second compound
Let molar mass of x is = X
Let molar mass of y is = Y
Moles of x in second compound = Mass / molar mass = 7 / X
Moles of y in second compound = Mass / molar mass = 4.5 / Y
For second compound
7 / X : 4.5/ Y = 1:1
Therefore
X / Y = 7/4.5
Y / X = 4.5/ 7
The mass of x in first compound = 14g
moles of x in first compound = 14/X
Mass of y in first compound = 3
moles of y in first compound = 3 / Y
14 / X : 3/ Y = 14Y / 3X = 14 X 4.5 / 3 X 7 = 3 :1
Thus molar ratio in first compound = moles of x / Moles of y = 3:2
Formula = x3y
Answer:
1.6 grams
Explanation:
We need to prepare 100 mL (0.100 L) of a 0.10 M CuSO₄ solution. The required moles of CuSO₄ are:
0.100 L × 0.10 mol/L = 0.010 mol
The molar mass of CuSO₄ is 159.61 g/mol. The mass corresponding to 0.010 moles is:
0.010 mol × (159.61 g/mol) = 1.6 g
We should use 1.6 grams of CuSO₄.
It is an element. Aluminun foil is aluminum prepared in thin leaves.
You are correct, but you needn't worry about the signs so much. Just remember that the negative sign is used to denote a loss of energy; since the water is hotter, it will be losing energy (-Q) and the iron will gain energy (Q). Now, we substitute the values:
-149.3 * 4.184 * (T - 95) = 412 * 0.44 * (T - 5)
Solving this equation for T,
T = 74.8 °C