All categories

4 years ago

In the chemical laboratory, what does a mortar and pestle function as?. .

Chemistry

2 answers:

likoan [24]4 years ago

5 0

B. it is used to crush substances therefore it would be decreasing the surface area

Eduardwww [97]4 years ago

3 0

In a chemical laboratory, a mortar and pestle function is to decrease surface area of a solute. The correct answer is B.

You might be interested in

What properties do atoms of the same element share? Select all that apply

Shalnov [3]

Answer: your answer is A

Explanation: the atoms has the same properties of the same element as the atomic number.

8 0

3 years ago

Which statement best describes the composition of the ancient supercontinent Pangaea?

irina [24]

Answer:

C. Pangaea contained all of Earth's continental crust.

Explanation:

Pangaea is the name given to the supercontinent that contained almost all of the landmass on Earth. The remaining area was ocean.

4 0

3 years ago

Humans exhale carbon dioxide when they breathe. if the concentration of carbon dioxide in the air that people breathe gets too h

Finger [1]

The maximum number of people the shuttle can safely carry for one mission is 4.

3 0

3 years ago

Read 2 more answers

Methane (CH4) is well-known as the key to reduce global warming. Why is that? How CO2 and methane are related anyhow to the glob

Zina [86]

Methane gas is the key element of Natural gas.
Natural gas is the Environment saver
It stops air pollution.
On the other hand CO2 is an air pollutant.
The Carbon dioxide is a key facter of Environment corrosion

5 0

3 years ago

Read 2 more answers

A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiSO4. (a) Write the balanced chemical equ

NISA [10]

Answer:

a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂

b) Ni(OH)₂

c) KOH

d) 0.927 g

e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M

Explanation:

a) The equation is:

2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂ (1)

b) The precipitate formed is Ni(OH)₂

c) The limiting reactant is:

$n_{KOH} = V*M = 100.0 \cdot 10^{-3} L*0.200 mol/L = 0.020 moles$

$n_{NiSO_{4}} = V*M = 200.0 \cdot 10^{-3} L*0.150 mol/L = 0.030 moles$

From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:

$n = \frac{2}{1}*0.030 moles = 0.060 moles$

Hence, the limiting reactant is KOH.

d) The mass of the precipitate formed is:

$n_{Ni(OH)_{2}} = \frac{1}{2}*n_{KOH} = \frac{1}{2}*0.020 moles = 0.010 moles$

$m = n*M = 0.010 moles*92.72 g/mol = 0.927 g$

e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:

$C_{K^{+}} = \frac{2*\frac{1}{2}*n_{KOH}}{V} = \frac{0.020 moles}{0.300 L} = 0.067 M$

$C_{SO_{4}^{2-}} = \frac{\frac{1}{2}*n_{KOH + (0.03 - 0.01)}}{V} = \frac{0.030 moles}{0.300 L} = 0.1 M$

$C_{Ni^{2+}} = \frac{0.020 moles}{0.300 L} = 0.067 M$

I hope it helps you!

5 0

4 years ago