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3 years ago

Molar mass of propan-1-ol

Chemistry

2 answers:

Lady_Fox [76]3 years ago

8 0

Answer:

60g/mol.

Explanation:

TO FIND :-

Molar mass of Propan-1-ol

FACTS TO KNOW BEFORE FINDING ANSWER :-

Chemical formula of Propan-1-ol = CH₃CH₂CH₂OH [or] C₃H₈O
Molar mass of C = 12g/mol.
Molar mass of H = 1g/mol.
Molar mass of O = 16g/mol.

SOLUTION :-

Molar mass of C₃H₈O = 3(Molar mass of C) + 8(Molar mass of H) + 1(Molar mass of O)

⇒ Molar mass of C₃H₈O = (3×12) + (8×1) + 16 = 36 + 8 + 16 = 60g/mol.

Ira Lisetskai [31]3 years ago

4 0

Answer:

60.0952 g/mol

Formula: C₃H₈O

Density: 803 kg/m³

Melting point: -126 °C

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When calcium carbonate reacts with hydrochloric acid, calcium ions are transferred to the solution. Write only one net ionic equ

nika2105 [10]

Answer:

In conclusion, the net ionic equation for the reaction between calcium carbonate and hydrochloric acid is CaCO3 solid plus two H+ aqueous react to form Ca2+ aqueous plus CO2 gas plus H2O liquid.

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2 years ago

A compound is found to contain 38.65 % carbon, 16.25 % hydrogen, and 45.09 % nitrogen by mass. what is the empirical formula for

iren [92.7K]

The molar mass of carbon is 12, hydrogen is 1, and nitrogen is 14, hence the ratio are:

C = 38.65 / 12 = 3.22

H = 16.25 / 1 = 16.25

N = 45.09 / 14 = 3.22

Divide the three by the lowest ratio which is 3.22:

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5 0

3 years ago

How many moles of ScCl3 can be produced when 10.00 mol Sc react with 9.00 mol Cl2

Nuetrik [128]

Answer:

Moles of $ScCl_3$ = 6 moles

Explanation:

The reaction of $Sc$ and $Cl_2$ to make $ScCl_3$ is:

$2Sc+3Cl_2$ ⇒ $2ScCl_3$

The above reaction shows that 2 moles of Sc can react with 3 moles of $Cl_2$ to form $ScCl_3.$

Mole Ratio= 2:3

For 10 moles of Sc we need:

Moles of $Cl_2$ = $Moles of Sc *\frac{3 moles of Cl_2}{2 Moles of Sc}$

Moles of $Cl_2$ = $10 *\frac{3 moles of Cl_2}{2 Moles of Sc}$

Moles of $Cl_2$ =15 moles

So 15 moles of $Cl_2$ are required to react with 10 moles of $Sc$ but we have 9 moles of $Cl_2$ , it means $Cl_2$ is limiting reactant.

$Moles of ScCl_3=Given\ Moles\ of\ Cl_2 *\frac{2\ Moles\ o\ fScCl_3}{3\ Moles\ of\ Cl_2}$