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dmitriy555 [2]
3 years ago
10

In your own words can yall give me a example of air pollution/

Chemistry
2 answers:
raketka [301]3 years ago
3 0
Sulfur dioxide is an example of air pollution.
It is the main component of acid rain. It is released into the air as a result of the burning of fossil fuels.

dimulka [17.4K]3 years ago
3 0

One example of air pollution is smoking. Smoking isn't good for you or for our air.

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How many grams in a kilogram,how many quarts are in a pint and if how many ml of h20 = 1g
vodomira [7]

Answer:

1 g  +  10 g

Explanation:

5 0
3 years ago
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What caused the bubbles to form when you added the catalase to the hydrogen peroxide and water mixture at 40 °c?
Dahasolnce [82]
<span>What caused the bubbles to form when you added the catalyses to the hydrogen peroxide and water mixture at 40 °C? A. Catalyses activity heated the solution to its boiling point. B. Hydrogen gas formed during the formation of hydrogen peroxide. C. Oxygen gas formed during the decomposition of hydrogen peroxide.

This would be the water, mixture.</span>
7 0
3 years ago
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The reducing agent in this catalytic hydrogenation reaction was molecular hydrogen (H2), which was produced in situ (in the reac
creativ13 [48]

Answer:

a)  After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) 4NaBH_{4} + 2HCl + 7H_{2}O ⇒ 16H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7} c) No H_{2} was not the limiting reactant.

Explanation:

Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.

a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.

b)  4NaBH_{4} + 2HCl + 7H_{2}O ⇒ 16H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}

c) H_{2} was not the limiting reactant based on the mol to mol ratio of H_{2} and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of H_{2} would also be produced.

4 0
3 years ago
A 25.0-mL sample containing Cu2+ gave an instrument signal of 25.2 units (corrected for a blank). When exactly 0.500 mL of 0.027
irga5000 [103]

Answer:

The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.

Explanation:

The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:

C = \frac{n}{V} \\0.0275 = \frac{n}{0.0005} \\

n = 1.375x10⁻⁵ mol

The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):

1.375x10⁻⁵ mol _________ 19.9 units

        x              _________  25.2 units

x = 1.741x10⁻⁵mol

Finally, we can calculate the Cu²⁺ concentration :

C = 1.741x10⁻⁵mol / 0.025 L

C = 6.964x10⁻⁴ M

7 0
3 years ago
A chemist has one solution that is 80% acid and another solution that is 30% acid. how much of the second (30%) solution is need
Aleksandr [31]

messed up oops !!!!!!!!!


8 0
3 years ago
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