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Alborosie
3 years ago
9

HELP HELP ME PLSS

Chemistry
1 answer:
Akimi4 [234]3 years ago
6 0

Answer:

the average answer in maybe 60

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How many atoms of iodine are in 12.75g of CaI2
dexar [7]

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.

8 0
3 years ago
ASAP PLEASE
marishachu [46]

Answer:

A

Explanation:

the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work.

8 0
3 years ago
Read 2 more answers
Scientific notation is a way of writing very large or very small
damaskus [11]

Answer:

C

Explanation:

4 0
2 years ago
A buffer solution contains 0.20 mol of propionic acid (CH3CH2COOH) and 0.25 mol of sodium propionate (CH3CH2COONa) in 1.50 dm3.
const2013 [10]

Answer:

I don't know how to do it the subject

6 0
3 years ago
What mass of KNO3 would be needed to produce 18.4 liters of oxygen gas, measured at 1.50 x 10^3 kPa and 15 degrees Celsius?
Brut [27]

Answer:-

2328.454 grams

Explanation:-

Volume V = 18.4 litres

Temperature T = 15 C + 273 = 288 K

Pressure P = 1.5 x 10^ 3 KPa

We know universal Gas constant R = 8.314 L KPa K-1 mol-1

Using the relation PV = nRT

Number of moles of oxygen gas n = PV / RT

Plugging in the values

n = (1.5 x 10^3 KPa ) x ( 18.4 litres ) / ( 8.314 L KPa K-1 mol-1 x 288 K)

n = 11.527 mol

Now the balanced chemical equation for this reaction is

2KNO3 --> 2KNO2 + O2

From the equation we can see that

1 mol of O2 is produced from 2 mol of KNO3.

∴ 11.527 mol of O2 is produced from 2 x 11.527 mol of KNO3.

= 23.054 mol of KNO3

Molar mass of KNO3 = 39 x 1 + 14 x 1 + 16 x 3 = 101 grams / mol

Mass of KNO3 = 23.054 mol x 101 gram / mol

= 2328.454 grams

7 0
3 years ago
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