Since your multiplying by the same number, 104, both terms, you can group them like this:
104(1.26+2.50)
104(3.76)
391.04
Knowing that scientific notation needs to be a number x, that’s greater or equal to 1, and less then 10.
3.9104 * 10^2
Answer:
Seven
Explanation:
The 3 means that you must multiply everything inside the parentheses by 3.
If you write the formula as AlO₃H₃, it is easier to count the atoms.
A:l 1 atom
O: 3
H: <u>3 atoms</u>
7 atoms
There are seven atoms in one formula unit of Al(OH)₃.
Answer: the density of the rock
55.91g/ml
Explanation:
The density of the rock after it had being placed in the cylinder of water so the calculation should look like this:
Volume of water substract the mass of the rock:
And that is 142.5 ml - 86.59g =
Answer 55.91 g/ml
So the density of the rock is 55.91g/ml
1) Chemical reaction:
2(CH₃COO)₃Fe + 3MgCrO₄ → Fe₂(CrO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgCrO₄) = 10,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol.
n(MgCrO₄) = m(MgCrO₄) ÷ M(MgCrO₄).
n(MgCrO₄) = 10 g ÷ 140,3 g/mol.
n(MgCrO₄) = 0,071 mol; limiting reagens.
From chemical reaction: n(MgCrO₄) : n((CH₃COO)₂Mg) = 3 : 3.
n((CH₃COO)₂Mg) = 0,071 mol.
m((CH₃COO)₂Mg) = 0,071 mol · 142,4 g/mol.
n((CH₃COO)₂Mg) = 10,11 g.
2) Chemical reaction:
2(CH₃COO)₃Fe + 3MgSO₄ → Fe₂(SO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgSO₄) = 15,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol; limiting ragens.
n(MgSO₄) = m(MgSO₄) ÷ M(MgSO₄).
n(MgSO₄) = 15 g ÷ 120,36 g/mol.
n(MgSO₄) = 0,125 mol; limiting reagens.
From chemical reaction: n(CH₃COO)₃Fe) : n((CH₃COO)₂Mg) = 2 : 3.
n((CH₃COO)₂Mg) = 0,064 mol · 3/2.
n((CH₃COO)₂Mg) = 0,096 mol.
m((CH₃COO)₂Mg) = 0,096 mol · 142,4 g/mol.
m((CH₃COO)₂Mg) = 13,7 g.
<span>Part A:
Molar mass luminol = 177g/mol
10g = 10/177 = 0.0565 mol Luminol
Dissolved in 75mL solution = 0.075L
Molarity = 0.0565/0.075 = 0.753M solution
Part B.
There should be no problem with this part If .0L contains 3.0*10^-2 mol Luminol, then 2.0L will contain 2*3.0*10^-2 = 6.0*10^-2 mol
Part C:
Use equation:
M1V1 = M2V2
0.753 *V1 = (3.0*10^-2) * 1000
V1 = 30/0.753
V1 = 39.8 mL of stock solution contains 3.0*10^-2 mol Luminol.</span>
