Answer:
Explanation:
L
=
1.10
L
of solution
Explanation:
The Molarity
M
is calculated by the equation comparing moles of solute to liters of solution
M
=
m
o
l
L
For this question we are given the Molarity 0.88M
We are told the solute is a 25.2 gram sample of LiF, Lithium Fluoride
We can convert the mass of LiF to moles by dividing by the molar mass of LiF
Li = 6.94
F = 19.0
LiF = 25.94 g/mole
25.2
g
r
a
m
s
x
1
m
o
l
25.94
g
r
a
m
s
=
0.97
moles
Now we can take the the molarity and the moles and calculate the Liters of solution
M
=
m
o
l
L
M
L
=
m
o
l
L
=
m
o
l
M
L
=
0.97
m
o
l
0.88
M
L
=
1.10
L
of solution i just did look at my papaer
Answer:
1. 25 moles water.
2. 41.2 grams of sodium hydroxide.
3. 0.25 grams of sugar.
4. 340.6 grams of ammonia.
5. 4.5x10²³ molecules of sulfur dioxide.
Explanation:
Hello!
In this case, since the mole-mass-particles relationships are studied by considering the Avogadro's number for the formula units and the molar mass for the mass of one mole of substance, we proceed as shown below:
1. Here, we use the Avogadro's number to obtain the moles in the given molecules of water:

2. Here, since the molar mass of NaOH is 40.00 g/mol, we obtain:

3. Here, since the molar mass of C6H12O6 is 180.15 g/mol:

4. Here, since the molar mass of ammonia is 17.03 g/mol:

5. Here, since the molar mass of SO2 is 64.06 g/mol:

Best regards!
Answer:
If one of the reactants is a solid, only the particles at the surface can partake in the reaction. Breaking the reactant into smaller pieces increases the surface and more particles are exposed to the reaction mixture. This results in an increased frequency of collisions and therefore a faster rate of reaction
Answer:
7.5 moles of CaBr2 are produced
Explanation:
Based on the equation:
2AlBr3 + 3CaO → Al2O3 + 3CaBr2
<em>2 moles of AlBr3 produce 3 moles of CaBr2 if CaO is in excess.</em>
<em />
Using this ratio: 2 moles AlBr3 / 3 moles CaBr2. 5 moles of AlBr3 produce:
5 moles AlBr3 * (3 moles CaBr2 / 2 moles AlBr3) =
<h3>7.5 moles of CaBr2 are produced</h3>
<em />