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Serhud [2]
3 years ago
15

128g of sulphur are burned in excess oxygen. What mass of sulphur dioxide forms? S + O2 --> SO2 *

Chemistry
1 answer:
Rasek [7]3 years ago
5 0

Mass of Sulphur dioxide : 256 g

<h3>Further explanation</h3>

Given

Reaction

S + O2 --> SO2 *

Required

Mass of Sulphur dioxide

Solution

mol of Sulphur (Ar=32 g/mol) :

mol = mass : Ar

mol = 128 : 32

mol = 4

From the equation, mol ratio S : SO2 = 1 : 1, so mol SO2 = 4

Mass of SO2 :

mass = mol x MW SO2

mass = 4 x 64

mass = 256 g

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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
NeTakaya

Answer:

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

Explanation:

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}

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

The units of liters cancel.

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

\frac {50.25\  atm }{14.5 }=P_2

3.53448276 \ atm = P_2

The original values of volume and pressure have 3 significant figures, so our answer must have the same.

For the number we found, that is the hundredth place.

  • 3.53448276

The 4 in the thousandth place (in bold above) tells us to leave the 3 in the hundredth place.

3.53 \ atm \approx P_2

The new pressure is approximately <u>3.53 atmospheres.</u>

8 0
3 years ago
A 1.000 gram sample of an alcohol was burned in oxygen to produce 1.913 g of CO2 and 1.174 g of H20. The alcohol contained only
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Assuming the conditions of the reaction are maintained and appropriate for the reaction to still occur, the reaction rate can be affected by increasing the concentration of the reagents used in a reaction. It will speed it up.
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