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3 years ago

Can someone explain how I would solve this?

Chemistry

1 answer:

ValentinkaMS [17]3 years ago

4 0

Answer:

sorry I don't know about this yet

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Compound that contains a terminal carbonyl?

malfutka [58]

Many compunds have a terminal carbonyl

Aldehyde, Ketone, Carboxylic acid, Amide, Imide, Acid anhydride are the first that come to my mind.

7 0

3 years ago

For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox

ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: $m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

The reaction

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

Moles of nitrogen monoxide

Molecular weight: $M_{NO}=30 g/mol$

$n_{NO}=\frac{m_{NO}}{M_{NO}}$

$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

Molecular weight: $M_{H2}=2 g/mol$

$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) Maximun ammount of nitrogen gas => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

$m_{N2}=4.274 g$

8 0

3 years ago

In this reaction, _____.

ella [17]

Answer:

AB + CD ----> AC + BD

Explanation:

If you think this reaction:

AB + CD ----> AC + BD

(Reactants) (Products)

All the statements are true.

7 0

3 years ago

What are some uses of radioactive dating? HELPP PLEASEEEE!!!

Irina-Kira [14]

Answer:

Radioactive dating is a method of dating rocks and minerals using radioactive isotopes. This method is useful for igneous and metamorphic rocks, which cannot be dated by the stratigraphic correlation method used for sedimentary rocks. Over 300 naturally-occurring isotopes are known.

4 0

2 years ago

The rate law for the decomposition of phosphine (PH3) is below. It takes 120. s for 1.00 M PH3 to decrease to 0.250 M. How much

Lubov Fominskaja [6]

Answer: 588.8 seconds

Explanation:

1.00 - 0.25 = 0.75M

4.00 - 0.320 = 3.68M

Therefore,

0.75M in 120 seconds

3.68M in x seconds

Cross multiply:

0.75*x = 120*3.68

x = 120*3.68/0.75

x = 588.8 seconds

8 0

3 years ago