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Rasek [7]
1 year ago
12

A chemist combined 0.440 L of an unknown calcium solution with an excess of ammonium chromate. This resulted in the precipitatio

n of calcium chromate. The mass of the precipitate was 346.7 mg. What was the molar concentration of Ca2 in the original sample
Chemistry
1 answer:
Rina8888 [55]1 year ago
6 0

The concentration of the original calcium ions is 0.005 M

<h3>What is concentration?</h3>

The term concentration has to do with the amount of substance in solution. We know that the concentration can be measured in a lot of units such as mole/litre, grams per litre, percentage and so on.

As such we have the equation;

Ca^2+(aq) + (NH4)2CrO4(aq) --------> CaCrO4(s) + 2NH4^+(aq)

Number of moles of the precipitate = 346.7 * 10^-3 g/156 g/mol

= 0.0022 moles

Now;

1 mole of Ca^2+ produces 1 mole of CaCrO4 hence 0.0022 moles of CaCrO4 was produced by  0.0022 moles of CaCrO4.

Given that the volume of the solution is  0.440 L, the concentration of the solution is;   0.0022 moles/0.440 L

= 0.005 M

Learn more about molarity:brainly.com/question/8732513

#SPJ1

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a researcher obtains a sample of 0.070 M nitrate solition. A 20.0 mL aliquote of the nitrate solution is added to 10.0 mL of amm
katovenus [111]

Answer:

Concentration of nitrate in the new solution = 0.007 M

Explanation:

Given:

Concentration nitrate solution = 0.070 m

Volume of aliquote of the nitrate solution is add = 10.0 ml

Total volume = 100 ml

Find:

Concentration of nitrate in the new solution

Computation:

Number of M. mole = 0.070 m x 10.0 ml

Number of M. mole = 0.7 m-moles

Concentration of nitrate in the new solution = 0.7 m-moles / 100 ml

Concentration of nitrate in the new solution = 0.007 M

6 0
2 years ago
What property do atoms of these elements have that helps make the molecules polar
DENIUS [597]
You have not mention here about those elements but the general concept for this is the:
uneven distribution ------> polar
even distribution -------> non-polar
4 0
3 years ago
How many moles of ammonium nitrate will be produced 110.0g of ammonium carbonate
marshall27 [118]

Answer:

                     2.288 Moles of NH₄NO₃

Explanation:

The Balance chemical equation is as follow:

                  (NH₄)₂CO₃ + 2 HNO₃  →  2 NH₄NO₃ + H₂O + CO₂

To solve this problem we will do following steps:

Finding moles of Ammonium Carbonate:

As we know,

                        Moles  =  Mass / M.Mass

So,

                        Moles  =  110 g  /  96.08 g/mol

                        Moles  =  1.144 moles

Calculating moles of Ammonium Nitrate:

According to balance chemical equation;

                      1 mole of (NH₄)₂CO₃ produces  =  2 moles of NH₄NO₃

So,

          1.144 moles of (NH₄)₂CO₃ will produce  =  X moles of NH₄NO₃

Solving for X,

                     X =  2 moles × 1.144 moles ÷ 1 mole

                     X  =  2.288 moles of NH₄NO₃

8 0
3 years ago
An element has 8 protons and 8 neutrons in its nucleus and 8 electrons revolving around the nucleus. What is the atomic number a
Molodets [167]
C.

Protons is 8 so the atomic number will be 8.

Number of neutrons= atomic mass - atomic number

16-8 = 8.
5 0
3 years ago
Consider the reaction 4PH3(g) → P4(g) + 6H2(g) At a particular point during the reaction, molecular hydrogen is being formed at
Svetllana [295]

Answer:

The rate at which P_4 is being produced is 0.0228 M/s.

The rate at which PH_3 is being consumed is 0.0912 M/s.

Explanation:

4PH_3\rightarrow P_4(g)+6H_2(g)

Rate of the reaction : R

R=\frac{-1}{4}\frac{d[PH_3]}{dt}=\frac{1}{6}\frac{d[H_2]}{dt}=\frac{1}{1}\frac{d[P_4]}{dt}

The rate at which hydrogen is being formed = \frac{d[H_2]}{dt}=0.137 M/s

R=\frac{1}{6}\frac{d[H_2]}{dt}

R=\frac{1}{6}\times 0.137 M/s=0.0228 M/s

The rate at which P_4 is being produced:

R=\frac{1}{1}\frac{d[P_4]}{dt}

0.0228 M/s=\frac{1}{1}\frac{d[P_4]}{dt}

The rate at which PH_3 is being consumed :

R=\frac{-1}{4}\frac{d[PH_3]}{dt}

0.0228 M/s\times 4=\frac{-1}{1}\frac{d[PH_3]}{dt}

\frac{-1}{1}\frac{d[PH_3]}{dt}=0.912 M/s

6 0
3 years ago
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