Answer:
option B.
Explanation:
Given,
V₁ = 156 L
P₁ =2 atm
Now, in the cylinder
P₂ = ?
V₂ = 36
Using relation between pressure and volume
![\dfrac{P_1}{V_2}=\dfrac{P_2}{V_1}](https://tex.z-dn.net/?f=%5Cdfrac%7BP_1%7D%7BV_2%7D%3D%5Cdfrac%7BP_2%7D%7BV_1%7D)
![\dfrac{2}{36}=\dfrac{P_2}{156}](https://tex.z-dn.net/?f=%5Cdfrac%7B2%7D%7B36%7D%3D%5Cdfrac%7BP_2%7D%7B156%7D)
![P_2 = 8.67\ atm](https://tex.z-dn.net/?f=P_2%20%3D%208.67%5C%20atm)
Hence, pressure is equal to 8.67 atm.
Hence, the correct answer is option B.
The First 2 statements stated above were false whereas the third one is a true statement.
Explanation:
- The viscosity of bitumen is about 100 times greater than the viscosity of water - False
Reason - The viscosity of bitumen is about not 100 times greater than the viscosity of water, it is actually 100, 000 times greater.
- Oil from oil sand deposits is only obtained by first heating the sands at high temperatures is False.
Reason- Oil from oil sand deposits is not obtained by first heating the sands at high temperatures but by using steams
- Oil sands contain sand, water, and light crude oil is true.
Answer: Atoms lose energy as a gas changes to a solid.
Explanation:
Answer:
26.981539 u
Explanation:
One mole of Al atoms has a mass in grams that is numerically equivalent to the atomic mass of aluminum. The periodic table shows that the atomic mass (rounded to two decimal points) of Al is 26.98, so 1 mol of Al atoms has a mass of 26.98 g.
Answer:
![V_{base}=24.04mL](https://tex.z-dn.net/?f=V_%7Bbase%7D%3D24.04mL)
Explanation:
Hello!
In this case, according to the chemical reaction by which HBr reacts with Ba(OH)2:
![2HBr+Ba(OH)_2\rightarrow BaBr_2+2H_2O](https://tex.z-dn.net/?f=2HBr%2BBa%28OH%29_2%5Crightarrow%20BaBr_2%2B2H_2O)
We can see there is a 2:1 mole ratio between the acid and the base; thus, at the equivalent point we can write:
![2M_{base}V_{base}=M_{acid}V_{acid}](https://tex.z-dn.net/?f=2M_%7Bbase%7DV_%7Bbase%7D%3DM_%7Bacid%7DV_%7Bacid%7D)
Therefore, for is to compute the volume of the used base, we proceed as shown below:
![V_{base}=\frac{M_{acid}V_{acid}}{2M_{base}}](https://tex.z-dn.net/?f=V_%7Bbase%7D%3D%5Cfrac%7BM_%7Bacid%7DV_%7Bacid%7D%7D%7B2M_%7Bbase%7D%7D)
And we plug in to obtain:
![V_{base}=\frac{0.319M*50.8mL}{2*0.337M}\\\\V_{base}=24.04mL](https://tex.z-dn.net/?f=V_%7Bbase%7D%3D%5Cfrac%7B0.319M%2A50.8mL%7D%7B2%2A0.337M%7D%5C%5C%5C%5CV_%7Bbase%7D%3D24.04mL)
Best regards!