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tamaranim1 [39]
3 years ago
5

What would the world be like without elements? please​

Chemistry
2 answers:
kifflom [539]3 years ago
5 0

Answer:

Nothing

Explanation:

Everything we have in our world is made up of elements. Without them, we would have nothing.

Lisa [10]3 years ago
3 0

Answer:

It would be nothing. Quite literally nothing. No Oxygen, no dirt, no anything.

Explanation:

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DedPeter [7]

Answer:

Lithium has 3 protons, 4 neutrons, 3 electrons, and its mass is 6.941 u

4 0
3 years ago
What’s the answer pls hrlp
arsen [322]

Answer:

The answer is A

Explanation:

5 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:

<u>The reaction </u>

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) <u>Maximun ammount of nitrogen gas</u> => 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) <u>Limiting reagent</u>: NO

3) <u>Ammount of excess reagent</u>:

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
How are radio waves transmitted
kotegsom [21]
Attractive antennas that pull the waves tword it.
7 0
3 years ago
The solubility of copper chloride at 20 °C is 73 g/100 g of water. Kiera adds 100 g of copper chloride to 100 g of water and sti
OLEGan [10]

Answer:

m_{undissolved}=27g

Explanation:

Hello there!

In this case, according to the given information of the solubility of copper chloride, as the maximum amount of this salt one can dissolve without having a precipitate, we infer that since just 73 grams are actually dissolved, the following amount will remain solid as a precipitate:

m_{undissolved}=100g-73g\\\\m_{undissolved}=27g

Best regards!

3 0
2 years ago
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