Answer:
Density = 4.191 gm/L
Explanation:
Given:
Molar mass = 93.89 g/mol
Volume(Missing) = 22.4 L (Approx)
Find:
Density at STP
Computation:
Density = Mass/Volume
Density = 93.89 / 22.4
Density = 4.191 gm/L
Answer:
3.861x10⁻⁹ mol Pb⁺²
Explanation:
We can <u>define ppm as mg of Pb²⁺ per liter of water</u>.
We<u> calculate the mass of lead ion in 100 mL of water</u>:
- 100.0 mL ⇒ 100.0 / 1000 = 0.100 L
- 0.100 L * 0.0080 ppm = 8x10⁻⁴ mg Pb⁺²
Now we <u>convert mass of lead to moles</u>, using its molar mass:
- 8x10⁻⁴ mg ⇒ 8x10⁻⁴ / 1000 = 8x10⁻⁷ g
- 8x10⁻⁷ g Pb²⁺ ÷ 207.2 g/mol = 3.861x10⁻⁹ mol Pb⁺²
To determine the time it takes to completely vaporize the given amount of water, we first determine the total heat that is being absorbed from the process. To do this, we need information on the latent heat of vaporization of water. This heat is being absorbed by the process of phase change without any change in the temperature of the system. For water, it is equal to 40.8 kJ / mol.
Total heat = 40.8 kJ / mol ( 1.50 mol ) = 61.2 kJ of heat is to be absorbed
Given the constant rate of 19.0 J/s supply of energy to the system, we determine the time as follows:
Time = 61.2 kJ ( 1000 J / 1 kJ ) / 19.0 J/s = 3221.05 s
Explanation:
a. 0.0093
Number of significant figures = 2
All zero’s preceding the first integers are never significant
b. 120.9
Number of significant figures = 4
All zero’s between integers are always significant.
c. 1,000
Number of significant figures = 1
All zeroes used solely for spacing the decimal point are not significant.
d. 1.008
Number of significant figures = 4
All zero’s between integers are always significant.
All zero’s after the decimal point are always significant.
e. 670
Number of significant figures = 2
All zeroes used solely for spacing the decimal point are not significant.
f. 0.184
Number of significant figures = 3
All zero’s after the decimal point are always significant.
g. 1.30
Number of significant figures = 3
All zero’s after the decimal point are always significant.
