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o-na [289]
3 years ago
15

How many atoms of iodine are in 12.75g of CaI2

Chemistry
1 answer:
dexar [7]3 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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3 years ago
Calculate the number of ammonia molecules in 3.9 g.
disa [49]
•3.9g of ammonia
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1st you have to covert grams to moles by dividing the mass of ammonia with the molar mass:

(3.9 g)/ (17.03g/mol) = 0.22900763mols

Then convert the moles to molecules by multiplying it with Avogadro’s number:

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6 0
3 years ago
A sample of nitrogen is initially at a pressure of 1.7 kPa, a temperature of -10 C and a volume of 7.5 m3. Then the volume is de
zhannawk [14.2K]

Answer:

\boxed{\text{2.6 kPa}}

Explanation:

To solve this problem, we can use the Combined Gas Laws:

\dfrac{p_{1}V_{1} }{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}

Data:

p₁ = 1.7 kPa; V₁ = 7.5 m³;  T₁ =   -10 °C

p₂ = ?;          V₂ = 3.8 m³; T₂ = 200  K

Calculations:

(a) Convert temperature to kelvins

T₁ = (-10   + 273.15) K = 263.15 K

(b) Calculate the pressure

\begin{array}{rcl}\dfrac{1.7 \times 7.5 }{263.15} & = & \dfrac{p_{2} \times 3.8}{200}\\\\0.0485 & = & 0.0190p_{2}\\p_{2} & = & \textbf{2.6 kPa}\\\end{array}\\\text{The new pressure of the gas is \boxed{\textbf{2.6 kPa}}}

7 0
3 years ago
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