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atroni [7]
3 years ago
14

Calculate the molarity of 0.700 mol of na2s in 1.05 l of solution.

Chemistry
1 answer:
slamgirl [31]3 years ago
6 0
Molarity is the amount of solute molecule (in moles) per 1L of solvent. In this case, the solute is 0.7mol Na2S and the solvent volume is 1.05L. Since the unit in this problem is already mol and L then you don't need to do any conversion of the units. The calculation would be:

molarity = mol of solute / (1L/ volume of solvent)
molarity = 0.7 mol/ (1L/ 1.05L)= 0.67M
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<h3><u>Answer;</u></h3>

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7 0
3 years ago
An element's atomic number is 60. How many protons would an atom of this element have? __protons
Lady_Fox [76]
If an an element has an atomic number of 60, then there would me 60 protons.
Something that helps me is APE.
The atomic number is the same as the number of protons and electrons.
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3 years ago
Write the action of monohydroxy benzene with aqueous bromine​
Romashka-Z-Leto [24]

The reaction yields the product compound 2,4,5-tribromophenol.

<h3>Bromination of monohydroxy benzene</h3>

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Learn more about benzene: brainly.com/question/14525517

5 0
3 years ago
You add 7.8 g of iron to 20.70 mL of water and observe that the volume of iron and water together is 21.69 mL . Calculate the de
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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

d)

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
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