Latent heat of melting is the energy that a solid absorbs to change its phase as its liquid. During this process, since all energy is used to change the phase, the temperature is constant.
Here the latent energy of melting for 1 g of ice is 80 calories and that 1 g of ice only absorbed 60 calories. hence the phase is not changed because it requires more 20 calories to melt.
Hence 1 g of ice remains as its solid phase (ice).
The percentage of yield was 777.78%
<u>Explanation:</u>
We have the equation,
Be
[s] + 2
HCl
[aq] → BeCl
2(aq] +
H
2(g] ↑ Be
(s] +
2
HCl
[aq] → BeCl
2(aq] +
H
2(g]
↑
To find the percent yield we have the formula
Percentage of Yield= what you actually get/ what you should theoretically get x 100
=3.5 g/0.45 g 100
= 777.78 %
The percentage of yield was 777.78%
Answer:
C)52g KCl in 100g water at 80°C
Explanation:
A saturated solution is one that contains as much solute as it can dissolve in the presence of excess solute at that particular temperature.
A solutibility curve is a graph that shows the variability with temperature of the solubility of a solute in a given solvent. A solutibility curve can provide information of whether a solution formed frommthe solute and solvent are saturated or not at a given temperature.
From the solubility curve in the attachment below:
A) A saturated solution of NH₄Cl will contain about 52 g solute per 100 g sat 50 °C. Thus, a solution of 40 g NH₄Cl in 100 g water at 50 °C is an unsaturated solution.
B) A saturated solution of SO₂ at 10°C will contain about 70 g of solute in 100 g of water. Thus a solution of 2g SO₂ in 100g water at 10°C is an unsaturated solution.
C) A saturated solution of KCl at 80 °C will contain about 52 g of solute in 100 g of water. Thus, a solution of 52g KCl in 100g water at 80°C is a saturated solution.
D) A saturated solution of Kl at 20 °C will contain about 145 g of solute in 100 g of water. Thus, a solution of 120g KI in 100g water at 20°C is an unsaturated solution.
Carbon carbon triple bonds
Answer:
0.006 48 km/s
Explanation:
1. Convert miles to kilometres
14.5 mi × (1.609 km/1 mi) = 23.33 km
2. Convert hours to seconds
1 h × (60 min/1h) × (60 s/1 min) = 3600 s
3. Divide the distance by the time
14.5 mi/1 h = 23.3 km/3600 s = 0.006 48 km/s