We do a heat balance to solve this:
(m cp ΔT)water = -(m cp ΔT)metal
100.8 (4.18) (27 - 22) = -65 (cp)(27-100)
cp = 100.8 (4.18) (27 - 22) / (-65 (27-100))
cp = 0.44 J/ (°C × g)
The specific heat of the metal is 0.44 J/ (°C × g)
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.
Answer: 162.8 grams
Explanation:
Magnesium nitrate has a chemical formula of Mg(NO3)2.
Given that:
Number of moles of Mg(NO3)2 = 1.1 moles
Mass in grams of Mg(NO3)2 = ?
For Molar mass of Mg(NO3)2, use atomic mass of magnesium = 24g, nitrogen = 14g, oxygen = 16g
Mg(NO3)2 = 24g + (14g + 16gx3) x 2
= 24g + (14g + 48g) x 2
= 24g + (62g) x 2
= 24g + 124g
= 148g/mol
Now, apply the formula:
Number of moles = Mass in grams / molar mass
1.1 moles = Mass / 148g/mol
Mass = 1.1 moles x 148g/mol
Mass = 162.8 grams
Thus, there are 162.8 grams of magnesium nitrate.
