To create the liquid and superfluid states you cool down helium gas to a few degrees above absolute zero
Answer:
copper has 2+ ions in a sample of copper
The grams that would be produced from 7.70 g of butanoic acid and excess ethanol is 7.923grams
calculation
Step 1: write the chemical equation for the reaction
CH3CH2CH2COOH + CH3CH2OH → CH3CH2CH2COOCH2CH3 +H2O
step 2: find the moles of butanoic acid
moles= mass/ molar mass
= 7.70 g/ 88 g/mol=0.0875 moles
Step 3: use the mole ratio to determine the moles of ethyl butyrate
moles ratio of CH3CH2CH2COOH :CH3CH2CH2COOCH2CH3 is 1:1 therefore the moles of CH3CH2CH2COOCH2CH3 = 0.0875 x78/100=0.0683moles
step 4: find mass = moles x molar mass
= 0.0683 moles x116 g/mol=7.923grams
Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C
