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2 years ago

mass of 6.584 g. After it is heated, it has a mass of 4.194 g. What is the percentage by mass of water in the hydrate?

1 answer:

8 0

M₁=6.584 g
m₂=4,194 g

m(H₂O)=m₁-m₂

w(H₂O)=m(H₂O)/m₁

w(H₂O)=(m₁-m₂)/m₁

w(H₂O)=(6.584-4.194)/6.584=0.3630 (36.30%)

the percentage by mass of water in the hydrate 36.30

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Answer: 7 g/cm to the third power
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3 years ago

For the following reaction, calculate how many moles of NO2 forms when 0.356 moles of the reactant completely reacts.

Nesterboy [21]

Answer:

0.712 mol of NO₂ are formed .

Explanation:

For the reaction , given in the question ,

2 N₂O₅ ( g ) → 4 NO₂ ( g ) + O₂ ( g )

From the above balanced reaction ,

2 mol of N₂O₅ reacts to give 4 mol of NO₂

Applying unitary method ,

1 mol of N₂O₅ reacts to give 4 / 2 mol of NO₂

From the question , 0.356 mol of N₂O₅ are reacted ,

<u>now, using the above equation , to calculate the moles of the NO₂ , as follow -</u>

Since ,

1 mol of N₂O₅ reacts to give 4 / 2 mol of NO₂

0.356 mol of N₂O₅ reacts to give 4 / 2 * 0.356 mol of NO₂

Calculating ,

0.712 mol of NO₂ are formed .

7 0

3 years ago

Would the water in a pond or the water in a full bathtub have more thermal energy? Explain why.

Mekhanik [1.2K]

Thermal energy is defined as the total kinetic energy of all particles in an object. Even though the cup of water has a higher temperature, the bathtub has more thermal energy because it contains much more mass of water compared to the cup.

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2 years ago

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iogann1982 [59]

If you need to translate, the answer is
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6 0

3 years ago

How many molecules are in 7.62 L of CH4, at 87.5°C and 722 torr

pickupchik [31]

Answer: There are $1.469 \times 10^{23}$ molecules present in 7.62 L of $CH_4$ at $87.5^{o}C$ and 722 torr.

Explanation:

Given : Volume = 7.62 L

Temperature = $87.5^{o}C = (87.5 + 273) K = 360.5 K$

Pressure = 722 torr

1 torr = 0.00131579

Converting torr into atm as follows.

$722 torr = 722 torr \times \frac{0.00131579 atm}{1 torr}\\= 0.95 atm$

Therefore, using the ideal gas equation the number of moles are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

$PV = nRT\\0.95 atm \times 7.62 L = n \times 0.0821 L atm/mol K \times 360.5 K\\n = \frac{0.95 atm \times 7.62 L}{0.0821 L atm/mol K \times 360.5 K}\\= \frac{7.239}{29.59705}\\= 0.244 mol$

According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms. Hence, number of atoms or molecules present in 0.244 mol are calculated as follows.

$0.244 mol \times 6.022 \times 10^{23}\\= 1.469 \times 10^{23}$

Thus, we can conclude that there are $1.469 \times 10^{23}$ molecules present in 7.62 L of $CH_4$ at $87.5^{o}C$ and 722 torr.

5 0

2 years ago