How many moles is 1200 grams of Mn3(SO4)7?

Chemistry

2 answers:

irga5000 [103]3 years ago

6 0

no.of moles= mass/ Mr

=1200/(54.9*3)+7[32.1+(16*4)]

=1.433

pav-90 [236]3 years ago

5 0

1.64 moles

grams to mole

add all of them together 54.938+672.427=727.365

divide 1200/727.365=1.64979068

Just learned this in chemistry got a 100 on test.

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A 25.0 L tank of nitrogen gas is at 25 oC and 2.05 atm . If the temperature stays at 25 oC and the volume is decreased to 14.5 L

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Answer:

$\boxed {\boxed {\sf P_2 \approx 3.53 \ atm}}$

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In this problem, the temperature stays constant. The volume and pressure change, so we use Boyle's Law. This states that the pressure of a gas is inversely proportional to the volume. The formula is:

$P_1V_1=P_2V_2$

Now we can substitute any known values into the formula.

Originally, the gas has a volume of 25.0 liters and a pressure of 2.05 atmospheres.

$25.0 \ L * 2.05 \ atm = P_2V_2$

The volume is decreased to 14.5 liters, but the pressure is unknown.

$25.0 \ L * 2.05 \ atm = P_2 * 14.5 \ L$

Since we are solving for the new pressure, or P₂, we must isolate the variable. It is being multiplied by 14.5 liters and the inverse of multiplication is division. Divide both sides by 14.5 L .

$\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}=\frac{P_2 *14.5 \ L}{14.5 \ L}$