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o-na [289]
4 years ago
15

How many atoms of iodine are in 12.75g of CaI2

Chemistry
1 answer:
dexar [7]4 years ago
8 0

Answer:

0.5  × 10²³ atoms of iodine

Explanation:

Given data:

Mass of calcium iodide = 12.75 g

Number of atoms of iodine = ?

Solution:

First of all we will calculate the number of moles of calcium iodide.

Number of moles = mass/ molar mass

Number of moles = 12.75 g/ 293.9 g/mol

Number of moles = 0.04 mol

In one mole of calcium iodide there are two moles of iodine.

Thus in 0.04 moles:

0.04 mol × 2 = 0.08 moles of iodine

Now we will use the Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

0.08 moles of iodine × 6.022 × 10²³ atoms / 1 mol

0.5  × 10²³ atoms of iodine.

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Answer:

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products: H2O

Explanation:

The simple reaction that produces a water molecule from H2 and O would be the one written above, even though there are 2 hydrogen molecules, they will form an H2 molecule rather than 2 individual H molecules (almost never seen) the reactants would be your hydrogen and oxygen molecules individually before they bond to form a molecule of water (H2O) which is the product

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What is the volume of 33.25g of butane gas at 293 C and 10.934 kPa?
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Answer:

V = 240.79 L

Explanation:

Given data:

Volume of butane = ?

Temperature = 293°C

Pressure = 10.934 Kpa

Mass of butane = 33.25 g

Solution:

Number of moles of butane:

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V = nRT/P

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