Answer:
2.29*10^24 molecules of water
Explanation:
1 molof water =6.022*10^23 molecules of water
therefore
molecules of water = 3.8*6.022*10^23
Answer:
81.26% is the percent yield
Explanation:
Based on the reaction:
CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃
<em>Where 1 mole of CaCl₂ in excess of sodium carbonate produces 1 mole of calcium carbonate.</em>
<em />
To solve this question we must find the moles of CaCl2 added = Moles CaCO₃ produced (Theoretical yield). The percent yield is:
Actual yield (0.366g) / Theoretical yield * 100
<em>Moles CaCl₂ = Moles CaCO₃:</em>
0.0500L * (0.0900moles / L) = 0.00450 moles of CaCO₃
<em>Theoretical mass -Molar mass CaCO₃ = 100.09g/mol-:</em>
0.00450 moles of CaCO₃ * (100.09g / mol) = 0.450g of CaCO₃
Percent yield = 0.366g / 0.450g * 100
81.26% is the percent yield
Answer:
28.01g
Explanation:
Given the weight of one mole of Cabon as 12.01g and that of oxygen as 16.00g.
The molecular weight of a compound can be gotten by adding the molar weights of the elements that constitutes the compound .
The molecular weight of the compound CO is therefore
equal to the sum of the weight of both elements.
That’s = 12.01g + 16.00g
= 28.01g
Therefore, the molecular weight of CO is 28.01g
Answer : The final concentration of 
is, 2.9 M
Explanation :
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = 
t = time passed by the sample = 3.5 min
a = initial concentration of the reactant = 3.0 M
a - x = concentration left after decay process = ?
Now put all the given values in above equation, we get
Thus, the final concentration of is, 2.9 M
