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ozzi
3 years ago
7

A 25.00 mL solution of 0.150 M NaCl is combined with 10.00 mL of a 0.0750 M CaCl2 solution. Assuming a total volume of 35.00 mL,

determine the concentration of chloride ion in the combined solution.
Chemistry
1 answer:
Rudiy273 years ago
5 0

Answer:

[Cl⁻] = 1,5x10⁻⁴M

Explanation:

First of all, let's determinate the mole of each salt.

Molarity . volume = Mole

Volume must be in L, cause molarity is mol/L

NaCl  → Na⁺  + Cl⁻

Ratio is 1:1

0.15 mol/L . 0.025L = 3.75x10⁻³ mole

As ratio is 1:1, from 3.75x10⁻³ mole of salt, I have 3.75x10⁻³ mole of chloride

CaCl₂ → Ca²⁺  +  2Cl⁻

Ratio is 1:2 so, from 1 mol of salt I'll get the double of mole of chloride

0.075 mol/L . 0.010 L = 7.5x10⁻⁴ mol

7.5x10⁻⁴ mol . 2 = 1.5x10⁻³ mole

Total mole of Cl⁻:  3.75x10⁻³ +  1.5x10⁻³  = 5.25x10⁻³

This 5.25x10⁻³ mole are present in a total volume of 35 mL.

Let's convert 35 mL in L → 0.035L   (35/1000)

Molarity is mol/L → 5.25x10⁻³ mol / 0.035L = 1,5x10⁻⁴M

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2 years ago
A hypothetical element consists of four isotopes. Use the information below to calculate the average atomic volume mass of the e
Keith_Richards [23]

Answer:

104.969 amu.

Explanation:

From the question given above, the following data were obtained:

Isotope A:

Mass of A = 107.977 amu

Abundance (A%) = 0.1620%

Isotope B:

Mass of B = 106.976 amu

Abundance (B%) = 1.568%

Isotope C:

Mass of C = 105.974 amu

Abundance (C%) = 47.14%

Isotope D:

Mass of D = 103.973 amu

Abundance (D%) = 51.13%

Average atomic mass =?

The average atomic mass of the element can be obtained as follow:

Average atomic mass = [(Mass of A × A%) /100] + [(Mass of B × B%) /100] + [(Mass of C × C%) /100] + [(Mass of D × D%) /100]

Average atomic mass = [(107.977 × 0.1620)/100] + [(106.976 × 1.568)/100] + [(105.974 × 47.14)/100] + [(103.973 × 51.13)/100]

= 0.175 + 1.677 + 49.956 + 53.161

= 104.969 amu

Therefore, the average atomic mass of the element is 104.969 amu.

8 0
3 years ago
Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the fi
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The question is incomplete, the complete question is;

Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the first heating, your crucible and contents weighs 17.51 g. After the second heating, your crucible and contents weighs 17.50 g.

What is the theoretical yield of sodium carbonate?

What is the experimental yield of sodium carbonate?

What is the percent yield for sodium carbonate?

Which errors could cause your percent yield to be falsely high, or even over 100%?

Answer:

See Explanation

Explanation:

We have to note that water is driven away after the second heating hence we are concerned with the weight of the pure dry product.

Hence;

From the reaction;

2 NaHCO3 → Na2CO3(s) + H2O(l) + CO2(g)

Number of moles of  sodium bicarbonate = 18.56 - 15.98 = 2.58 g/87 g/mol

= 0.0297 moles

2 moles of sodium bicarbonate yields 1 mole of sodium carbonate

0.0297 moles of 0.015 moles  sodium bicarbonate yields 0.0297 * 1/2 = 0.015 moles

Theoretical yield of sodium carbonate = 0.015 moles * 106 g/mol = 1.59 g

Experimental yield of sodium bicarbonate = 17.50 g - 15.98 g = 1.52 g

% yield = experimental yield/Theoretical yield * 100

% yield = 1.52/1.59 * 100

% yield = 96%

The percent yield may exceed 100% if the water and CO2 are not removed from the system by heating the solid product to a constant mass.

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