Answer:
0.15g
Explanation:
Given parameters:
Number of molecules of water = 1.2 x 10²¹ molecules
Unknown:
Mass of SnO₂ = ?
Solution:
To solve this problem, we have to work from the known to the unknown specie;
SnO₂ + 2H₂ → Sn + 2H₂O
Ensure that the equation given is balanced;
Now,
the known species is water;
6.02 x 10²³ molecules of water = 1 mole
1.2 x 10²¹ molecules of water = 
= 0.2 x 10⁻²moles
Number of moles of water = 0.002moles
From the balanced chemical equation:
2 mole of water is produced from 1 mole of SnO₂
0.002 moles of water will be produced from 
= 0.001moles
To find the mass;
Mass = number of moles x molar mass
Molar mass of SnO₂ = 118.7 + 2(16) = 150.7g/mol
Mass = 0.001 x 150.7 = 0.15g
1245 liters because you move the decimal place over to the right six times
Answer:
Percent yield = 89.1%
Explanation:
Based on the equation:
Cl₂ + 2KI → 2KCl + I₂
<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>
<em />
To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:
<em>Moles Cl₂:</em>
8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles
<em>Moles KI -Molar mass: 166.0028g/mol-</em>
25g * (1mol / 166.0028g) = 0.15 moles
Here, clarely, the KI is the limiting reactant
As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:
0.15 moles * (74.5513g / mol) =
11.2g KCl
Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100
<h3>Percent yield = 89.1%</h3>
Answer:
[Ca²⁺] = 1M
[NO₃⁻] = 2M
Explanation:
Calcium nitrate dissociates in water as follows:
Ca(NO₃)₂ ⇒ Ca²⁺ + 2NO₃⁻
The moles of Ca²⁺ can be found using the molar relationship between Ca(NO₃)₂ and Ca²⁺
(0.100mol Ca(NO₃)₂) (Ca²⁺ /Ca(NO₃)₂) = 0.100 mol Ca²⁺
The concentration of Ca²⁺ is then:
[Ca²⁺] = n/V = (0.100mol)/(100.0mL) x (1000ml)/(1L) = 1M
Similarly, moles of NO₃⁻ can be found using the molar relationship between Ca(NO₃)₂ and NO₃⁻:
(0.100mol Ca(NO₃)₂) (2NO₃⁻/Ca(NO₃)₂) = 0.200 mol NO₃⁻
The concentration of NO₃⁻ is then:
[NO₃⁻] = (0.200mol)/(100.0mL) x (1000ml)/(1L) = 2M
Answer:
3.02 X1023 atoms Ag limol. - - 0.50 1 moles. 6.02241023 atoms.
