I'm sure that to calculate the freezing point depression <span>subtract</span> solution's freezing point and the freezing point of it's pure solvent. According to the formula.
Answer:
c =0.2 J/g.°C
Explanation:
Given data:
Specific heat of material = ?
Mass of sample = 12 g
Heat absorbed = 48 J
Initial temperature = 20°C
Final temperature = 40°C
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 40°C -20°C
ΔT = 20°C
48 J = 12 g×c×20°C
48 J =240 g.°C×c
c = 48 J/240 g.°C
c =0.2 J/g.°C
If you'd like the full working, here it is:
I calculated this by using the formula triangle.
Mass
Number Formula
Of moles Mass
To calculate the number if moles in a substance, you need to divide the Mass by the Formula mass. You get the formula mass by adding the atomic masses of the elements in the compound together. In this situation, H2O, it would be two hydrogen molecules plus one oxygen molecule which is 2 + 16. This is because the atomic mass of Hydrogen is 1 and the atomic mass of Oxygen is 16.
Now that we have the Formula mass we can go ahead and do the calculation since we already have the Mass. You do as follows:
Mass divided by Formula mass which is in this case - 25 divided by 18
By doing this calculation you will get the answer which is 1.38 moles which can be rounded to 1.4
Hope this helps :)
Answer: i don't know
Explanation:
u gave no information on what you're asking
Moles of PF₃ : 4
<h3>Further explanation</h3>
A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.
Reaction
1.25 moles of P₄(s) is reacted with 6 moles of F₂(g)
Limiting reactant : the smallest ratio (mol divide by coefficient)
P₄ : F₂ =
mol PF₃ based on mol of limiting reactant(F₂), so mol PF₃ :
