Answer:
158.35 K
Explanation:
Using the general gas law
P₁V₁/T₁ = P₂V₂/T₂ where P₁ = 150.4 kPa, V₁ = 12.2 L, T₁ = 22.7 °C = 273 + 22.7 = 295.7 K, P₂ = 101.3 kPa, V₂ = 9.7 L and T₂ = unknown.
Making T₂ subject of the formula, we have
T₂ = P₂V₂T₁/P₁V₁
Substituting the values of the variables into the equation, we have
T₂ = P₂V₂T₁/P₁V₁
T₂ = 101.3 kPa × 9.7 L × 295.7 K/150.4 kPa × 12.2 L
T₂ = 290,557.777 K/1,834.88
T₂ = 158.35 K
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Answer : The molar heat of solution of KBr is 19.9 kJ/mol
Explanation :
Mass of KBr = 7.00 g
Molar mass of KBr = 119 g/mole
Heat capacity = 2.72 kJ/K
Change in temperature = 0.430 K
First we have to calculate the moles of KBr.
Now we have to calculate the heat of the reaction.
where,
q = amount of heat = ?
= heat capacity =
= change in temperature = 0.430 K
Now put all the given values in the above formula, we get:
Now we have to calculate the molar heat of solution of KBr.
where,
n = number of moles of KBr
Therefore, the molar heat of solution of KBr is 19.9 kJ/mol
Answer:
The answer is 5.10
Explanation:
<h3><u>Given</u>;</h3>
<h3>
<u>To </u><u>Find</u>;</h3>
We know that
pH + pOH = 7
pOH = 7 – pH
pOH = 7 – 1.90
pOH = 5.10
Thus, The pOH of the solution is 5.10