Answer:
2, strong acid
Explanation:
Data obtained from the question. This includes:
[H+] = 0.01 M
pH =?
pH of a solution can be obtained by using the following formula:
pH = –Log [H+]
pH = –Log 0.01
pH = 2
The pH of a solution ranging between 0 and 6 is declared to be an acid solution. The smaller the pH value, the stronger the acid.
Since the pH of the above solution is 2, it means the solution is a strong acid.
D-number of electrons in the outer energy level.
Answer:
a) 231.9 °C
b) 100% Sn
c) 327.5 °C
d) 100% Pb
Explanation:
This is a mixture of two solids with different fusion point:


<u>Given that Sn has a lower fusion temperature it will start to melt first at that temperature. </u>
So the first liquid phase forms at 231.9 °C and because Pb starts melting at a higher temperature, that phase's composition will be 100% Sn.
The mixture will be completely melted when you are a the higher melting temperature of all components (in this case Pb), so it will all in liquid phase at 327.5 °C.
At that temperature all Sn was already in liquid state and, therefore, the last solid's composition will be 100% Pb.
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
Answer : The concentration of NaOH is, 0.336 M
Explanation:
To calculate the concentration of base, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:


Thus, the concentration of NaOH is, 0.336 M