Answer:
Molecular formula for the gas is: C₄H₁₀
Explanation:
Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g
At STP → 1 atm and 273.15K
1 atm . 0.0336 L = n . 0.082 . 273.15 K
n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)
n = 1.500 × 10⁻³ moles
Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m
Now we propose rules of three:
If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C
58 g of gas (1mol) would have:
(58 g . 0.480) / 0.580 = 48 g of C
(58 g . 0.100) / 0.580 = 10 g of H
48 g of C / 12 g/mol = 4 mol
10 g of H / 1g/mol = 10 moles
Answer:
D. 1,950 g
Explanation:
I completed the test and it is the correct answer!
First compute the number of grams of manganese metal required to make 21.7 grams of H2.
21.7 g H2 x (1 mole H2/ 2 g H2) x (1 mole Mn/1 mol H2) x (55 grams Mn/1 mol Mn) = 596.75 grams
Now density = mass/volume
7.43 = 596.75/volume
volume = 596.75/7.43 = 80.31 mL
80.31 mL is the amount of manganese needed.
For part 1, just copy them off of the periodic table. For example, element 1 is Hydrogen, and its symbol is H
Elements on the left usually lose electrons and elements on the right tend to gain them. Noble gases have no charge.
