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3 years ago

Determine the molecular weight for Mg(NO3)2

1 answer:

4 0

The molecular weight is the same as molecular mass. You find this by using the periodic table
The molar mass of Mg(NO3)2 is 148.3

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3) How many grams are in 0.79 moles of HCI?<br> 1<br> UOL

snow_tiger [21]

Depending how you round and if you’re counting sig figs, it could be 28.8 or 29

3 0

3 years ago

Differentiate between elements and mixtures

Ksenya-84 [330]

Answer:

ELEMENTS MIXTURE

Elements are made up of one kind of atoms. Mixtures are made up of two or more kinds of Compounds .

Elements cannot be broken down into simpler substances by any physical or chemical method. The various constituents are seperated by simple physical means

3 0

2 years ago

Read 2 more answers

Calculate by (a)% weight and (b) %mole each of the elements present in sugar

Musya8 [376]

Explanation:

Molecular mass of sugar = $C_{12}H_{22}O_{11}$ : = 432 g/mol

Atomic mass of carbon atom = 12 g/mol

Atomic mass of hydrogen atom = 1 g/mol

Atomic mass of oxygen atom = 16 g/mol

a) Percentage of an element in a compound:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of compound}}\times 100$

Percentage of carbon by weight in $C_{12}H_{22}O_{11}$ :

$\frac{12\times 12 g/mol}{342 g/mol}\times 100=42.10\%$

Percentage of hydrogen by weight in $C_{12}H_{22}O_{11}$ :

$\frac{22\times 1g/mol}{342 g/mol}\times 100=6.43\%$

Percentage of oxygen by weight in $C_{12}H_{22}O_{11}$ :

$\frac{11\times 16g/mol}{342 g/mol}\times 100=51.46\%$

b) Percentage of mole each of the elements present in sugar:

= $\frac{\text{Moles of atoms of an element}}{\text{total moles of all types of atoms}}\times 100$

In mole of sugar we have 12 moles of carbon atom , 22 moles of hydrogen atoms and 11 moles of oxygen atoms.

Percentage of carbon by mole in $C_{12}H_{22}O_{11}$ :

$\frac{12 mol}{45 mol}\times 100=26.66\%$

Percentage of hydrogen by mole in $C_{12}H_{22}O_{11}$ :

$\frac{22 mol}{45 mol}\times 100=48.88\%$

Percentage of oxygen by mole in $C_{12}H_{22}O_{11}$ :

$\frac{11 mol}{45 mol}\times 100=24.44\%$

7 0

4 years ago

C. Balance these fossil-fuel combustion reactions. (1 point)

storchak [24]

The question is incomplete. The complete question is :

C. Balance these fossil-fuel combustion reactions. (1 point)

C8H18(g) + 12.5O2(g) → ____CO2(g) + 9H2O(g) + heat

CH4(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat

C3H8(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat

C6H6(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat

Solution :

C8H18(g) + 12.5O2(g) → __8__CO2(g) + 9H2O(g) + heat

When 1 part of octane reacts with 12.5 parts of oxygen, it gives 8 parts of carbon dioxide and 9 parts of water along with liberation of energy.

CH4(g) + __2__O2(g) → __1__CO2(g) + __2__H2O(g) + heat

When 1 part of methane reacts with 2 parts of oxygen, it gives 1 part of carbon dioxide and 2 parts of water along with liberation of energy.

C3H8(g) + __5__O2(g) → __3__CO2(g) + __4__H2O(g) + heat

When 1 part of propane reacts with 5 parts of oxygen, it gives 3 part of carbon dioxide and 4 parts of water along with liberation of energy.

C6H6(g) + __1/2__O2(g) → __6__CO2(g) + __3__H2O(g) + heat

When 1 part of propane reacts with 1/2 parts of oxygen, it gives 6 part of carbon dioxide and 3 parts of water along with liberation of energy.

5 0

3 years ago

How many mL of 1.50 M hydrochloric acid will neutralize 25 mL of 2.00 sodium hydroxide?

Lostsunrise [7]

Answer: The Answer is 18.7ml.

Explanation: Solved in the attached picture.

5 0

4 years ago