Using the specific heat capacity formula:
q = mc ∆ t
60.0 J = (6g)(x)(11*C)
x = 0.9 J/g*C
Aluminum
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Answer:
m H2(g) = 2.241 g H2(g)
Explanation:
- 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)
limit reagent:
∴ Mw Al = 26.982 g/mol
∴ Mw H2SO4 = 98.0785 g/mol
⇒ n Al = (20 g Al)×(mol/26.982 g) = 0.7412 mol Al
⇒ n H2SO4 = ( 115 g H2SO4 )×(mol/98.0785 g) = 1.173 mol H2SO4
⇒ n H2 = (0.7412 mol Al)×(3 mol H2/ 2 mol Al) = 1.112 mol H2
∴ Mw H2 = 2.016 g/mol
⇒ g H2 = (1.112 mol H2)×(2.016 g/mol) = 2.241 g H2
Answer:
b. 3.66x10²³ atoms of chromium.
Explanation:
First we calculate how many moles are there in 31 grams of chromium, using its molar mass:
- Molar Mass of Chromium = 51 g/mol (This can be found on any periodic table)
- 31 g ÷ 51 g/mol = 0.608 mol
Then we <u>calculate how many atoms are there in 0.608 moles</u>, using <em>Avogadro's number</em>:
- 0.608 mol * 6.023x10²³ atoms/mol = 3.66x10²³ atoms
The correct answer is thus option b. 3.66x10²³ atoms of chromium.