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horsena [70]
3 years ago
9

If concentrated sulfuric acid is spilled during the experiment, what is the proper procedure for cleaning the spill?

Chemistry
1 answer:
Komok [63]3 years ago
5 0

Sulfuric acid (H2SO4) is a strong acid which should be neutralized with a base. Ideally, sodium bicarbonate (NaHCO3) is used to treat acid spills. The reaction is as follows:

H2SO4 + 2NaHCO3 → Na2SO4 + 2H2O + 2CO2

Bubbles will be formed as CO2 is released. Once all H2SO4 has been neutralized the bubbles will stop. The pH of the spill after neutralization can be checked with a pH and should be between 6-9. If optimal pH is observed the waste can be disposed into a proper container and the work station can be cleaned with a paper towel.

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2

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to separate objects or ideas into group based on ways they are alike

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The mass of a proton is equal to the mass of?
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7 0
2 years ago
An increase in temperature results in A) a decrease in the required activation energy while the reaction rate remains constant.
N76 [4]

Answer:

C) an increase in rate of reaction because reactant molecules collide with greater energy

Explanation:

Temperature is one of the factors that affect the rate of a reaction. The rate of a reaction increases with an increase in temperature and vice versa. When the temperature of a reaction increases, the kinetic energy of the reactant molecules increases causing them to react at a faster rate.

The reactant molecules respond to an increase in temperature by colliding at a faster rate due to an increased kinetic energy between the reactant molecules.

7 0
3 years ago
What mass of Cu(s) is electroplated by running 24.5A of current through a Cu2+(aq)solution for 4.00 h?Express your answer to thr
Masja [62]

Answer: 116 g of copper

Explanation:

Q=I\times t

where Q= quantity of electricity in coloumbs

I = current in amperes = 24.5A

t= time in seconds =  4.00 hr = 4.00\times 3600s=14400s  (1hr=3600s)

Q=24.5A\times 14400s=352800C

Cu^{2+}+2e^-\rightarrow Cu

2\times 96500C=193000C  of electricity deposits 63.5 g of copper.

352800 C of electricity deposits = \frac{63.5}{193000}\times 352800=116g of copper.

Thus 116 g of Cu(s) is electroplated by running 24.5A of current

Thus  remaining in solution = (0.1-0.003)=0.097moles

8 0
3 years ago
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