The number of moles for co2=mass(g)/molar mass
n=.22/44=.005 mole of CO2
from the equation we see the relationship between the moles of co2 and O2 and we find that they have the same number of moles
So we need .005mole of O2
Multiple the number of moles with avogadro’s number to know the number of O2molecules
.005x6.022 x10^23
Answer:
97 000 g Na
Explanation:
The absortion (or liberation) of energy in form of heat is expressed by:
q=m*Cp*ΔT
The information we have:
q=1.30MJ= 1.30*10^6 J
ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)
Cp=30.8 J/(K mol Na)
If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.
To do so, we use the molar mass of Na= 22.99g/mol

Now, we are able to solve for m:
=97 000 g Na
Answer:
1,033.56 grams of carbon dioxide was emitted into the atmosphere.
Explanation:
Energy absorbed by pork,E =
(assuming)
Total energy produced by barbecue = Q
Percentage of energy absorbed by pork = 10%


Since, it is a energy produced in order to indicate the direction of heat produced we will use negative sign.
Q = 
Moles of propane burnt to produce Q energy =n


According to reaction , 1 mol of propane gives 3 moles of carbon dioxide. then 7.83 moles of will give:
carbon dioxide gas.
Mass of 23.49 moles of carbon dioxide gas:
23.49 mol × 44 g/mol =1,033.56 g
1,033.56 grams of carbon dioxide was emitted into the atmosphere.
Answer: that all thre water cycle and C is vaporation
Explanation:
Answer:
The stronger conjugate base will be the weaker acid; i.e., the acid with the smaller Ka-value.
Explanation:
Given conjugate base CN⁻ => weak acid => HCN => Ka =4.9 x 10⁻¹⁰
Given conjugate base OCN⁻ => weak acid=> HOCN => Ka = 3.5 x 10⁻⁴
Ka(HCN) << Ka(HOCN) => CN⁻ is a much stronger conjugate base than OCN⁻