Answer:
John Dalton, an English chemist and meteorologist, is credited with the first modern atomic theory based on his experiments with atmospheric gases.
Shade only feels cooler because you are avoiding solar radiation. In reality, the temperature in the sun is the same as the temperature in the shade. ... When in the shade, your skin is not being “heated” by the sun's rays, so your skin and your body feel a more comfortable temperature.
Answer:
0.425 moles of P₂O₅.
Explanation:
As per given balanced equation, four moles of phosphorus reacts with five moles of oxygen to give two moles of P₂O₅.
As given that the initial moles of phosphorus taken = 0.97 moles
moles of phosphorus left after reaction = 0.12 moles
moles of phosphorus reacted = 0.97-0.12 = 0.85 moles
When four moles of P reacts they give two moles of P₂O₅.
when one mole of P will react to give = ![\frac{2}{4}molP_{2}O_{5}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B4%7DmolP_%7B2%7DO_%7B5%7D)
0.85 moles of P will react to give =
P₂O₅
Answer:
e. 81.0%
Explanation:
Morphine has Molar mass = 285.34 g/mol
Also, number of males of Morphine = Number of moles of HCl
(At equivalence point )
Now, number of moles of HCl = Molarity × volume
=0.01 mol × 2.84×10^(-3) L
= 2.84×10^(-5) mol
therefore number of moles of morphine= 2.85×10^(-5)
Mass of Morphine= No. of moles × molar mass
= 2.84×10^(-5)×285.34 g/mol
= 0.0081 g
= 8.1 mg
therefore
percentage of morphine = mass of morphine/ mass of powder×100%
= 8.1/10×100% = 81.0%
hence option e is correct.
Answer:
![Q=16.62kJ](https://tex.z-dn.net/?f=Q%3D16.62kJ)
Δ![_CH=48.6kJ/g](https://tex.z-dn.net/?f=_CH%3D48.6kJ%2Fg)
Explanation:
Hello,
A. In this part, we must include the heat effect for pentane and water:
![Q=1000gH_2O*4.184J/(g^0C)*\frac{1kJ}{1000J} *(22.82^0C-20.22^0C)+2.21kJ/^0C*(22.82^0C-20.22^0C)\\Q=16.62kJ](https://tex.z-dn.net/?f=Q%3D1000gH_2O%2A4.184J%2F%28g%5E0C%29%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%20%2A%2822.82%5E0C-20.22%5E0C%29%2B2.21kJ%2F%5E0C%2A%2822.82%5E0C-20.22%5E0C%29%5C%5CQ%3D16.62kJ)
B. In this part, we just consider the previously computed heat and divide it by the sample of pentane as follows:
Δ![_CH=\frac{Q}{m}=\frac{16.6kJ}{0.3423g}= 48.6kJ/g](https://tex.z-dn.net/?f=_CH%3D%5Cfrac%7BQ%7D%7Bm%7D%3D%5Cfrac%7B16.6kJ%7D%7B0.3423g%7D%3D%2048.6kJ%2Fg)
Best regards.