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3 years ago

Find the mass percentage of water in ZnSO4 · 7H20 PLEASE HELP

Chemistry

1 answer:

netineya [11]3 years ago

8 0

Answer:

Hello, I can help!

Explanation:

Molecular Formula: ZnSO4. 7H2O or H14O11SZn

Molecular Weight: 287.6 g/mol

<h2>Computing molar mass (molar weight)</h2>

Functional groups: D, Ph, Me, Et, Bu, AcAc, For, Ts, Tos, Bz, TMS, tBu, Bzl, Bn, Dmg

Common compound names.

Examples of molar mass computations: NaCl, Ca(OH)2, K4[Fe(CN)6], CuSO4*5H2O, water, nitric acid, potassium permanganate, ethanol, fructose.

-Molecular mass (molecular weight) is the mass of one molecule of a substance and is expressed in the unified atomic mass units (u). (1 u is equal to 1/12 the mass of one atom of carbon-12)

-Molar mass (molar weight) is the mass of one mole of a substance and is expressed in g/mol.

God bless you!

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Within an atom, electrons A. are in motion inside the nucleus. B. are in motion outside the nucleus. C. are inside the nucleus a

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3 years ago

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A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

natali 33 [55]

Answer:

There are 0.93 g of glucose in 100 mL of the final solution

Explanation:

In the first solution, the concentration of glucose (in g/L) is:

15.5 g / 0.100 L = 155 g/L

Then a 30.0 mL sample of this solution was taken and diluted to 0.500 L.

30.0 mL equals 0.030 L (Because 30.0 mL ÷ 1000 = 0.030 L)

The concentration of the second solution is:

$155 \frac{g}{L} *\frac{0.030L}{0.500L}=9.3\frac{g}{L}$

So in 1 L of the second solution there are 9.3 g of glucose, in 100 mL (or 0.1 L) there would be:

1 L --------- 9.3 g

0.1 L--------- Xg

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8 0

3 years ago

Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of c

Anna [14]

The question is incomplete. The complete question is

Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of calcium nitrate. A double displacement reaction occurs. What mass of each of the following substances is present when the reaction stops. A) potassium phosphate remaining B) calcium nitrate g remaining C) calcium phosphate formed D) potassium nitrate g formed

Answer:

a)84.91g

b)8.20g

c)316.4g

d)616.73g

Explanation:

The equation of the reaction:

2K3PO4(aq) + 3Ca(NO3)2 (aq)-------> 6KNO3(aq) + Ca3(PO4)2(s)

Molar mass of potassium phosphate= 212.27 g/mol

Amount of potassium phosphate= 500/212.27= 2.4 moles

Molar mass of calcium nitrate= 164.088 g/mol

Amount of calcium nitrate= 500/164.088=3.05moles

a) amount of potassium phosphate reacted according to reaction equation= 2 moles

Amount of potassium phosphate remaining= 2.4-2=0.4 moles

Mass of potassium phosphate remaining= 0.4×212.27=84.91g

b) Amount of calcium nitrate reacted according to reaction equation=3

Amount of calcium nitrate remaining=3.05-3= 0.05

Mass of calcium nitrate remaining= 0.05×164.088= 8.20g

c) since calcium nitrate is the limiting reactant, we use to estimate the mass of products formed.

From the reaction equation,

3 moles of calcium nitrate yields 1 mole of calcium phosphate

3.05 moles of calcium nitrate yields 3.05/3 = 1.02 moles of calcium phosphate

Molar Mass of calcium phosphate= 310.18 g/mol

Mass of calcium phosphate produced= 1.02×310.18= 316.4g

3 moles of calcium nitrate yields 6 moles of potassium nitrate

3.05 moles of calcium nitrate yields 3.05×6/3= 6.1 moles of potassium nitrate

Molar mass of potassium nitrate = 101.1032 g/mol

Mass of potassium nitrate formed= 6.1× 101.1032= 616.73g

6 0

3 years ago