Answer:
1.63ₓ10⁻⁶ g of U
139.03 g of H
0.385 g of O
141.8 g of Pb
Explanation:
In first place, we need to convert the number of atoms to moles, as we know that 1 mol of anything occupies 6.02×10²³ particles
Therefore:
4.12×10¹⁵ atoms of U . 1 mol / 6.02×10²³ atoms = 6.84×10⁻⁹ moles of U
8.37×10²⁵ atoms of H . 1 mol /6.02×10²³ atoms = 139.03 moles of H
1.45×10²² atoms of O . 1 mol /6.02×10²³ atoms = 0.0241 moles of O
4.12×10²³ atoms of Pb . 1 mol /6.02×10²³ atoms = 0.684 moles of Pb
Moles . Molar mass = Mass (g)
6.84×10⁻⁹ moles of U . 238.03 g/mol = 1.63ₓ10⁻⁶ g of U
139.03 moles of H . 1 g/mol = 139.03 g of H
0.0241 moles of O . 16 g/mol = 0.385 g of O
0.684 moles of Pb . 207.2 g/mol = 141.8 g of Pb
Explanation:
What will the question be ?
Answer:
The pH value of the mixture will be 7.00
Explanation:
Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,
![pH=pK_{a} + log(\frac{[Base]}{[Acid]})](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%20%2B%20log%28%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%29)
According to the given conditions, the equation will become as follow
![pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%20%2B%20log%28%5Cfrac%7B%5BNa_%7B2%7DHPO_%7B4%7D%20%5D%7D%7B%5BNaH_%7B2%7DPO_%7B4%7D%5D%7D%29)
The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.
Placing all the given data we obtain,
Answer:
Pb3O4
Explanation:
According to this question, 3.425g of lead oxide was reduced to form 3.105g of lead in an experiment. Since lead oxide contains both lead (Pb) and oxygen (O) element,
Mass of lead oxide = 3.425g
Mass of lead = 3.105g
Mass of oxygen = (3.425g - 3.105g) = 0.320g
Next, we convert each mass value to mole by dividing by respective molar mass
Pb = 3.105g ÷ 207.2 = 0.0149mol
O = 0.320g ÷ 16 = 0.02mol
Next, we divide each mole value by the smallest (0.0149)
Pb = 0.0149mol ÷ 0.0149mol = 1
O = 0.02mol ÷ 0.0149mol = 1.342
Multiply each ratio value by 3 to get:
Pb = 1 × 3 = 3
O = 1.342 × 3 = 4.026
The whole number ratio, approximately, of Pb and O is 3:4, hence, their empirical formula is Pb3O4.
