<h3>
Answer:</h3>
0.90J/g°C
<h3>
Explanation:</h3>
We are given:
Mass of Aluminium = 10 g
Quantity of heat = 677 Joules
Change in temperature = 125°C - 50°C
= 75°C
We are required to calculate the specific heat capacity of Aluminium
But, Quantity of heat = Mass × specific heat × Change in temperature
Q = mcΔt
Rearranging the formula;
c = Q ÷ mΔt
= 677 J ÷ (10 g × 75°C)
= 677 J ÷ 750g°C
= 0.903 J/g°C
= 0.90J/g°C
Thus, the specific heat capacity of Aluminium is 0.90J/g°C
They flow through the inslation of the house .
they both help things travel
Answer:
0.0100M of AgNO3 contains 0.0100M of Ag+
Explanation:
AgNO3 when ionized yields Ag+ and NO3-. This means that the amount of AgNO3 in solution is equivalent to the amount of Ag+ and NO3- in that same solution.
1M of AgNO3 solution produces 1M of Ag+
1M of AgNO3 solution produces 1M of NO3-
This occurs because of the complete ionization of AgNO3 in solution, allowing complete dissolution of the compound.
Empirical formula is the simplest ratio of whole numbers of components in a compound.
Assuming for 100 g of the compound
Cu As S
mass 48.41 g 19.02 g 32.57 g
number of moles 48.41 / 63.5 g/mol 19.02 / 75 g/mol 32.57 / 32 g/mol
= 0.762 mol = 0.2536 mol = 1.018 mol
divide by the least number of moles
0.762 / 0.2536 0.2536 / 0.2536 1.018 / 0.2536
= 3.00 = 1.00 = 4.01
once they are rounded off
Cu - 3
As - 1
S - 4
therefore empirical formula is Cu₃AsS₄