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m_a_m_a [10]
2 years ago
13

A teacher places a cup of coffee onto an electronic balance at the front of the science laboratory. The teacher then adds three

large teaspoons of sugar to the coffee. The volume of the coffee does not appear to increase. What happens to the mass displayed on the front of the balance?
Chemistry
2 answers:
tamaranim1 [39]2 years ago
5 0

Answer:

The mass on the balance stays the same

Explanation:

.

motikmotik2 years ago
4 0

Answer:

it will be same as before

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Determine the molarity for each of the following solution solutions:
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Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of H_2SO_4 solution is, 0.00525 mole/L

(c)The molarity of Al(NO_3)_3 solution is, 0.0612 mole/L

(d)The molarity of CuSO_4.5H_2O solution is, 7.61 mole/L

(e)The molarity of Br_2 solution is, 0.0565 mole/L

(f)The molarity of C_2H_5NO_2 solution is, 0.0113 mole/L

Explanation :

<u>(a) 1.457 mol of KCl in 1.500 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Solute is KCl.

\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L

The molarity of KCl solution is, 0.9713 mole/L

<u>(b) 0.515 gram of H_2SO_4, in 1.00 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Solute is H_2SO_4

Molar mass of H_2SO_4 = 98 g/mole

\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L

The molarity of H_2SO_4 solution is, 0.00525 mole/L

<u>(c) 20.54 g of Al(NO_3)_3 in 1575 mL of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Solute is Al(NO_3)_3

Molar mass of Al(NO_3)_3 = 213 g/mole

\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L

The molarity of Al(NO_3)_3 solution is, 0.0612 mole/L

<u>(d) 2.76 kg of CuSO_4.5H_2O in 1.45 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Solute is CuSO_4.5H_2O

Molar mass of CuSO_4.5H_2O = 250 g/mole

\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L

The molarity of CuSO_4.5H_2O solution is, 7.61 mole/L

<u>(e) 0.005653 mol of Br_2 in 10.00 ml of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

Solute is Br_2.

\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L

The molarity of Br_2 solution is, 0.0565 mole/L

<u>(f) 0.000889 g of glycine, C_2H_5NO_2, in 1.05 mL of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Solute is C_2H_5NO_2

Molar mass of C_2H_5NO_2 = 75 g/mole

\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L

The molarity of C_2H_5NO_2 solution is, 0.0113 mole/L

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