Answer:
1. The pressure will be 32 atm, twice the initial pressure.
2. The pressure will be 1.83 atm, one third of the initial pressure.
Explanation:
Boyle's law is one of the gas laws that relates the volume and pressure of a certain quantity of gas kept at a constant temperature.
This law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.
Boyle's law is expressed mathematically as:
Pressure * Volume = constant
or P * V = k
Ahora es posible suponer que tienes un cierto volumen de gas V1 que se encuentra a una presión P1 al comienzo del experimento. Si varias el volumen de gas hasta un nuevo valor V2, entonces la presión cambiará a P2, y se cumplirá:
P1*V1=P2*V2
1. In this case:
- P1= 16 atm
- V1
- P2= ?
- V2= V1÷2=
because the volume is halved.
So:
16 atm*V1= P2*
Solving:
=P2
16 atm*2= P2
32 atm= P2
<u><em>The pressure will be 32 atm, twice the initial pressure.</em></u>
2. Now
- P1= 5.5 atm
- V1
- P2= ?
- V2= V1*3 because the volume is tripled.
So:
5.5 atm*V1= P2* V1*3
Solving:
=P2
= P2
1.83 atm= P2
<u><em>The pressure will be 1.83 atm, one third of the initial pressure.</em></u>
It is D: a spinning cloud of dust...
Answer:
- Take 3.3 mL of 3.0-M hydrochloric acid and subsequently add 76.7 mL of water to complete the 100.00 mL.
- Take 11.7mL of 6.0-M hydrochloric acid and subsequently add 88.3 mL of water to complete the 100.00 mL
Explanation:
Hello,
In this case, given that the dilutions are preparedfrom 3.0-M and 6.0-M hydrochloric acid, we must proceed as follows:
- 3.0-M stock: when using this stock, the aliquot you must take is computed as shown below:
It means that you must take 23.3 mL of 3.0-M hydrochloric acid and subsequently add 76.7 mL of water to complete the 100.00 mL.
- 6.0-M stock: when using this stock, the aliquot you must take is computed as shown below:
It means that you must take 11.7mL of 6.0-M hydrochloric acid and subsequently add 88.3 mL of water to complete the 100.00 mL.
Regards.
The initial sample has a molecular formula of MnSO₄·H₂O. This molecule is a hydrate as it has a unit of water within its structure for every molecule of MnSO₄. This sample is being dehydrated to remove the water to give.
MnSO₄·H₂O → MnSO₄ + H₂O
MnSO₄·H₂O has a molecular mass of 169.02 g/mol. While MnSO₄ has a molecular mass of 151 g/mol. Water has a molecular mass of 18.02 g/mol. We now can use the ratio of the mass of water to the mass of the initial sample to determine the percentage of each component by mass.
% water by mass:
18.02/169.02 x 100% = 10.7% Water by mass.
% MnO₄ by mass:
151/169.02 x 100% = 89.3% MnSO₄ by mass.
Water makes up 10.7% of the initial mass of MnSO₄·H₂O.
Answer:
A
Explanation:
Carbon dioxide in the air
