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erastova [34]
4 years ago
13

Using the unbalanced reaction below:

Chemistry
1 answer:
Vladimir [108]4 years ago
5 0

1.16 g of UF₄ can be formed in this reaction.

<u>Explanation:</u>

We have to write the balanced equation of the reaction between Uranium oxide and hydrogen fluoride to form Uranium tetra fluoride and water as,

UO₂(s) +4 HF(aq) → UF₄ + 4 H₂O(l)

We can find the moles of UF₄ and then multiplying molar mass of UF₄ with the moles of UF₄, we will get the mass of UF₄ in grams.

Moles of UF₄ =   $  \frac{ 1 mol of UO_{2}\times 1 mol of UO_{2\times 1 mol of UF_{4} }  }{270.03 g \times 1 mol of UO_{2} }

                       = 0.00370 mol UF₄

Grams of UF₄ =0.00370 mol × 314.02 g/mol = 1.16 g of UF₄

So mass of UF₄ formed in the reaction is 1.16 g.

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The answer is 2 electrons.


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Calcium is a metal. When metals react with non-metals, electrons are transferred from the metal atoms to the non-metal atoms forming ions. The resulting compound is known  as an ionic compound.

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4 years ago
Read 2 more answers
How many grams of S are in 475 g of SO2?
enot [183]

There are 237. 5 g of Sulfur,S in 475 g of SO2?

<h3 />

<h3>Calculation of grams of Sulfur</h3>

From the question, we can say that

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  • The molar mass of oxygen = 16 g/mol

Therefore,

The molar mass for SO2 = 32 + (16 × 2) g/mol  = 64 g/mol

Now,

If 1 mole of SO2 contains 1 mole of S

Then 64 g of SO2, will contain 32g of S;

Such that

475 g of SO2 will give { \frac{ 32g of S) ( 475 g of SO2)}{64 of SO2} }  = 237. 5 g of Sulfur.

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Let P and V represent the pressure and volume of the Xe(g) in the container in diagram 3. If a piston is used to reduce the volu
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Answer:

2p

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To solve this question, we can use Boyle's Law, which states that:

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p\propto \frac{1}{V}

where

p is the pressure of the gas

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The equation can be rewritten as

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where in this problem we have:

p_1 = p is the initial pressure of the Xe(g) gas

V_1=V is the initial volume of the Xe(g) gas

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Solving for p2, we find the final pressure of the gas:

p_2=\frac{p_1 V_1}{V_2}=\frac{pV}{V/2}=2p

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