Density, Volume and Mass
3. A metal weighing 7.101 g is placed in a graduated cylinder containing 33.0 mL of water. The water
level rose to the 37.4 mL mark.
a) Calculate the density of the metal (in g/mL).
b) If you were to do this with an equal mass of aluminum (d = 2.7 g/mL), how high would the water rise?
All of these are compounds except oxygen because a compound is two or more different elements bonded together.
Answer:
energy subshell
Explanation:
electron are mostly found in energy subshell.
Answer:
48.84mL
Explanation:
H2SO4 + 2KOH → K2SO4 + 2H2O
From the question:
nA = 1
nB = 2
From the question given we obtained the following information:
Ma = 0.43M
Va =?
Mb = 0.35M
Vb = 120mL
Using MaVa / MbVb = nA/nB, we can easily find the volume of the acid required. This is illustrated below:
MaVa / MbVb = nA/nB
0.43 x Va / 0.35 x 120 = 1/2
Cross multiply to express in linear form
2 x 0.43 x Va = 0.35 x 120
Divide both side by the (2 x 0.43)
Va = (0.35 x 120) /(2 x 0.43)
Va = 48.84mL
Therefore, the volume of H2SO4 required is 48.84mL
Answer:
The third one makes the most sense