The objects mass I took physical science
Answer:
my answer is b oti d I am not sure
Answer:
25 mL
Explanation:
Step 1: Given data
- Concentration of the concentrated solution (C₁): 2 M
- Volume of the concentrated solution (V₁): ?
- Concentration of the diluted solution (C₂): 0.1 M
- Volume of the diluted solution (V₂): 0.500 L
Step 2: Calculate the volume of the concentrated NaCl solution
We will use the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.1 M × 0.500 L / 2 M
V₁ = 0.025 L = 25 mL
Answer:
36.55kJ/mol
Explanation:
The heat of solution is the change in heat when the KNO3 dissolves in water:
KNO3(aq) → K+(aq) + NO3-(aq)
As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.
To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:
<em>Moles KNO3 -Molar mass: 101.1032g/mol-</em>
10.6g * (1mol/101.1032g) = 0.1048 moles KNO3
<em>Change in heat:</em>
q = m*S*ΔT
<em>Where q is heat in J,</em>
<em>m is the mass of the solution: 10.6g + 251.0g = 261.6g</em>
S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-
And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C
q = 261.6g*4.184J/g°C*3.5°C
q = 3830.87J
<em>Molar heat of solution:</em>
3830.87J/0.1048 moles KNO3 =
36554J/mol =
<h3>36.55kJ/mol</h3>
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<u>Answer:</u> The correct statement is low temperature only, because entropy decreases during freezing.
<u>Explanation:</u>
The relationship between Gibb's free energy, enthalpy, entropy and temperature is given by the equation:
Where,
= change in Gibb's free energy
= change in enthalpy
T = temperature
= change in entropy
It is given that freezing of methane is taking place, which means that entropy is decreasing and 
is becoming negative. It is also given that the reaction is an exothermic reaction, this means that the is also negative.
For a reaction to be spontaneous, must be negative.
![-ve=-ve-[T(-ve)]\\\\-ve=-ve+T](https://tex.z-dn.net/?f=-ve%3D-ve-%5BT%28-ve%29%5D%5C%5C%5C%5C-ve%3D-ve%2BT)
From above equations, it is visible that will be negative only when the temperature will be low.
Hence, the correct statement is low temperature only, because entropy decreases during freezing.
