Krypton-84 is the most abundant type of Krypton. The answer is C.
If you need this to be explained, I will do my best it's a bit difficult to say how to find it out, but I will if you need me to.
Answer:
Explanation:
Given
485 L
Required
Determine the measurement not equal to 485L
<em>From standard unit of conversion;</em>
1 KL = 1000 L
<em>Multiply both sides by 485</em>
<em>Divide both sides by 1000</em>
---- This is equivalent
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<em>From standard unit of conversion;</em>
1000 mL = 1 L
<em>Multiply both sides by 485</em>
Convert to standard form
Hence; is not equivalent to 485L
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<em>From standard unit of conversion;</em>
100 cL = 1 L
<em>Multiply both sides by 485</em>
Convert to standard form
---- This is equivalent
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μL
<em>From standard unit of conversion;</em>
1000000 μL = 1 L
<em>Multiply both sides by 485</em>
Convert to standard form
---- This is equivalent
From the list of given options;
is not equivalent to 485L
Symbiosis describes close<span> interactions between two or more different species. There are four main types of symbiotic relationships: mutualism, commensalism, parasitism and competition.</span>
...because this notation increases the convenience in using the numbers
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
