What is the name of this molecule ?

Find the number of moles of water that can be formed if you have 126 mol of hydrogen gas and 58 mol of oxygen gas.

Elden [556K]

Since a water molecule is H2O, you would divide 126 hydrogen molecules by 2, and you would get 63. That means you have 63 double hydrogen molecules, and 58 oxygen molecules to pair up with them. So that means you could have 58 molecules of water, with 5 double hydrogen molecules, so basically 10 extra molecules of hydrogen along with the H2O molecules. Hope I helped! :)

6 0

3 years ago

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How many moles of 0.225 M CaOH2 are present in 0.350 L of solution?

weeeeeb [17]

Answer : The number of moles of solute $Ca(OH)_2$ is, 0.0788 moles.

Explanation : Given,

Molarity = 0.225 M

Volume of solution = 0.350 L

Formula used:

$\text{Molarity}=\frac{\text{Moles of }Ca(OH)_2}{\text{Volume of solution (in L)}}$

Now put all the given values in this formula, we get:

$0.225M=\frac{\text{Moles of }Ca(OH)_2}{0.350L}$

$\text{Moles of }Ca(OH)_2=0.0788mol$

Therefore, the number of moles of solute $Ca(OH)_2$ is, 0.0788 moles.

7 0

3 years ago

What is the molarity (M) of the following solutions?

Dennis_Churaev [7]

Answer:

The molarity (M) of the following solutions are :

A. M = 0.88 M

B. M = 0.76 M

Explanation:

A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.

Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)

= 27 + 3(16 + 1)

= 27 + 3(17) = 27 + 51

= 78 g/mole

$Al(OH)_3$ = 78 g/mole

Given mass= 19.2 g/mole

$Mole = \frac{Given\ mass}{Molar\ mass}$

$Mole = \frac{19.2}{78}$

Moles = 0.246

$Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}$

Volume = 280 mL = 0.280 L

$Molarity = \frac{0.246}{0.280)}$

Molarity = 0.879 M

Molarity = 0.88 M

B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr

Molar mass of KBr = 119 g/mole

Given mass = 235.9 g

$Mole = \frac{235.9}{119}$

Moles = 1.98

Volume = 2.6 L

$Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}$

$Molarity = \frac{1.98}{2.6)}$

Molarity = 0.762 M

Molarity = 0.76 M

4 0

2 years ago

How many grams of calcium phosphate (Ca3(PO4)2) re theoretically produced if we start with 3.40 moles of Ca(NO3)2 and 2.40moles

sattari [20]

1) Balance the chemical equation.

$3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2$

2) List the known and unknown quantities.

Reactant 1: Ca(NO3)2.

Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

3) Which is the limiting reactant?

3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4$

We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. This is the excess reactant.

3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2$

We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

$mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2$

5) Mass of Ca3(PO4)2 produced.

The molar mass of Ca3(PO4)2 is 310.1767 g/mol.

$g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}$ $g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2$

The mass of Ca3(PO4)2 produced is 351 g Ca3(PO4)2.

Option D.

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8 0

1 year ago

1. What is the balanced chemical equation for the reaction that takes place between bromine and sodium iodine?

barxatty [35]

1 is B (Just remember to have the same number of atoms on both sides)
2 is B (A precipitate is a solid forming from 2 liquids)

5 0

3 years ago

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