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3 years ago

What is the ratio of Al 3+ ions to S 2- ions in a neutral compound?

2 answers:

4 0

In a neutral ionic compound, you can determine its sub-scripts by simply flipping the ionic charges and dropping the signs: so AlS would be Al2S3

mr_godi [17]3 years ago

4 0

Answer:

The ratio of aluminium to sulfur is 2:3

Explanation:

The ratio of the elements of the neutral compound can be computed from the ionic charges that makes up the compound. The element that makes up the compound is aluminium and sulfur. From the ions aluminium is the cation with a charge of 3+ . The aluminium ion is the cations because it loses 3 electrons from the bonding between it and sulfur. The sulfur is the anion as it gains electrons from the cations.

Aluminium has a charge of 3+ and sulfur has a charge of 2- . The atom of element that makes up the neutral compound can be computed when you cross multiply the charges.

Al3+ and S2-

cross multiply the charges

Al2S3.

The ratio of aluminium to sulfur is 2:3

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Which of the following vary for isotopes of atoms of the same element?

cupoosta [38]

Answer:

The answer is B. Atomic Mass

7 0

2 years ago

Read 2 more answers

How many moles are there in 7.5 L Of H2

True [87]

So multiply number of moles x number of atoms/mole = 1.8066 x 10^24 atoms of H2. One mole of any gas at STP has a volume of 22.4 L. So first determine the number of moles of gas you have.
for example do 7
$7.5 \div 22.4$
that 's what I think

3 0

2 years ago

An unknown compound is found to have a molar mass of 392.16 g/mol. If the empirical formula is C2H5PF2, what is the molecular fo

notsponge [240]

Answer:

C8H20P4F8

Explanation:

Molecular formula is based off a ratio of the molecular formula's molar mass divided by the empirical formula's molar mass.

The molar mass of the empirical formula C2H5PF2 is 98.02g. We find this by adding the molar masses of all elements in the formula, multiplied by their subscripts.

2(12.01) + 5(1.01) + 30.97 + 2(18.99) = 98.02

We then divide the molecular molar mass by the empirical molar mass.

392.16/98.02 = 4

This tells us that the molecular formula has 4 times the mass of the empirical formula. Because mass comes from the elements in the formula, we multiply all the subscripts by 4 to get the molecular formula.

2x4 = 8

5x4 = 20

1x4 = 4

2x4 = 8

So the molecular formula is C8H20P4F8

8 0

2 years ago

Which set of coefficients will balance this chemical equation?

Katarina [22]

Answer:

B 1,3

Explanation:

Count the atoms on each side of the arrow, they have to be equal

After arrow

2x C

6x O

4x H

Before the arrow

2x C - correct, as it is the only compound with C then there must only be 1 mole of it.

This is also the correct amount of H

We need 6x O total so 6/2 = 3

So there are 3 moles of O

7 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

2 years ago