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In-s [12.5K]
3 years ago
5

A solution is made by dissolving 26.42g of (NH4)2so4 in enough h2o to make 50.00ml of solution

Chemistry
2 answers:
abruzzese [7]3 years ago
7 0
Molarity = moles of solute/liters of solution

We first need to convert grams of ammonium sulfate to moles.
To convert from grams to moles, we need to divide by it's molecular mass.

26.42 g (NH₄)₂SO₄ * (1 mole (NH₄)₂SO₄ / 132.14 g (NH₄)₂SO₄) = 0.1999 mole (NH₄)₂SO₄

0.1999 mol/0.05000L = 3.998 M

Since there 4 significant digits in both the volume and mass, we are limited to 4 significant digits.

Your final answer is 3.998 M (NH₄)₂SO₄.
DochEvi [55]3 years ago
6 0
Moles of ammonium sulfate = 26.42/molar mass of (NH4)2SO4
                                             =   26.42/132.14 = 0.19 mole.

Molarity = moles of ammonium sulfate/volume of solution
             =                0.19/50x10^-3
             =                    3.8M

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Which term is de ned as the region in an atom where an electron is most likely to be located?
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3 years ago
Help needed Assignment is due Justify that H2SO4 is Arrhenius acid and KOH is Arrhenius base.
irakobra [83]

Answer:

<u><em>Arrhenius Acid:</em></u>

According to Arrhenius concept, Acids are proton donors.

Since H₂SO₄ have a proton (H⁺ ion) and it can donate it to be made a sulphate ion, So it is an Arrhenius acid.

See the following reaction =>

<u><em>H₂SO₄ + H₂O => HSO₄ + H₃O⁺</em></u>

<u><em>Arrhenius Base:</em></u>

An Arrhenius base is a a proton acceptor.

KOH accepts the proton to to made to KOH₂ and a proton acceptor.

See the following reaction =>

<u><em>KOH + H₂o => KOH₂ + OH⁻</em></u>

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6 0
3 years ago
Which of the following is true regarding the law of conservation of mass?
nexus9112 [7]
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8 0
3 years ago
0.254g lead(ii)ethanoate, on adding excess K2CrO4 solution, gave 0.130g of lead(ii)chromate precipitate. what is the percentage
Alborosie

Answer:

32.8%

Explanation:

All of the Pb⁺² species precipitated as lead(II) cromate, PbCrO₄ (we know this as excess K₂CrO₄ was used).

First we convert 0.130 g of PbCrO₄ into moles, using its molar mass:

  • 0.130 g ÷ 323 g/mol = 4.02x10⁻⁴ mol PbCrO₄

There's 1 Pb⁺² mol per PbCrO₄ mol, so in total 4.02x10⁻⁴ moles of Pb⁺² were in the ethanoate sample.

We <u>convert those 4.02x10⁻⁴ moles of Pb into grams</u>:

  • 4.02x10⁻⁴ mol * 207 g/mol = 0.083 g Pb

Finally we calculate the percentage composition of Pb:

  • 0.083 g Pb / 0.254 g salt * 100% = 32.8%
3 0
3 years ago
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