Answer:
3.14
Explanation:
A student was comparing two samples with an equal number of carbon atoms. One sample contained only Carbon-12 atoms. One sample contained only Carbon-14 atoms, which contain two more neutrons than Carbon-12 atoms. The student measured the mass of each sample and testing the reactivity of each sample.
Required:
What would best describe the results of the investigation?
Answer:
The correct answer is option A.
Explanation:
Equilibrium is a state when rate of forward reaction is equal to the rate of backward reaction. The concentration of reactants and products becomes constant at this state.
The ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric coefficients is termed as Equilibrium constant. It is denoted by 
.
aA + bB 
cC
![K_{eq}=\frac{[C]^c}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BC%5D%5Ec%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
According to the chemical equation, the reaction ratio between O2 and CO2 is 2:1, which mean for every 2 moles of O2 reacted there is 1 mole of CO2 formed.
Use the molar mass and mass of O2 to find out the moles of O2: moles of O2 = mass of O2/molar mass of O2 = 8.94g/32.00g/mol = 0.2794 mole. Therefore, the moles of CO2 that formed is 0.2794moles/2 = 0.1397 mole
Use the moles and molar mass CO2 to find out the mass of CO2:
Mass of CO2 = moles of CO2 * molar mass of CO2 = 0.1397 mole * .44.01g/mole = 6.15 g.
So the answer is B 6.15g.
Answer:
T2 = 2843.1 oK. This is a huge temperature. Check it for errors.
Explanation:
Remark
This is the same question as the other one I've answered. Only the numbers have been altered.
Givens
v1 = 56 mL
P1 = 1 atm
T1 = 273o K
v2 = 162
P2 = 3.6 atm
T2 = ?
Formula
Vi * P1 / T1 = V2 * P2/T2
Solution
Rearrange the formula so T2 is on the left
T2 = V2 P2 * T1 / (V1 * P1) Now just put the numbers in.
T2 = 162 * 3.6* 273 / (56 *1)
T2 = 159213.6/56
T2 = 2843.1
Answer:
2 mol NO2
Explanation:
3NO2(g)+H2O(l)→2HNO3(l)+NO(g)
from reaction 3 mol 1 mol
given 11 mol 3 mol
for 3 mol NO2 ----- 1 mol H2O
for x mol NO2 ----- 3 mol H2O
3:x = 1:3
x = 3 *3/1 = 9 mol NO2
So, for 3 mol H2O are needed only 9 mol NO2.
But we have 11 mol NO2. So, NO2 is in excess, and
11 mol NO2 - 9 mol NO2 = 2 mol NO2 will be left after reaction.
