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Andreas93 [3]
3 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]3 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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Blababa [14]
First, we write the balanced equation for this reaction:

2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂

From this equation, we see that there are 2 moles of potassium iodide required for each mole of lead (II) nitrate. Moreover, we may use the formula:

Moles = volume (in L) * molarity

We find the molar relation ship for KI : Pb(NO₃)₂ to be 2 : 1. So:

M₁V₁ = 2M₂V₂

V₁ = 2M₂V₂/M₁
V₁ = 2 * 0.112 * 0.155 / 0.2
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1. A 25 g rock is placed in a graduated cylinder with water. The volume of the liquid rises from 18.3 mL to 21.4 mL Calculate th
MaRussiya [10]

Answer:

1. \rho=8.06g/cm^3

2. H=10cm

Explanation:

Hello,

1. In this case, since the volume of the rock is obtained via the difference between the volume of the cylinder with the water and the rock and the volume of the cylinder with the water only:

V=21.4mL-18.3mL=3.1mL

Thus, the density turns out:

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2. In this case, given the density and mass of aluminum we can compute its volume as follows:

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Moreover, as the volume is also defined in terms of width, height and length:

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The height is computed to be:

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3 years ago
You need to prepare a solution with a specific concentration of Na+Na+ ions; however, someone used the end of the stock solution
babymother [125]

Explanation:

Ionic equation

NaCl(aq) --> Na+(aq) + Cl-(aq)

Na2SO4(aq) --> 2Na+(aq) + SO4^2-(aq)

In NaCl solution, 1 mole of Na+ is dissociated in 1 liter of solution while in Na2SO4, 2 moles of Na+ is dissociated in 1 liter of solution.

Molecular weight of NA2SO4 = (23*2) + 32 + (16*4)

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Molecular weight of NaCl = 23 + 35.5

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% Mass of NA+ in NaCl = mass of Na+/total mass of NaCl * 100

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