Answer:
6.67 moles
Explanation:
Given that:-
Moles of hydrogen gas produced = 10.0 moles
According the reaction shown below:-
3 moles of hydrogen gas are produced when 2 moles of aluminium undergoes reaction.
Also,
1 mole of hydrogen gas are produced when 
moles of aluminium undergoes reaction.
So,
10.0 moles of hydrogen gas are produced when 
moles of aluminium undergoes reaction.
<u>Moles of Al needed = moles = 6.67 moles</u>
Answer: c = 710 J/kg°C or 0.71 J/g°C
Explanation: Heat is expressed in the formula Q = mc∆T. Derive to find the specific heat c. So the formula will become c = Q / m∆T
c = Q / m∆T
= 42600 J / 2 kg ( 55°C - 25°C )
= 710 J /kg°C
Or can be expressed by converting kg to g.
c = 0.71 J /g°C
Answer:
Explanation:
<u>1) Data:</u>
a) V = 93.90 ml
b) T = 28°C
c) P₁ = 744 mmHg
d) P₂ = 28.25 mmHg
d) n = ?
<u>2) Conversion of units</u>
a) V = 93.90 ml × 1.000 liter / 1,000 ml = 0.09390 liter
b) T = 28°C = 28 + 273.15 K = 301.15 K
c) P₁ = 744 mmHg × 1 atm / 760 mmHg = 0.9789 atm
d) P₂ = 28.5 mmHg × 1 atm / 760 mmHg = 0.0375 atm
<u>3) Chemical principles and formulae</u>
a) The total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas. Hence, the partical pressure of the hydrogen gas collected is equal to the total pressure less the vapor pressure of water.
b) Ideal gas equation: pV = nRT
<u>4) Solution:</u>
a) Partial pressure of hydrogen gas: 0.9789 atm - 0.0375 atm = 0.9414 atm
b) Moles of hygrogen gas:
pV = nRT ⇒ n = pV / (RT) =
n = (0.9414 atm × 0.09390 liter) / (0.0821 atm-liter /K-mol × 301.15K) =
n = 0.00358 mol (which is rounded to 3 significant figures) ← answer
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Answer:
= 15.51 mL
Explanation:
Here's is the reaction:
2HgO(s) ⇒ 2 Hg(s)+O₂(g)
In this reaction 2mol HgO = 1mol O₂
The molecular weight of HgO = 216.59g
so, 3.0g HgO = 3.0g x 1.00molHgO/216.59gHgO
= 0.0138511 molHgO
The amount of Oxygen follows:
0.0138511 molHgOx1/2= 0.00692555 mol O₂
Now, volume of 1 any gas = 22400mL
so, 0.00692555 mol O₂ x22400mLO₂/1mol O₂
= 15.513232mL O₂
