I think the answer is A but I could be wrong
L = 0 to n-1. Hence, it can have values 0, 1, 2.
<span>m = -l to +l. Hence, it can have values -2, -1, 0, 1, 2 </span>
<span>s = +1/2 and -1/2</span>
Answer:
The molarity of the solution: 0,27M
Explanation:
First , we calculate the weight of 1 mol of NaCl:
Weight 1mol NaCl= Weight Na + Weight Cl= 23 g+ 35, 5 g= 58, 5 g/mol
58,5 g---1 mol NaCl
64 g--------x= (64 g x1 mol NaCl)/58,5 g= 1, 09 mol NaCl
A solution molar--> moles of solute in 1 L of solution:
4 L-----1,09 mol NaCl
1L----x0( 1L x1,09 mol NaCl)/4L =0,27moles NaCl--->0,27M
Answer:Phase changes require either the addition of heat energy (melting, evaporation, and sublimation) or subtraction of heat energy (condensation and freezing). ... Changing the amount of heat energy usually causes a temperature change.
Explanation:
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