Answer:
45.2%
Explanation:
To calculate the percent of an element in a compound we divide the molar mass of the element by the compound and multiply that by 100
First lets find the molar mass of Fluoride
Looking at the periodic table Fluoride has a molecular mass of 18.998 g
Now we need to find the molecular mass of NaF
Looking at a periodic table, Sodium (Na) has a molecular mass of 22.990g and Fluoride has a molecular mass of 18.998 so NaF has a molecular mass of 22.990(1) + 18.998(1) = 41.988g
Now we divide the mass of fluoride by the mass of sodium fluoride and multiply that by 100 to find the percentage of fluoride that is present in NaF
Mass of Fluoride = 18.998g
Mass of Sodium Fluoride = 41.988g
Percentage of fluoride present in NaF = (18.998g / 41.988g) * 100 = 45.2%
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Answer:
the particles start to spread out and cause it to get warmer
Explanation:
Answer #1 is "there is 2.5 grams of solute in every 100 g of solution."
We calculate for 2.5% by mass solution by dividing the mass of the solute by the mass of the solution and then multiply by 100.
Answer #2 is "that mass ratio would be 2.5/100 or 2.5 grams of solute/100 grams of solution."
We weigh out 2.5 grams of solute and then add 97.5 grams of solvent to make a total of 100 gram solution, that is,
mass of solute / mass of solution = 2.5g solute / (2.5g solute + 97.5g solvent)
= 2.5g solute / 100g solution
Answer#3 is "a solution mass of 1 kg is 10 times greater than 100 g, thus one kilogram (1 kg) of a 2.5% ki solution would contain 25 grams of ki."
We multiply 10 to each mass so that 100 grams becomes 1000grams since 1000 grams is equal to 1 kg:
mass of solute / mass of solution = 2.5g*10/[(2.5g*10) + (97.5g*10)]
= 25g solute/(25g solute + 975g solvent)
= 25g solute/1000g solution
= 25g solute/1kg solution
Answer:N
2
+ 3
H
2
-----> 2N
H
3
Explanation:
N
2
+
H
2
-----> N
H
3
Let us balance this equation by counting the number of atoms on both sides of the arrow.
N
2
+
H
2
-----> N
H
3
N=2 , H=2 N=1, H=3
To balance the number of N atom on Right Hand Side (RHS) , I will add one molecule of N
H
3
on RHS
N
2
+
H
2
-----> 2N
H
3
N=2 , H=2 N=2 , H= 6
To balance the number of H atoms on Left Hand Side (LHS) , I will add two molecules of
H
2
on LHS
N
2
+ 3
H
2
-----> 2N
H
3
N=2 , H=6 N=2 , H= 6
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