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3 years ago

What is acid rain? Please answer it as fast as possible....I really need it......

Chemistry

2 answers:

Volgvan3 years ago

8 0

Acid rain is rainfall made from atmospheric pollution.

Ne4ueva [31]3 years ago

3 0

Accid rain is essentially rain but with not water but pollution from the atmosphere. Hope it helped. :)

You might be interested in

What is endothermic process?

LiRa [457]

Answer:

An endothermic process is any process with an increase in the enthalpy H (or internal energy U) of the system. In such a process, a closed system usually absorbs thermal energy from its surroundings, which is heat transfer into the system. It may be a chemical process, such as dissolving ammonium nitrate in water, or a physical process, such as the melting of ice cubes.

7 0

3 years ago

Neon-20 and neon-22 are called

natita [175]

Isotopes:) hope this helps

7 0

3 years ago

To what pressure must a gas be compressed in order to get into a 3.00 cubic foot

Troyanec [42]

Answer:

133.3atm

Explanation:

Given parameters:

Initial volume = 400ft³

Final volume = 3ft³

Initial pressure = 1atm

Unknown:

Final pressure = ?

Solution:

To solve this problem, we apply Boyle's law which states that "the volume of a fixed mass of a gas varies inversely as the pressure changes if the temperature is constant".

Mathematically;

P₁V₁ = P₂V₂

P and V are pressure and volume

1 and 2 are initial and final states;

Insert parameters and solve;

1 x 400 = P₂ x 3

P₂ = 133.3atm

6 0

4 years ago

The center of hemoglobin is made of which of the following?

sveticcg [70]

The center of hemoglobin is made of which of the following?

amino acids

5 0

3 years ago

Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

4 years ago