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3 years ago

If the mass of the products of a chemical reaction was 100 g, what was

1 answer:

4 0

If the mass of all of the products in a chemical reaction is equal to 100g then the mass of the reactants in that same reaction had to have had a mass of 100g this is due to the law of conservation of matter stating matter cannot be created or destroyed in a chemical reaction.

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1. Silver nitrate will react with aluminum metal, yielding aluminum nitrate and silver metal. If you start with 0.223 moles of a

Archy [21]

Answer:

Explanation:

Given data:

Number of moles of aluminium = 0.223 mol

Mass of silver produced = ?

Solution:

Chemical equation:

3AgNO₃ + Al → 3Ag + Al(NO₃)₃

Now we will compare the moles of Al with silver.

Al : Ag

1 : 3

0.223 : 3×0.223= 0.669 mol

Grams of silver:

Mass = number of moles × molar mass

Mass = 0.669 mol × 107.87 g/mol

Mass = 72.2 g

Given data:

Number of moles of mercury(II) oxide produced = 3.12 mol

Mass of mercury = ?

Solution:

Chemical equation:

2Hg + O₂ → 2HgO

Now we will compare the moles of mercury with mercury(II) oxide.

HgO : Hg

2 : 2

3.12 : 3.12

Mass of Hg:

Mass = number of moles × molar mass

Mass = 3.12 mol × 200.59 g/mol

Mass = 625.84 g

Given data:

Number of moles of dinitrogen pentoxide = 12.99 mol

Mass of oxygen = ?

Solution:

Chemical equation:

2N₂ + 5O₂ → 2N₂O₅

Now we will compare the moles of N₂O₅ with oxygen.

N₂O₅ : O₂

2 : 5

12.99 : 5/2×12.99 = 32.48 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 32.48 mol × 32 g/mol

Mass = 1039.36 g

Given data:

Number of moles of benzene = 0.103 mol

Mass of carbon dioxide = ?

Solution:

Chemical equation:

2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

Now we will compare the moles of N₂O₅ with oxygen.

C₆H₆ : CO₂

2 : 12

0.103 : 12/2×0.103 = 0.618 mol

Mass of carbon dioxide:

Mass = number of moles × molar mass

Mass = 0.618 mol × 44 g/mol

Mass = 27.192 g

3 0

3 years ago

What are the phases of the moons all of them not just 3

levacccp [35]

Number 1

~ The New Moon

Number 2

~ Waxing Crescent

Number 3

~ First Quarter

Number 4

~ Waxing Gibbous

Number 5

~ Full Moon

Number 6

~ Waning Gibbous

Number 7

~ Last Quarter

Number 8

~ Waning Crescent

7 0

3 years ago

A 3.0-L sample of helium was placed in container fitted with a porous membrane. Half of the helium effused through the membrane

zhenek [66]

Explanation:

It is known that rate of effusion of gases are inversely proportional to the square root of their molar masses.

And, half of the helium (1.5 L) effused in 24 hour. So, the rate of effusion of He gas is calculated as follows.

$\frac{1.5 L}{24 hr}$

= 0.0625 L/hr

As, molar mass of He is 4 g/mol and molar mass of $O_{2}$ is 32 g/ mol.

Now,

$\frac{\text{Rate of He}}{\text{Rate of Oxygen}} = \sqrt{\frac{32}{4}$

= 2.83

or, rate of $O_{2} = \frac{\text{Rate of He}}{2.83}$

= $\frac{0.0625 L/hr}{2.83}$

Rate of $O_{2}$ = 0.022 L/hr.

This means that 0.022 L of $O_{2}$ gas effuses in 1 hr

So, time taken for the effusion of 1.5 L of $O_{2}$ gas is calculated as follows.

$\frac{1.5 L}{0.022 L/hr}$

= 68.18 hour

Thus, we can conclude that 68.18 hours will it take for half of the oxygen to effuse through the membrane.

3 0

3 years ago

How many atoms are present in a 1.0 mole sample of potassium perchlorate (KClO3)?

alekssr [168]

Okay, a mole of potassium perchlorate contains 6.02x1023 formula units of potassium perchlorate, but you're asking about individual atoms. So, let's look at the formula: KClO3. That's 1 potassium, 1 chlorine, and 3 oxygens, for a total of 5 atoms per formula unit. Now, multiple 5 by Avogadro's number above, to get 30.1x1023, which simplifies to 3.01x1024 atoms.

5 0

3 years ago

Read 2 more answers

Which is smallest to biggest glucose,starch,protein,carbon,water,amino,oxygen

likoan [24]

Answer:

Smallest to biggest order is as follows.

C < O < Amino < Water < Glucose < Starch < protein.

Explanation:

The given atoms or molecules and their atomic masses is as follows.

Glucose - $C_{6}H_{12}O_{6}$ - 180 g/mol