The mass of a sample of alcohol is found to be = m = 367 g
Hence, it is found out that by raising the temperature of the given product, the mass of alcohol would be 367 g.
Explanation:
The Energy of the sample given is q = 4780
We are required to find the mass of alcohol m = ?
Given that,
The specific heat given is represented by = c = 2.4 J/gC
The temperature given is ΔT = 5.43° C
The mass of sample of alcohol can be found as follows,
The formula is c = 
We can drive value of m bu shifting m on the left hand side,
m = 
mass of alcohol (m) = 
m = 367 g
Therefore, The mass of the given sample of alcohol is
m = 367g
It requires 4780 J of heat to raise the temperature by 5.43 C in the process which yields a mass of 367 g of alcohol.
1) Chemical equation
2Al + 6 HCl ---> 2Al Cl3 + 3 H2
2) molar ratios
2 mol Al : 3 moles H2
3) Proportion
2 mol Al / 3mol H2 = x / 9 mol H2
4) Solve for x
x = 9 mol H2 * 2 mol Al / 3 mol H2 = 6 mol Ag
Answer: 6 moles
Answer:
The volume is 1.2L
Explanation:
Initial volume (V1) = 700mL = 0.7L
Initial temperature (T1) = 7°C = (7 + 273.15)K = 280.15K
Initial pressure = 106.6kPa = 106600Pa
Final temperature (T2) = 27°C = (27 + 273.15)K = 300.15K
Final pressure (P2) = 66.6kPa = 66600Pa
Final volume (V2) = ?
To solve this question, we need to use combined gas equation which is a combination of Boyle's law, Charles Law and pressure law.
(P1 × V1) / T1 = (P2 × V2) / T2
solve for V2 by making it the subject of formula,
P1 × V1 × T2 = P2 × V2 × T1
V2 = (P1 × V1 × T2) / (P2 × T1)
V2 = (106600 × 0.7 × 300.15) / (66600 × 280.15)
V2 = 22397193 / 18657990
V2 = 1.2L
The final volume of the gas is 1.2L
Answer:
NH3
Explanation:
2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)
So for two moles of NH3 we need one mole of CO2. So let's count moles for each reagent.
n(NH3)=m(NH3)/M(NH3)=135700/17,03=7968.29 mol
n(CO2)=m(CO2)/M(CO2)=211400/44.01=4803.45 mol
From equation we have to divide n(NH3) by 2 because we need two equivalent per one CO2. That will be 3984.145. So the limiting agent is NH3 because it's not enough of it to react with all CO2
Answer:
From the periodic table:
mass of carbon = 12 grams
mass of hydrogen = 1 grams
mass of oxygen = 16 grams
molar mass of surcose = 12(12) + 22(1) + 11(16) = 342 grams
number of molecules = number of moles x Avogadro's number
number of moles = number of molecules / Avogadro's number
number of moles = (2.2x10^17) / (6.02x10^23) = 3.6544 x 10^-7 moles
number of moles = mass / molar mass
mass = number of moles x molar mass
= 1.7 x 10^17/6.022 x 10^23.
