35g Mg x 1mol / 24g = 840 mol
Answer:
3.10×10¯⁵ ft³.
Explanation:
The following data were obtained from the question:
Density (D) of lead = 11.4 g/cm³
Mass (m) of lead = 10 g
Volume (V) of lead =.?
Density (D) = mass (m) / Volume (V)
D = m/V
11.4 = 10 / V
Cross multiply
11.4 × V = 10
Divide both side by 11.4
V = 10 / 11.4
V = 0.877 cm³
Finally, we shall convert 0.877 cm³ to ft³. This can be obtained as follow:
1 cm³ = 3.531×10¯⁵ ft³
Therefore,
0.877 cm³ = 0.877 cm³ × 3.531×10¯⁵ ft³ /1 cm³
0.877 cm³ = 3.10×10¯⁵ ft³
Thus, 0.877 cm³ is equivalent to 3.10×10¯⁵ ft³.
Therefore, the volume of the lead in ft³ is 3.10×10¯⁵ ft³.
Answer:
Mg(OH)2 + 2HCI ⇒ MgCI2 + 2H2O
Explanation:
In order to balance a chemical equation you need to make sure both sides of the equations are equal.
Mg(OH)2 + 2HCI = MgCI2 + H2O
Mg = 1
Oh = 2
HCI = 2
Products:
Mg = 1
CI = 2
H = 2
O = 1
2H20 = 1 × 2 = 2
2 × 2 = 4
2HCI
1 × 2 = 2
Mg(OH)2 + 2HCI ⇒ MgCI2 + 2H2O
Hope this helps.
