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lisov135 [29]
3 years ago
10

sketch and label a helium-4 atom and helium-5 atom to include the protons, neutrons, and electrons. how are they alike and how a

re they different?
Chemistry
2 answers:
Doss [256]3 years ago
8 0

Answer:

Protons-2 for both

neutrons-2 for helium-4 and neutrons-3 for helium-5

electrons-1 for both

Explanation:

draw a ring and put protons and neutrons in the middle and the electrons on the ring

Naily [24]3 years ago
4 0

Answer:

U cant sketch on here mate

Explanation:

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1.65g of zinc is used to make 8g of zinc iodide. How much iodine is required for this reaction?
creativ13 [48]

Answer:

6.45 g of iodine, I₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

Zn + I₂ —> ZnI₂

Next, we shall determine the mass of Zn and I₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of Zn = 65 g/mol

Mass of Zn from the balanced equation = 1 × 65 = 65 g

Molar mass of I₂ = 127 × 2 = 254 g/mol

Mass of I₂ from the balanced equation = 1 × 254 = 254 g

SUMMARY:

From the balanced equation above,

65 g of Zn reacted with 254g of I₂.

Finally, we shall determine the mass of f I₂ needed to react with 1.65 g of Zn. This can be obtained as follow:

From the balanced equation above,

65 g of Zn reacted with 254g of I₂.

Therefore, 1.65 g of Zn will react with = (1.65 × 254)/65 = 6.45 g of I₂.

Thus, 6.45 g of iodine, I₂ is needed for the reaction.

4 0
3 years ago
What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?
AlladinOne [14]

The question is incomplete, here is the complete question:

The given chemical reaction is:

2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3

What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?

<u>Answer:</u> The theoretical yield of copper is 44.48 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of aluminium = 12.6 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium}=\frac{12.6g}{27g/mol}=0.467mol

The given chemical equation follows:

2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of copper

So, 0.467 moles of aluminium will produce = \frac{3}{2}\times 0.467=0.7005mol of copper

Now, calculating the mass of copper  from equation 1, we get:

Molar mass of copper = 63.5 g/mol

Moles of copper = 0.7005 moles

Putting values in equation 1, we get:

0.7005mol=\frac{\text{Mass of copper}}{63.5g/mol}\\\\\text{Mass of copper}=(0.7005mol\times 63.5g/mol)=44.48g

Hence, the theoretical yield of copper is 44.48 grams

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