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3 years ago

What must happen to uranium before it can be used as a fuel source?

Chemistry

1 answer:

kiruha [24]3 years ago

4 0

Explanation:

Before it can be used in a reactor for electricity generation, however, it must undergo a series of processes to produce a useable fuel. For most of the world's reactors, the next step in making the fuel is to convert the uranium oxide into a gas, uranium hexafluoride (UF6), which enables it to be enriched.

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3 years ago

Consider this chemical reaction. What chemical is reduced?<br> CH3OH + NAD --> CH2O + NADH

nikklg [1K]

Answer:

CH3OH and NADH

Explanation:

The given chemical reaction is an redox reaction in which reduction and oxidation take place.

In the process of oxidation: electrons are loss while in the process of reduction: electrons are gained.

In the given redox reaction: CH3OH + NAD --> CH2O + NADH

NAD is reduced to NADH as NADH gains one hydrogen electron while CH3OH (methanol) is oxidized to CH2O (methanal) by losing electrons.

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3 years ago

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2 years ago

how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate

Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

$3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3$

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

$0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})$

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

$63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})$

= $95.7gNaClO_3$

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate and then the grams are converted to kg.

$95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})$

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

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3 years ago