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3 years ago

Which part of an atom is gained, lost, or shared during a chemical reaction?

Chemistry

2 answers:

Serga [27]3 years ago

8 0

Answer: electrons

Explanation: A chemical bond is when there is a transfer (movement) of electrons from one atom to another.

Dovator [93]3 years ago

7 0

Answer:

Electrons

Explanation:

during chemical change when atoms trying to balanced out electrons is like the wall that protect them and it's the one that gain, lose, and shared :)

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Blank! An organism whose presence or absence reveals the environmental condition?

Ira Lisetskai [31]

An indicator species is an organism whose presence, absence, and abundance reveals a specific environmental condition, whether it be a positive or negative condition.

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4 years ago

Can you identify the end arrangement found in fatty acids as wells amino acids

3241004551 [841]

In amino acids the end terminal is NH2 - the amine group and in fatty acids it is COOH the carboxyl group.

Explanation:

Fatty acids are merely long chain radical acids that contains one carboxyl and a protracted hydrophobic organic compound chain. Whereas, amino acids are the compounds that contain both a carboxyl at and an chemical group hooked up to identical atom.

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3 years ago

Figure

Artyom0805 [142]

Answer:

Explanation:

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3 years ago

Decane (C10H22) is used in diesel. The combustion for decane follows the equation: 2 C10H22 + 31 O2 à 20 CO2 + 22 H2O. Calculate

creativ13 [48]

The mass of water produced is 792 grams by the combustion of 568 grams of decane.

Given:

Combustion of 568 grams of decane with 2979 grams of oxygen.

$2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O$

To find:

The mass of water produced by combustion of 568 grams of decane.

Solution:

Mass of decane = 568 g

Moles of decane :

= $\frac{568 g}{142 g/mol}=4 mol$

Mass of oxygen gas = 2976 g

Moles of oxygen gas:

= $\frac{2976 g}{32 g/mol}=93 mol$

$2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O$

According to reaction, 2 moles of decane reacts with 31 moles of oxygen, then 4 moles of decane will react with:

$=\frac{31}{2}\times 4mol=62\text{ mol of}O_2$

But according to the question, we have 93.0 moles of oxygen gas which is more than 62 moles of oxygen gas.

So, this means that oxygen gas is present in an excessive amount. Which simply means:

Oxygen gas is an excessive reagent.
Decane is a limiting reagent.
Decane being limiting reagent will be responsible for the amount of water produced after the reaction.

According to reaction, 22 moles of water is produced from 2 moles of decane, then 4 moles of decane will produce:

$=\frac{22}{2}\times 4mol=44\text{mol of }H_2O$