Answer:
2192.64 PSI.
Explanation:
- From the general law of ideal gases:
<em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the container in L (V = 1650 L).
n is the no. of moles of the gas in mol (n = 9750 mol).
R is the general gas constant (R = 0.082 L.atm/mol.K).
T is the temperature of the gas in (T = 35°C + 273 = 308 K).
∴ P = nRT/V = (9750 mol)(0.082 L.atm/mol.K)(308 K)/(1650 L) = 149.2 atm.
- <u><em>To convert from atm to PSI:</em></u>
1 atm = 14.696 PSI.
<em>∴ P = 149.2 atm x (14.696 PSI/1.0 atm) = 2192.64 PSI.</em>
Answer:
The answer is option 3, C5H12 + 8O2 → 5CO2 + 6H2O.
Explanation:
In an exothermic reaction, the energy change(ΔH) will always be a negative value.
For endothermic reaction, the energy change's value is positive.
In the options above, option 1 and 2 are endothermic reaction.
Answer:
a, g, c
Explanation:
The conversion of the stable cyclopentane into Trans-1, 2dibromocyclopentane will require three step reactions.
The first is to convert the compound into a cyclopentene, through the addition of Bromine water under heat and photons (light). So option A is the first in the order. This will generate 1 bromocyclopentane through halogenation of the alkane. Secondly, a hot and strong base should be added like the NaOEt, EtOH to remove the added bromine and one atom of hydrogen from the resulting 1 bromocyclopentane in the previous reaction. This will yield cyclopentene, thus making the compound more electrophilic. So option g is required. Thirdly, bromine molecules will be added (C) to take up their places at the two electrophilic regions of the compound to produce Trans-1, 2dibromocyclopentane.
Explanation:
Depression in Freezing point
= Kf × i × m
where m is molality , i is Van't Hoff factor, m = molality
Since molality and Kf remain the same
depression in freezing point is proportional to i
i= 2 for CuSO4 ( CuSO4----------> Cu+2 + SO4-2
i=1 for C2h6O
i= 3 for MgCl2 ( MgCl2--------> Mg+2+ 2Cl-)
So the freezing point depression is highest for MgCl2 and lowest for C2H6O
so freezing point of the solution = freezing point of pure solvent- freezing point depression
since MgCl2 has got highest freezing point depression it will have loweest freezing point and C2H6O will have highest freezing point
Rutherford's result can not be explained on the basis of the plum pudding model because of the fact that, since some alpha particles were deflected, the atom must contain a small region with a strong electric charge.
The empirical study of the atom led to the emergence of several models of the atom. In the Plum - pudding model, the atom was regarded as a positively charged sphere with embedded negative charges.
This model can not interpret the Rutherford experiment since alpha particles were deflected. The deflection of alpha particles means that, the atom must contain a small region with a strong electric charge.
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