How many uranium atoms are there in 6.1 g

32 gm of O2 to mole of O2

lana [24]

Molar mass of O2:-

$\\ \rm\longmapsto 2(16u)=32g/mol$

Now

$\boxed{\sf No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}}$

$\\ \rm\longmapsto No\:of\;moles=\dfrac{32}{32}$

$\\ \rm\longmapsto No\:of\;moles=1mol$

5 0

3 years ago

Classify which compounds will dissolve in water and which ones will not dissolve in water.

vredina [299]

Answer:

Dissolves in water: Acetone, 1-propanol, Methanol

Do not dissolve in water: Hexane, Decane

Explanation:

From the principle that like dissolves like,polar substances will dissolve in polar solutions, while non-polar substances will dissolve in non-polar substances.

Water is a polar solution, therefore, polar substances will dissolve in it.From the given options:

Acetone is a polar molecule, therefore, it will dissolve in water.

1-propanol is a polar molecule, therefore, it will dissolve in water.

Hexane is a non-polar molecule, therefore, it will not dissolve in water.

Methanol is a polar molecule, therefore, it will dissolve in water.

Decane is a non-polar molecule, therefore, it will not dissolve in water

6 0

3 years ago

Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. How many grams of di

pychu [463]

Answer:

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant. Explanation:

$Ca(NO_3)_2(aq)+2NH_4F(aq)\rightarrow CaF_2(aq)+2N_2O(g)+4H_2O(g)$

Moles of calcium nitrate = $\frac{31.3 g}{164 g/mol}=0.1908 mol$

Moles of ammonium fluoride = $\frac{38.7 g}{37 g/mol}=1.046 mol$

According to reaction , 2 moles of ammonium fluoride reacts with 1 mole of calcium nitrate.

Then 1.046 moles of ammonium fluoride will react with :

$\frac{1}{2}\times 1.046 mol=0.523 mol$ calcium nitarte .

This means that ammonium fluoride is in excess amount and calcium nitrate is in limiting amount.

Hence, calcium nitrate is a limiting reactant.

So, amount of dinitrogen monoxide will depend upon moles of calcium nitrate.

So, according to reaction , 1 mole of calcium nitarte gives 2 moles of dinitrogen monoxide gas .

Then 0.1908 moles of calcium nitrate will give:

$\frac{2}{1}\times 0.1908 mol=0.3816 mol$ of dinitrogen monoxide gas.

Mass of 0.03816 moles of dinitrogen monoxide gas:

0.03816 mol × 44 g/mol = 16.79 g

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant.

8 0

3 years ago

What creates an ionic bond?

Anna71 [15]

Answer:c

Explanation: it’s c

6 0

3 years ago

Write the orbital notation and electron configuration for the following elements. (Worksheet #2)

Mademuasel [1]

Answer:

the answer is B

Explanation:

4 0

3 years ago

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