A.glucose is a product of the photosynthesis reaction carbon dioxide is a reactant.
Answer:
The empirical formula for the compound is Na2O
Explanation:
Data obtained from the question include:
Sodium (Na) = 74.2g
Oxygen (O) = 25.8g
We can obtain the empirical formula for the compound as follow:
First, divide the above by their individual molar mass as shown below:
Na = 74.2/23 = 3.226
O = 25.8/16 = 1.613
Next, divide the above by the smallest number
Na = 3.226/1.613 = 2
O = 1.613/1.613 = 1
Therefore, the empirical formula is:
Na2O
Answer:
The equation: (NH₄)₂SO₄ = 2NH4(+) + SO4(-2)
The number of moles = 5 g / 132.14 g/mol = 0.038 mol
The number of molecules = 0.038 X 6.022x10^23 = 2.29x10^23
the number of positive ions present in the ammonium sulphate solution:
2 positive ions for every 1 molecule of (NH₄)₂SO₄
so 2 x 2.29x10^23 = 4.58x10^23
the number of negative ions present in the ammonium sulphate solution
1 negative ion for every 1 molecule of (NH₄)₂SO₄
so 1 x 2.29x10^23 = 2.29x10^23
the total number of ions present in the ammonium sulphate solution
4.58x10^23 + 2.29x10^23 = 6.87x10^23
Answer:
HOFO = (0, 0, +1, -1)
Explanation:
The formal charge (FC) can be calculated using the following equation:
<u>Where:</u>
V: are the valence electrons
N: are the nonbonding electrons
B: are the bonding electrons
The arrange of the atoms in the oxyacid is:
H - O₁ - F - O₂
Hence, the formal charge (FC) on each of the atoms is:
H: FC = 1 - 0 - 1/2*(2) = 0
O₁: FC = 6 - 4 - 1/2*(4) = 0
F: FC = 7 - 4 - 1/2*(4) = +1
O₂: FC = 6 - 6 - 1/2*(2) = -1
We can see that the negative charge is in the oxygen instead of the most electronegative element, which is the F. This oxyacid is atypical.
I hope it helps you!
6.0m(mol/kg) of HCl
125mL H2O = 0.125kg
6mol/kg = n mol/0.125kg, n = 0.75mol
When 0.75mol of HCl reacts, 0.75/2=0.375mol of H2 is produced. H2 = 2g/mol
So, 0.375mol H2 = 0.75g
