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3 years ago

How many moles of MgS2O3 are in 171 g of the compound?

Chemistry

1 answer:

Alexxx [7]3 years ago

8 0

Divide the 181g by the mol mass of the compound. Just add up the masses of the various elements.

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Ethanol, C2H6O, is most often blended with gasoline - usually as a 10 percent mix - to create a fuel called gasohol. Ethanol is

ch4aika [34]

Answer:

4054 kcal of heat is released during complete combustion of 354 g of octane.

Explanation:

Heat of combustion of 1 mol of octane is $1.308\times 10^{3}$ kcal

Molar mass of octane = 114.23 g/mol

We know, no. of moles = (mass)/(molar mass)

So, $1.308\times 10^{3}$ kcal of heat is released during complete combustion of 114.23 g of octane.

So, amount of heat is released during complete combustion of 354 g of octane = $(\frac{1.308\times 10^{3}}{114.23})\times 354$ kcal = 4054 kcal

Hence 4054 kcal of heat is released during complete combustion of 354 g of octane.

5 0

3 years ago

Phcl2 lewis structure

DiKsa [7]

Answer:

What do you want answered?

Explanation:

4 0

2 years ago

How does it? Please help

pashok25 [27]

How does it what. i don’t know if there’s a photo but can’t see it

5 0

3 years ago

Chemistry question.

grandymaker [24]

Hello!

Percentage error = (true value - measured value)/true value x 100%

Percentage error = (27.7-27.0)/27.7 x100% = 25.3%

8 0

3 years ago

A rigid tank that contains 2 kg of N2 at 25°C and 550 kPa is connected to another rigid tank that contains 4 kg of O2 at 25°C an

koban [17]

Answer:

The volume in the first tank = 0.32 $m^{3}$

The volume in the second tank = 2.066 $m^{3}$

The final pressure of the mixture = 203.64 K pa

Explanation:

<u>First Tank </u>

Mass = 2 kg

Pressure = 550 k pa

Temperature = 25 °c = 298 K

Gas constant for nitrogen = 0.297 $\frac{KJ}{Kg K}$

From the ideal gas equation

P V = m R T

550 × V = 2 × 0.297 × 298

V = 0.32 $m^{3}$

This is the volume in the first tank.

<u>Second tank</u>

Mass = 4 kg

Pressure = 150 K pa

Temperature = 25 °c = 298 K

Gas constant for oxygen = 0.26 $\frac{KJ}{Kg K}$

From the ideal gas equation

P V = m R T

150 × V = 4 × 0.26 × 298

V = 2.066 $m^{3}$

This is the volume in the second tank.

This is the iso thermal mixing. i.e.

$P_{3} V_{3} = P_{1} V_{1} + P_{2} V_{2}$ ----- (1)

$V_{3} = V_{1} + V_{2}$

$V_{3} = 0.32 + 2.066$

$V_{3} = 2.386 \ m^{3}$

Put this value in equation (1)

$P_{3}$ × 2.386 = 550 × 0.32 + 150 × 2.066

$P_{3}$ = 203.64 K pa

Therefore the final pressure of the mixture = 203.64 K pa

3 0

3 years ago