Density (d) which is the quotient when mass (m) is divided by volume (v) is usually reported in terms of g/mL.
d = m /v
Substituting the known values,
d = (1.62 kg) x (1000 g/ 1 kg) / (205 mL)
The answer would be approximately equal to 7.9 g/mL.
We have that the molecular weight (3sf) of the compound (g/mol)
From the question we are told
A solution made by mixing 20.0 g of a non-volatile compound with 125 mL of water at 25°C has a vapor pressure of 22.67 torr. What is the molecular weight (3sf) of the compound (g/mol).
Generally the equation for the Rouault's law is mathematically given as
P=P_0 N
Therefore
The molecular weight (3sf) of the compound (g/mol)
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Answer:
Solubility, Volatility, Viscosity and Surface Tension.
Answer:
C) 8
Explanation:
Total number of carbon atoms = 3
Number of single bonds = 3
So, each carbon is bonded to the next carbon with the single bond, Number of unshared electrons left with the terminate carbons are 3 and with the intermediate carbon is 2.
Thus, the two terminate carbon have 3 hydrogen each and the intermediate will have 2.
<u>Total - 8</u>
Answer:
142.82 g
Explanation:
The following data were obtained from the question:
Volume of water = 12 mL
Volume of water + gold = 19.4 mL
Density of gol= 19.3 g/cm³
Mass of gold =.?
Next, we shall determine the volume of the gold. This can be obtained as follow:
Volume of water = 12 mL
Volume of water + gold = 19.4 mL
Volume of gold =.?
Volume of gold = (Volume of water + gold) – (Volume of water)
Volume of gold = 19.4 – 12
Volume of gold = 7.4 mL
Finally, we shall determine the mass of the gold as follow:
Note: 1 mL is equivalent to 1 cm³
Volume of gold = 7.4 mL
Density of gol= 19.3 g/cm³ = 19.3 g/mL
Mass of gold =?
Density = mass /volume
19.3 = mass of gold /7.4
Cross multiply
Mass of gold = 19.3 × 7.4
Mass of gold = 142.82 g
Therefore, the mass of the gold pebble is 142.82 g
