Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.
Explanation :
First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.
As, 207.2 g of lead contains 
atoms
So, 5.5 g of lead contains 
atoms
and,
As, 118.71 g of lead contains atoms
So, 94.5 g of lead contains 
atoms
Now we have to calculate the percent composition of Pb and Sn in atom.
and,
Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.
<span>It rises confidence for the reason that the more times you conduct the similar experiment over and over should either demonstrate your hypothesis right and wrong and remove any random incidences that might touch your results. Meaning it permits to have a more accurate measure or conclusion.</span>
Answer:
0.11 mol
Explanation:
<em>This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): CH₃CO₂H. An analytical chemist has determined by measurements that there are 0.054 moles of oxygen in a sample of acetic acid. How many moles of hydrogen are in the sample?</em>
Step 1: Given data
- Formula of acetic acid: CH₃CO₂H
- Moles of oxygen in the sample of acetic acid: 0.054 moles
Step 2: Establish the appropriate molar ratio
According to the chemical formula of acetic acid, the molar ratio of H to O is 4:2.
Step 3: Calculate the moles of atoms of hydrogen
We will use the theoretical molar ratio for acetic acid.
0.054 mol O × (4 mol H/2 mol O) = 0.11 mol H
Answer:
Concentration, because the amounts of reactants and products remain constant after equilibrium is reached.
Explanation:
The rate of reaction refers to the amount of reactants converted or products formed per unit time.
As the reaction progresses, reactions are converted into products. This continues until equilibrium is attained in a closed system.
When equilibrium is attained, the rate of forward reaction is equal to the rate of reverse reaction, hence the concentration of reactants and products in the system remain fairly constant over time.
When deducing the rate of reaction, concentration of the specie of interest is plotted on the y-axis against time on the x-axis.
3Na2S2O3 + AgBr ------>Na5[Ag(S2O3) 3] +NaBr
from equation 3 mol 1 mol
given x mol 0.10 mol
x= (3*0.10)/1=0.30 mol Na2S2O3
Answer: 0.30 mol Na2S2O3
