O(1)

ΔH_r={\sum}{ \Delta }H_f(\text{product})-{\sum}{ \Delta }H_f(\text{reactant})

ΔH r =(2ΔH f(N 2 )+6ΔH f (H 2 O(l)))−(4ΔH f​ (NH 3 (g))+3ΔH f (O 2 (g)))

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Isotopes of the same element have different numbers of<br> protons<br> electrons<br> neutrons
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3 years ago
You are given solutions of hcl and naoh and must determine their concentrations. you use 27.5 ml of naoh to titrate 100.0 ml of
Dafna1 [17]
1) Start by standardizing the solution of NaOH by using the solution of H2SO4 whose concentration is known.

2) Equation:

2Na OH + H2SO4 --> Na2 SO4 + 2H2O

3) molar ratios

2 mol NaOH : 1 mol H2SO4

4) Number of moles of H2SO4 in 50.0 ml of 0.0782 M solution

M = n / V => n = M*V = 0.0782 M * 0.050 l = 0.00391 mol H2SO4

5) Number of moles of NaOH

2 moles NaOH / 1mol H2SO4 * 0.00391 mol H2SO4 = 0.00782 mol NaOH

6) Concentration of the solution of NaOH

M = n / V = 0.00782 mol / 0.0184 ml = 0.425 M

7) Standardize the solution of HCl

Chemical reaction:

NaOH + HCl --> NaCl + H2O

8) Molar ratios

1 mol NaOH : 1 mol HCl

9) Number of moles of NaOH in 27.5 ml

M = n / V => n = M * V = 0.425 M * 0.0275 l = 0.01169 moles NaOH

10) Number of moles of HCl

1 mol HCl / 1mol NaOH * 0.01169 mol NaOH = 0.01169 mol HCl

11) Concentration of the solution of HCl

M = n / V = 0.01169 mol / 0.100 l = 0.1169 M

Rounded to 3 significant figures = 0.117 M

Answers:

[NaOH] = 0.425 M
[HCl] = 0.117 M
3 0
3 years ago
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