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2 years ago

Calculate the mass of grams of 14.79ml of a substance. It'd density is 1.193gml

1 answer:

8 0

<span>D=m/v

Given:
d=1.193g/ml
v= 14.79ml

Solution
m=dv
m=1.193g/ml x 14.79ml

m= 17.644g

Density is the mass per volume of the material. Basically, it refers to the measurement of how crammed together is the matter. The formula to calculate density is rho = m/v wherein rho is the density, m stands for the mass and v stands for the volume.</span>

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3 years ago

What is the correct chemical equation for the reaction between methane and oxygen to produce carbon dioxide and water?

kodGreya [7K]

Answer:

The answer to your question is: CH₄ + 3/2 O₂ ⇒ CO₂ + 2 H₂O

Explanation:

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2 years ago

How many moles are present in 4.7 liters of argon gas as STP

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Answer:

0.21mol Ar (g)

Explanation:

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2 years ago

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2 years ago

N2H4 + N2O4 --> N2 + H2O

ahrayia [7]

Answer:

The limiting reagent is N2O4
14,09g

Explanation:

First, we adjust the reaction.

$2N_{2} H_{4}$ + $N_{2} O_{4}$ ⇄ $6N_{2} + 4H_{2}O$

Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using $N_{2} H_{4}$ to form $H_{2}O$

$molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4} }{32,04\frac{g}{mol} N_{2} H_{4} }$

$molH_{2} O = 1, 125 mol$

Using $N_{2} O_{4}$ to form $H_{2} O$

$molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4} }{92\frac{g}{mol} N_{2} O_{4} }$

$molH_{2} O = 0,783 mol$

The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.

This is the minimum measure can be formed of each product.

∴ $MassOfH_{2}O = 0,783mol . 18\frac{g}{mol}$

$MassOfH_{2}O = 14,09g$

5 0

3 years ago