16. What volume of water vapor gas would be produced from the combustion of 535.55 grams of propane (C3H8) with 2,665.06 grams o
1 answer:
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Answer:
80mg
Explanation:
Given parameters:
Density of Platinum metal = 21.4g/mL
Dimension of the metal = 3cm by 0.03m by 40mm
Unknown:
Weight of the metal in milligrams?
Solution:
Density is the mass per unit volume of a substance.
Mathematically;
Density =
Volume of the platinum metal = l x w x h
Let us convert the dimension of the metal to the same unit;
3cm x 0.03m x 40mm
convert to cm;
100m = 1cm
0.03m = = 0.0003cm
-- 40mm;
1mm = 0.1cm
40mm = 40 x 0.1 = 4cm
Volume of the metal = 3 x 0.0003 x 4 = 0.0036cm³
1mL = 1cm³; so volume = 0.0036mL
Mass of the metal = 21.4 x 0.0036 = 0.08g
1g = 1000mg
0.08g = 0.08 x 1000 = 80mg
Answer:
0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl
Explanation:
The balanced reaction is:
Mg + 2 HCl → MgCl₂ + H₂
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles react:
- Mg: 1 mole
- HCl: 2 moles
- MgCl₂: 1 mole
- H₂: 1 mole
Being:
- Mg: 24. 31 g/mole
- H: 1 g/mole
- Cl: 35.45 g/mole
the molar mass of the compounds participating in the reaction is:
- Mg: 24.31 g/mole
- HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
- MgCl₂: 24.31 g/mole + 2*35.45 g/mole= 95.21 g/mole
- H₂: 2*1 g/mole= 2 g/mole
Then, by stoichiometry of the reaction, the following quantities of mass participate in the reaction:
- Mg: 1 mole* 24.31 g/mole= 24.31 g
- HCl: 2 moles* 36.45 g/mole= 72.9 g
- MgCl₂: 1 mole* 95.21 g/mole= 95.21 g
- H₂: 1 mole* 2 g/mole= 2 g
Then you can apply the following rule of three: if by stoichiometry 24.31 grams of Mg form 1 mole of H₂, 3.25 grams of Mg how many moles of H₂ will they form?
moles of H₂= 0.134
<u><em>0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl</em></u>
Question:
A) 12
B) 29
C) 2.1 × 10⁻²
D) 8.7 × 10⁻²
E) 47
Answer:
The correct option is;
E) 47
Explanation:
Kc, which is the equilibrium constant of a chemical reaction is derived by finding the ratio between the product of the equilibrium concentration of the product raised to their respective coefficients to the product of the equilibrium concentration of the reactants also raised to their respective coefficients.
Here we have;
[H₂] = 0.14 M
[Cl₂] = 0.39 M
[HCl] = 1.6
The reaction is given as follows;
H₂ (g) + Cl₂ (g) ⇄ 2HCl (g)
The formula for Kc is given as follows;
Therefore, the Kc for the reaction is approximately equal to 47.