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Kisachek [45]
3 years ago
15

What happens when you shoot a neutron at a uranium-238 atom?

Chemistry
1 answer:
serious [3.7K]3 years ago
7 0

Answer: Scientist shoot a whole bunch of neutrons at uranium-235 atoms. When one neutron hits the nucleus the uranium becomes U-236. When it becomes 236, the uranium atoms wants to spilt apart. After it splits it gives off three neutrons and a lot of energy.

Explanation:

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The sun solar energy produce all the energy
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What are the moles of silver metal produced from 0.0999 mol of copper?
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1234567i9812345678912121212121

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Why a finely powdered sample should be used in a melting point measurement?
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because it a essential for good heat transfer

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Consider the chemical equation. 2H2 + O2 mc022-1.jpg 2H2O What is the percent yield of H2O if 87.0 g of H2O is produced by combi
ch4aika [34]

Answer:

88.50 %

The balance chemical equation is as follow,

2 H₂ + O₂ → 2 H₂O

Step 1: Find the limiting reactant;

According to eq.

4.032 g (2 mole) H₂ reacts with = 32 g (1 mole) of O₂

So,

11 g of H₂ will react with = X g of O₂

Solving for X,

X = (11 g × 32 g) ÷ 4.032 g

X = 87.301 g of O₂

Therefore, H₂ is the limiting reactant as O₂ is present in excess.

Step 2: Calculating %age Yield;

According to eq.

4.032 g (2 mole) H₂ produces = 36.032 g (1 mole) of H₂O

So,

11 g of H₂ will react with = X g of H₂O

Solving for X,

X = (11 g × 36.032 g) ÷ 4.032 g

X = 98.301 g of H₂O

So,

Actual Yield = 87 g

Theoretical Yield = 98.301 g

Using formula = Actual Yield / Theoretical Yield × 100

= 87 g / 98.301 × 100

= 88.50 %

4 0
4 years ago
Read 2 more answers
What is the empirical formula for a compound that contains 79.86 % iodine and 20.14 % oxygen by mass?
serg [7]

Answer:

IO₂

Explanation:

We have been given the mass percentages of the elements that makes up the compound:

Mass percentage given are:

Iodine = 79.86%

Oxygen = 20.14%

To calculate the empirical formula which is the simplest formula of the compound, we follow these steps:

> Express the mass percentages as the mass of the elements of the compound.

> Find the number of moles by dividing through by the atomic masses

> Divide by the smallest and either approximate to nearest whole number or multiply through by a factor.

> The ratio is the empirical formula of the compound.

Solution:

I O

% of elements 79.86 20.14

Mass (in g) 79.86 20.14

Moles(divide by

Atomic mass) 79.86/127 20.14/16

Moles 0.634 1.259

Dividing by

Smallest 0.634/0.634 1.259/0.634

1 2

The empirical formula is IO₂

7 0
3 years ago
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