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Veseljchak [2.6K]
3 years ago
13

Question 2

Chemistry
1 answer:
aev [14]3 years ago
7 0

touching a hot stove and burning your hand

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Mike mixes two chemicals in a container. The container quickly becomes
Vedmedyk [2.9K]

Answer:

it is an exothermic reaction

7 0
3 years ago
How should students prepare to use chemicals in the lab? Select one or more: Sort the lab chemicals in alphabetical order for qu
maksim [4K]

Answer:

Sort the lab chemicals in alphabetical order for quick access.

Become familiar with the chemicals to be used, including exposure or spill hazards.

Locate the spill kits and understand how they are used.

Explanation:

There are many chemicals in a laboratory hence they should be sorted out and arranged in alphabetical order so that theory can easily be identified and located whenever they are required.

The properties of each chemical should be known especially hazards connected to exposure or spill of the chemicals.

The students should also familiarize themselves with the contents of spill kits and how they are used.

7 0
3 years ago
3 points
vfiekz [6]

Answer:

density=6.74g/ml

:320g÷47.5ml

d=6.74g/ml

thank you

<em><u>I </u></em><em><u>hope</u></em><em><u> </u></em><em><u>this </u></em><em><u>is </u></em><em><u>helpful</u></em>

8 0
3 years ago
Calculate the molarity of a soln of 3 mols of NaOH in 2 L
Grace [21]
Molarity = 1.5

molarity = moles/liters
m= 3/2
m= 1.5
4 0
3 years ago
Calculate the pH of the solution that results when 20 mL of 0.2M HCOOH is mixed with 25mL of 0.2M NaOH solution.(post-equivalenc
Paul [167]

The pH of the solution : 12

<h3>Further explanation</h3>

Reaction

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

mol HCOOH =

\tt 20~ml\times 0.2~M=4~mlmol

mol NaOH =

\tt 25~ml\times 0.2~M=5~mlmol

Mol NaOH>mol HCOOH ⇒ at the end of the reaction there will be a strong base remains from mol NaOH, so that the pH is determined from [OH⁻]

ICE method :

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

4                          5

4                          4                     4                   4

0                          1                      1                    1

Concentration of [OH⁻] from NaOH :

\tt \dfrac{1~mlmol}{20+25~ml}=0.02

pOH=-log[OH⁻]

pOH=-log 10⁻²=2

pH+pOH=14

pH=14-2=12

3 0
3 years ago
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