The volume of 0. 250 mole sample of 
gas occupy if it had a pressure of 1. 70 atm and a temperature of 35 °C is 3.71 L.
Calculation,
According to ideal gas equation which is known as ideal gas law,
PV =n RT
- P is the pressure of the hydrogen gas = 1.7 atm
- Vis the volume of the hydrogen gas = ?
- n is the number of the hydrogen gas = 0.25 mole
- R is the universal gas constant = 0.082 atm L/mole K
- T is the temperature of the sample = 35°C = 35 + 273 = 308 K
By putting all the values of the given data like pressure temperature universal gas constant and number of moles in equation (i) we get ,
1.7 atm×V = 0.25 mole ×0.082 × 208 K
V = 0.25 mole ×0.082atm L /mole K × 308 K /1.7 atm
V = 3.71 L
So, volume of the sample of the hydrogen gas occupy is 3.71 L.
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Answer:
The answer to your question is: letter D
Explanation:
In a combustion reaction, the reactants are always a molecule with Carbon that reacts with oxygen and the products are carbon dioxide and water.
According to the explanation, the only possible solution is:
a) C₆H₁₂O₂(l) ⇒ 6 C(s) + 6 H₂(g) + O₂(g)
b) Mg(s) + C₆H₁₂O₂(l) ⇒ MgC₆H₁₂O₂(aq)
c) 6 C(s) + 6 H₂(g) + O₂(g) ⇒ C₆H₁₂O₂(l)
d) C₆H₁₂O₂(l) + 8 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(g)
e) None of the above represent the combustion of C₆H₁₂O₂.
I'm not so sure but I would say Answer Choice B
Answer:
We need 3910.5 joules of energy
Explanation:
Step 1: Data given
Mass of aluminium = 110 grams
Initial temperature = 52.0 °C
Final temperature = 91.5 °C
Specific heat of aluminium = 0.900 J/g°C
Step 2: Calculate energy required
Q = m*c*ΔT
⇒with Q = the energy required = TO BE DETERMINED
⇒with m = the mass of aluminium = 110 grams
⇒with c = the specific heat of aluminium = 0.900 J/g°C
⇒with ΔT = the change in temperature = T2 - T1 = 91.5 °C - 52.0 °C = 39.5 °C
Q = 110 grams * 0.900 J/g°C * 39.5
Q = 3910.5 J
We need 3910.5 joules of energy
