+4 and +6 I think not sure tho
Answer is: A. 1.1 3 1023 NiCl2 formula units.
m(NiCl₂) = 24.6 g; mass of nickel(II) chloride.
M(NiCl₂) = 129.6 g/mol; molar mass of nickel(II) chloride.
n(NiCl₂) = m(NiCl₂) ÷ M(NiCl₂).
n(NiCl₂) = 24.6 g ÷ 129.6 g/mol.
n(NiCl₂) = 0.19 mol; amount of nickel(II) chloride.
Na = 6.022·10²³ 1/mol; Avogadro constant.
N(NiCl₂) = n(NiCl₂) · Na.
N(NiCl₂) = 0.19 mol · 6.022·10²³ 1/mol.
N(NiCl₂) = 1.13·10²³; number of formula units.
Answer:
44 g oxygen are needed.
Explanation:
Given data:
Mass of oxygen needed = ?
Mass of ammonia = 18.2 g
Solution:
Chemical equation:
4NH₃ + 5O₂ → 4NO + 6H₂O
Now we will calculate the number of moles of ammonia:
Number of moles = mass/molar mass
Number of moles = 18.2 g/ 17 g/mol
Number of moles = 1.1 mol
Now we will compare the moles of ammonia with oxygen from balance chemical equation.
NH₃ : O₂
4 : 5
1.1 : 5/4×1.1 = 1.375 mol
Mass of oxygen needed:
Mass = number of moles × molar mass
Mass = 1.375 mol × 32 g/mol
Mass = 44 g
Answer:
The answer to your question is C₂HO₃
Explanation:
Data
Hydrogen = 3.25%
Carbon = 19.36%
Oxygen = 77.39%
Process
1.- Write the percent as grams
Hydrogen = 3.25 g
Carbon = 19.36 g
Oxygen = 77.39 g
2.- Convert the grams to moles
1 g of H ----------------- 1 mol
3,25 g of H ------------- x
x = (3.25 x 1) / 1
x = 3.25 moles
12 g of C ---------------- 1 mol
19.36 g of C ---------- x
x = (19.36 x 1) / 12
x = 1.61 moles
16g of O --------------- 1 mol
77.39 g of O --------- x
x = (77.39 x 1)/16
x = 4.83
3.- Divide by the lowest number of moles
Carbon = 3.25/1.61 = 2
Hydrogen = 1.61/1.61 = 1
Oxygen = 4.83/1.61 = 3
4.- Write the empirical formula
C₂HO₃
