It is known that 1 mol of a substance is equal to 6.02x10^3 particles. This is called the Avogadro's number. From this knowledge, we convert the mass of the substances given above into units of moles by using the molar mass.
12.01 g C ( 1 mol / <span>12.01 g) = 1 mol
44.08 g SiO2 ( </span>1 mol / <span>60.08 g ) = 0.73 mol
16 g of O3 (</span>1 mol / <span>48 g) = 0.33 mol
16.04 g CH4 (</span>1 mol / <span>16.05 g ) = 0.99 mol
So, the answers are carbon and </span><span>carbon tetrahydride</span>.
Product:
C: 1
O:3
H: 2
reactant:
C: 8
H: 18
O: 2
first you need to put probably an eight in front of the Carbon.
Next, add a coefficient of 9 to Hydrogen
finally, add a 3 INSTEAD of a 2 to Oxygen
should look like:
C8H18+O3>C8O2+H9O
Answer:
Mass = 713.4 ×10⁻⁴ g
Explanation:
Given data:
Number of moles of calcium phosphate = 2.3×10⁻⁴ mol
Mass of calcium phosphate = ?
Solution:
Formula:
Number of moles = mass/molar mass
Molar mass of calcium phosphate is 310.18 g/mol
by putting values,
2.3×10⁻⁴ mol = mass / 310.18 g/mol
Mass = 2.3×10⁻⁴ mol × 310.18 g/mol
Mass = 713.4 ×10⁻⁴ g
Answer:
A. C₃H₄N
Explanation:
- Firstly, we need to calculate the no. of moles of C, H, and N using the relation:
<em>no. of moles = mass/molar mass.</em>
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∴ no. of moles of C = mass/molar mass = (90.0 g)/(12.0 g/mol) = 7.5 mol.
∴ no. of moles of H = mass/molar mass = (11.0 g)/(1.0 g/mol) = 11.0 mol.
∴ no. of moles of N = mass/molar mass = (35.0 g)/(14.0 g/mol) = 2.5 mol.
- We should get the mole ratio of each atom by dividing by the lowest no. of moles (2.5 mol of N).
∴ the mole ratio of C: H: N = (7.5 mol/2.5 mol): (11.0 mol/2.5 mol): (2.5 mol/2.5 mol) = (3: 4.4: 1) ≅ (3: 4: 1).
- So, the empirical formula is: A. C₃H₄N.
Answer:
Molar solubility of AgBr = 51.33 × 10⁻¹³
Explanation:
Given:
Amount of NaBr = 0.150 M
Ksp (AgBr) = 7.7 × 10⁻¹³
Find:
Molar solubility of AgBr
Computation:
Molar solubility of AgBr = Ksp (AgBr) / Amount of NaBr
Molar solubility of AgBr = 7.7 × 10⁻¹³ / 0.150
Molar solubility of AgBr = 51.33 × 10⁻¹³
