All categories

2 years ago

In order to calculate the mass of an element in a given mass of a compound the mass of the compound is multiplied by the mass

Chemistry

1 answer:

Allushta [10]2 years ago

7 0

Molar mass is the mass of a given substance divided by the amount of that substance, measured in g/mol.

You might be interested in

Ethanol can react with hydrochloric acid to form chloroethane and water. When sulfuric acid is added, the

kaheart [24]

Answer: 1.Sulfuric acid is a catalyst

2. Vanadium(v) oxide is a catalyst

6 0

2 years ago

What are the names of the components (each shown with a leader and a line) of a circuit shown in the diagram? Please help!!!!

alekssr [168]

Maybe: Battery, wire, lamp and switch

6 0

2 years ago

which pair shares the same empirical formula. c2h2 and c6h6, c2h2 and c2h4, ch2 and c6h6, ch and c2h4

olasank [31]

Answer:

Answer: CH₃ and C₂H₆ have same empirical formula.

Explanation:

it just compares in that it its the same

7 0

2 years ago

How many liters of hydrogen are required to react completely with 2.4L of oxygen to form water? 2H2 + O2 --> 2H2O

RoseWind [281]

Answer:

2.4 mole of oxygen will react with 2.4 moles of hydrogen

Explanation:

As we know

1 liter = 1000 grams

2H2 + O2 --> 2H2O

Weight of H2 molecule = 2.016 g/mol

Weight of water = 18.01 gram /l

2 mole of oxygen react with 2 mole of H2

2.4 mole of oxygen will react with 2.4 moles of hydrogen

3 0

2 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

2 years ago