Answer:
Q = mcT ...you can either substitute the molar heat capacity of water in the place of c or the specific heat capacity of water.
Explanation:
Answer:
The final temperature of water is 54.5 °C.
Explanation:
Given data:
Energy transferred = 65 Kj
Mass of water = 450 g
Initial temperature = T1 = 20 °C
Final temperature= T2 = ?
Solution:
First of all we will convert the heat in Kj to joule.
1 Kj = 1000 j
65× 1000 = 65000 j
specific heat of water is 4.186 J /g. °C
Formula:
q = m × c × ΔT
ΔT = T2 - T1
Now we will put the values in Formula.
65000 j = 450 g × 4.186 J /g. °C × (T2 - 20°C )
65000 j = 1883.7 j /°C × (T2 - 20°C )
65000 j/ 1883.7 j /°C = T2 - 20°C
34.51 °C = T2 - 20°C
34.51 °C + 20 °C = T2
T2 = 54.5 °C
5.7% KCl is 94.3 % water.
Therefore, for 1000 g of water the mass of KCl will be (1000× 5.7)/94.3 = 60.445 grams.
1 mole of KCl is equal to 74.55 g,
therefore, 60.445 g will be 60.445/74.55 = 0.8108 mole of KCl
Hence, 0.8108 moles of KCl should release twice that number of moles 1.6216 moles ions.
Having 1.6216 moles of KCl ions dissolved in 1000g of water, gives us 1.6216 molar if solution.
Using the freezing point depression constant of water.
dT = Kf (molarity)
dT = (1.86 C/ molar) (1.6216 m)
dT = 3.016 C drop in freezing point
Therefore, it should freeze at - 3.016 Celsius
Answer:
NiO Does Not Dissolve In An Aqueous Solution Of NaNO3.
Explanation:
Answer:
Answers are in the explanation.
Explanation:
- The half‑life of A increases as the initial concentration of A decreases. order: <em>2. </em>In the half-life of second-order reactions, the half-life is inversely proportional to initial concentration.
- A three‑fold increase in the initial concentration of A leads to a nine‑fold increase in the initial rate. order: <em>2. </em>The rate law of second-order is: rate = k[A]²
- A three‑fold increase in the initial concentration of A leads to a 1.73‑fold increase in the initial rate. order: <em>1/2. </em>The rate law for this reaction is: rate = k √[A]
- The time required for [A] to decrease from [A]₀ to [A]₀/2 is equal to the time required for [A] to decrease from [A]₀/2 to [A]₀/4. order: <em>1. </em>The concentration-time equation for first-order reaction is: ln[A] = ln[A]₀ - kt. That means the [A] decreasing logarithmically.
- The rate of decrease of [A] is a constant. order: <em>0. </em>The rate law is: rate = k -<em>where k is a constant-</em>