7 A lead
B Gold
Csilver
8 Afeathrr
B water
C silver
Answer:
6.61 Pounds
Solution:
Step 1: Calculate Mass of Water as;
Density = Mass ÷ Volume
Solving for Mass,
Mass = Density × Volume ------ (1)
As,
Density of Water = 1 g.cm⁻³
And,
3 L of Water = 3000 cm³
Putting values in equation 1,
Mass = 1 g.cm⁻³ × 3000 cm³
Mass = 3000 g
Step 2: Convert Grams into Pounds;
As,
1 Gram = 0.002204 Pounds
So,
3000 Grams = X Pounds
Solving for X,
X = (3000 Grams × 0.002204 Pounds) ÷ 1 Gram
X = 6.61 Pounds
Answer:
D = 5.3 g/mL
Explanation:
Density = Mass over Volume
D = m/V
Step 1: Define
D = unknown
m = 16 g
v = 3.0 mL
Step 2: Substitute and Evaluate
D = 16 g / 3.0 mL
D = 5.333333333 g/mL
Step 3: Simplify
We have 2 sig figs.
5.333333333 g/mL ≈ 5.3 g/mL
Answer:
0.008 ÷ 51.3 = 0.00015594541910331380.00015594541910331382Round 0.0001559454191033138 → 0.0002 (Sig Figs: 1)
Explanation:
Supersaturation occurs with a chemical solution when the concentration of a solute exceeds the concentration specified by the value equilibrium solubility. Most commonly the term is applied to a solution of a solid in a liquid.
