Answer:
-4.59°C
Explanation:
Let's see the formula for freezing point depression.
ΔT = Kf . m . i
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Kf = Freezing constant. For water if 1.86°C/m
m = molality (moles of solute in 1kg of solvent)
i = Van't Hoff factor.
0°C - Freezing T° of solution = 1.86°C /m . 1.30m . 1.9
Freezing T° of solution = - (1.86°C /m . 1.30m . 1.9)
Freezing T° of solution = - (1.86°C/m . 1.30m . 1.9) → -4.59°C
NaCl → Na⁺ + Cl⁻
The quantity that you would measure in kg, is mass.
Answer:
3120.75J
Explanation:
So, we have the formula
. For this example, q is the heat energy in Joules, m is the mass in grams, c is the specific heat capacity in
, and
t is the change in temperature. In this case, m = 47.5g, c = 0.9
, and
t = 94-21 = 73°C. Plugging in the values, we get the joules of heat required to raise 47.5g of Al from 21°C to 94°C which is stated above. You can double check my answer but that should be it. An important thing to be aware of are the units. Sometimes, the heat capacity may not be
. I may be in Kelvin or something. Anyways, hope this helps.
Answer:
The percentage of Lake Parsons which has evaporated since it became isolated is 68.24%.
Explanation:
The concentration of salt dissolved in Lake Parson = 21.0 g/L
= In 1 L of lake Parson water = 21.0 g of salt
The concentration of salt in several nearby non-isolated lakes = 6.67 g/L
The salt concentration in Lake Parsons before it became isolated = 6.67 g/L
In 1 L of lake water = 6.67 g of salt
If 1 L of lake Parson water has 6.67 g of salt. Then 21.0 grams of salt will be in ;

Water evaporated = 3.15 L - 1 L = 2.15 L
The percentage of Lake Parsons which has evaporated since it became isolated:

Hey there!
B + O₂ → B₂O₃
Balance O.
2 on the left, 3 on the right. Add a coefficient of 3 in front of O₂ and a coefficient of 2 in front of B₂O₃.
B + 3O₂ → 2B₂O₃
Balance B.
1 on the left, 4 on the right. Add a coefficient of 4 in front of B.
4B + 3O₂ → 2B₂O₃
Our final balanced equation: 4B + 3O₂ → 2B₂O₃
Hope this helps!