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:
At 0.58 L of 0.540 M NaOH solution contain 12.5 g NaOH.
Explanation:
Given data:
At volume = ?
Mass of NaOH = 12.5 g
Molarity of solution = 0.540 M
Solution:
First of all we will calculate the number of moles of sodium hydroxide.
Number of moles = mass/molar mass
Number of moles = 12.5 g / 40 g/mol
Number of moles = 0.3125 mol
Volume of NaOH:
Molarity = number of moles / volume in L
Now we will put the values.
0.540 M = 0.3125 mol / volume in L
volume in L = 0.3125 mol / 0.540 mol/L
volume in L = 0.58 L
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Answer:
approximately 1.772 grams
Explanation:
molecular mass of AlCl3 is 132 g per mole and of Al(OH)3 is 78 g per mole
the reaction is
AlCl3 + 3 NaOH ---> Al(OH)3 + 3 NaCl
from the reaction it is clear that 1 mole AlCl3 makes 1 mole Al(OH)3
implies 132g AlCl3 gives 78g Al(OH)3
Implies 3g AlCl3 gives
