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3 years ago

How many grams in 1.000 moles of oxygen gas?

Chemistry

1 answer:

stich3 [128]3 years ago

6 0

1 mole is equal to 1 moles Oxygen, or 15.9994 grams.

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2 years ago

Which product is not made from a renewable resource? select one:

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3 years ago

Read 2 more answers

How many moles of magnesium, Mg, are there in 4.75 grams of magnesium?

Nina [5.8K]

Molar mass Mg = 24.3 g/mol

1 mole mg ------------ 24.3 g
?? moles mg --------- 4.75 g

4.75 x 1 / 24.3 => 0.195 moles of Mg

hope this helps!

3 0

3 years ago

For the chemical reaction CaI2+2AgNO3-> 2AgI+Ca(NO3)2 how many moles of silver iodide will be produced from 205g of calcium i

Colt1911 [192]

Answer:

Moles of silver iodide produced = 1.4 mol

Explanation:

Given data:

Mass of calcium iodide = 205 g

Moles of silver iodide produced = ?

Solution:

Chemical equation:

CaI₂ + 2AgNO₃ → 2AgI + Ca(NO₃)₂

Number of moles calcium iodide:

Number of moles = mass/ molar mass

Number of moles = 205 g/ 293.887 g/mol

Number of moles = 0.7 mol

Now we will compare the moles of calcium iodide with silver iodide.

CaI₂ : AgI

1 : 2

0.7 : 2×0.7 = 1.4

Thus 1.4 moles of silver iodide will be formed from 205 g of calcium iodide.

6 0

3 years ago

An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?

umka2103 [35]

<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

$K_b$ = molal boiling point elevation constant = 0.51°C/m

$m_{solute}$ = Given mass of solute (CsCl) = 8.00 g

$M_{solute}$ = Molar mass of solute (CsCl) = 168.4 g/mol

$W_{solvent}$ = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC$

Hence, the boiling point of solution is 100.53

6 0

3 years ago