Answer:
Not sure what the answer is
Explanation:
I did this a while ago and dont remember sorry
Answer:
<em>3.27·10²³ atoms of O</em>
Explanation:
To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.
The chemical formula for sodium sulfate is <em>Na₂SO₄, </em>and its molar mass is approximately 142.05
.
We will use stoichiometry to convert from our mass of <em>Na₂SO₄ </em>to moles of <em>Na₂SO₄</em>, and then from moles of <em>Na₂SO₄ </em>to moles of <em>O </em>using the mole ratio; then finally, we will convert from moles of <em>O </em>to atoms of <em>O </em>using Avogadro's constant.
19.3g <em>Na₂SO₄</em> · 
· 
·
After doing the math for this dimensional analysis, you should get a quantity of approximately <em>3.27·10²³ atoms of O</em>.
Answer:
1.822 g of magnesium hydroxide would be produced.
Explanation:
Balanced reaction: 
Compound Molar mass (g/mol)
NaOH 39.997
95.211
58.3197
So, 2.50 g of NaOH = 
mol of NaOH = 0.0625 mol of NaOH
4.30 g of = 
mol of = 0.0452 mol of
According to balanced equation-
2 mol of NaOH produce 1 mol of
So, 0.0625 mol of NaOH produce 
mol of NaOH or 0.03125 mol of NaOH
1 mol of produces 1 mol of
So, 0.0452 mol of produce 0.0452 mol of
As least number of moles of are produced from NaOH therefore NaOH is the limiting reagent.
So, amount of would be produced = 0.03125 mol
= 
g
= 1.822 g
Answer:
5
Explanation:
Given that the formula is;
1/λ= R(1/nf^2 - 1/ni^2)
λ = 93.7 nm or 93.7 * 10^-9 m
R= 1.097 * 10^7 m-1
nf = ?
ni = 1
From;
ΔE = hc/λ
ΔE = 6.63 * 10^-34 * 3* 10^8/93.7 * 10^-9
ΔE = 21 * 10^-19 J
ΔE = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)
21 * 10^-19 J = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)
21 * 10^-19/-2.18 * 10^-18 = (1/nf^2 - 1/1^2)
-0.963 = (1/nf^2 - 1)
-0.963 + 1 = 1/nf^2
0.037 = 1/nf^2
nf^2 = (0.037)^-1
nf^2 = 27
nf = 5
Q1)
the number of moles can be calculated as follows
number of moles = mass present / molar mass
number of moles is the amount of substance.
4.8 g of Ca was added therefore mass present of Ca is 4.8 g
molar mass of Ca is 40 g/mol
molar mass is the mass of 1 mol of Ca
therefore if we substitute these values in the equation
number of moles of Ca = 4.8 g / 40 g/mol = 0.12 mol
0.12 mol of Ca is present
q2)
next we are asked to calculate the number of moles of water present
again we can use the same equation to find the number of moles of water
number of moles = mass present / molar mass
3.6 g of water is present
sum of the products of the molar masses of the individual elements by the number of atoms
H - 1 g/mol and O - 16 g/mol
molar mass of water = (1 g/mol x 2 ) + 16 g/mol = 18 g/mol
molar mass of H₂O is 18 g/mol
therefore number of moles of water = 3.6 g / 18 g/mol = 0.2 mol
0.2 mol of water is present
