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Andreas93 [3]
2 years ago
5

What is the pH of a solution that contains 25 grams of hydrochloric acid (hcl) dissolved in 1.5 liters of water?

Chemistry
1 answer:
Darina [25.2K]2 years ago
4 0
First we calculate the concentration of HCl:

Moles = mass / Mr
= 25 / 36.5
= 0.685 mol

Concentration = 0.685/1.5 = 0.457 mol / dm³

For a strong monoprotic acid, the concentration of hydrogen ions is equal to the acid concentration.

pH = -log[H+]
pH = -log(0.457)
= 0.34
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How many moles of H2O2 are needed to react with 1.07 moles of N2H4?
I am Lyosha [343]

Answer:

2.14 moles of H₂O₂ are required

Explanation:

Given data:

Number of moles of H₂O₂ required = ?

Number of moles of N₂H₄ available = 1.07 mol

Solution:

Chemical equation:

N₂H₄  +   2H₂O₂       →   N₂ +  4H₂O

now we will compare the moles of H₂O₂ and N₂H₄

                          N₂H₄     :      H₂O₂  

                            1           :        2

                            1.07      :         2×1.07 = 2.14 mol

                   

6 0
3 years ago
The carbon-14 content of a wooden harpoon handle found in an Inuit archaeological site was found to be 61.9% of the carbon-14 co
astraxan [27]

Answer:

3,964 years.

Explanation:

  • It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
  • Half-life time is the time needed for the reactants to be in its half concentration.
  • If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
  • Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

  • The half-life of the element is 5,730 years.

  • For, first order reactions:

<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(5,730 years) = 1.21 x 10⁻⁴ year⁻¹.

  • Also, we have the integral law of first order reaction:

<em>kt = ln([A₀]/[A]),</em>

where, k is the rate constant of the reaction (k = 1.21 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of the sample ([A₀] = 100%).

[A] is the remaining concentration of the sample ([A] = 61.9%).

∴ t = (1/k) ln([A₀]/[A]) = (1/1.21 x 10⁻⁴ year⁻¹) ln(100%/61.9%) = 3,964 years.

7 0
3 years ago
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slega [8]

Answer:

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It could be Balance Mass

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