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3 years ago

What happens with the atomic numbers in nuclear equations?

Chemistry

1 answer:

topjm [15]3 years ago

8 0

Answer:

Conserved

Explanation:

In balancing nuclear reaction equations, there is conservation of mass numbers and atomic numbers, i.e the sum of the mass number on the product side must be equal to the sum of mass numbers on the reactant side.

Also, the sum of the atomic numbers on the product side must be equal to the sum of the atomic numbers on the reactant side.

Note: A nuclear reaction is a process in which a nucleus reacts with an elementary particle.

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What is a visual or mathematical representation of an object system or concept

sesenic [268]

The visual or mathematical representation of an object system or concept is known as a model. Hope this helps and mark as brainliest!

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4 years ago

Which of the following hydrocarbons will most likely be found around cows? ethane methane octane decane

Wewaii [24]

Methane
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3 years ago

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Marat540 [252]

Answer:

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3 years ago

For the following reaction, 7.43 grams of methane (CH4) are allowed to react with 27.6 grams of carbon tetrachloride. methane (C

NikAS [45]

Answer:

30.4 grams of CH2Cl2 will be formed

Explanation:

Step 1: data given

Mass of methane = 7.43 grams

Molar mass of methane (CH4) = 16.04 g/mol

Mass of CCl4 = 27.6 grams

Molar mass of CCl4 = 153.82 g/mol

Step 2: The balanced equation

CH4 + CCl4 → 2CH2Cl2

Step 3: Calculate moles

Moles = mass / molar mass

Moles methane = mass methane / molar mass methane

Moles methane = 7.43 grams / 16.04 g/mol

Moles methane = 0.463 moles

Moles CCl4 = 27.6 grams / 153.82 g/mol

Moles CCl4 = 0.179 moles

Step 4: Calculate limiting reactant

For 1 mol CH4 we need 1 mol CCl4 to produce 2 moles CH2Cl2

CCl4 is the limiting reactant. It will completely be consumed. (0.179 moles). Ch4 is in excess. There will react 0.179 moles. There will remain 0.463 - 0.179 = 0.284 moles

Step 5: Calculate moles CH2Cl2

For 1 mol CH4 we need 1 mol CCl4 to produce 2 moles CH2Cl2

For 0.179 moles CCl4 we'll have 2*0.179 = 0.358 moles CH2Cl2

Step 6: Calculate mass CH2Cl2

Mass CH2Cl2 = 0.358 moles* 84.93 g/mol

Mass CH2Cl2 = 30.4 grams

30.4 grams of CH2Cl2 will be formed

3 0

4 years ago

What mass of HCL, in grams, is required to react with 0.610 g of al(oh)3 ?

kompoz [17]

Answer: 0.8541 grams of HCl will be required.

Explanation: Moles can be calculated by using the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of $Al(OH)_3$ = 0.610 g

Molar mass of $Al(OH)_3$ = 78 g/mol

$\text{Number of moles}=\frac{0.610g}{78g/mol}$

Number of moles of $Al(OH)_3$ = 0.0078 moles

The reaction between $Al(OH)_3$ and HCl is a type of neutralization reaction because here acid and base are reacting to form an salt and also releases water.

Chemical equation for the above reaction follows:

$Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O$