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Elina [12.6K]
3 years ago
8

Fill in the missing symbol and name the reaction:

Chemistry
1 answer:
damaskus [11]3 years ago
5 0

Answer:

Give ur life to Jesus Christ before its to late!!!!!!!!!!

Explanation:

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True or false???????????
Leni [432]

Answer:

false

Explanation:

longer the wave length less frequency

7 0
3 years ago
Enter an equation showing how this buffer neutralizes added aqueous acid (HI). Express your answer as a chemical equation. Ident
andrey2020 [161]

Answer:

H^++NH_3\rightleftharpoons NH_4^+

Explanation:

Hello there!

In this case, since the buffer is not given, we assume it is based off ammonia, it means the ammonia-ammonium buffer, whereas the ammonia is the weak base and the ammonium ion stands for the conjugate acid. In such a way, when adding HI to the solution, the base of the buffer, NH3, reacts with the former to promote the following chemical reaction:

H^++NH_3\rightleftharpoons NH_4^+

Because the HI is totally ionized in solution so the iodide ion becomes an spectator one.

Best regards!

5 0
3 years ago
What happens to air pressure as altitude increases
gulaghasi [49]

Answer:

the air pressure decreases

explanation

as altitude increases air pressure decreases the surface of the Earth is bottom of the ocean of the are the layer on top increase pressure so that sea levels a body is under 14.7 pounds of pressure per square inch moving up in altitude decreases weight of the air that causes the pressure

4 0
3 years ago
Carbon dioxide is produced by the human body through
blsea [12.9K]

Answer:

Respiration

I hope this helps!

8 0
3 years ago
Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppo
Leto [7]

Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

CH_4+2O_2\rightarrow CO_2+2H_2O

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

left\ over=0g

Regards.

7 0
3 years ago
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