<span>C. 11.2 L
There are several different ways to solve this problem. You can look up the density of CO2 at STP and work from there with the molar mass of CO2, but the easiest is to assume that CO2 is an ideal gas and use the ideal gas properties. The key property is that a mole of an idea gas occupies 22.413962 liters. And since you have 0.5 moles, the gas you have will occupy half the volume which is
22.413962 * 0.5 = 11.20698 liters. And of the available choices, option "C. 11.2 L" is the closest match.
Note: The figure of 22.413962 l/mole is using the pre 1982 definition of STP which is a temperature of 273.15 K and a pressure of 1 atmosphere (1.01325 x 10^5 pascals). Since 1982, the definition of STP has changed to a temperature of 273.15 K and a pressure of exactly 10^5 pascals. Because of this lower pressure, one mole of an ideal gas will have the higher volume of 22.710947 liters instead of the older value of 22.413962 liters.</span>
Answer:
a) 1.248 x 10⁷ kg
b) 1.248 x 10⁴ Mg
c) 1.248 x 10¹³ mg
d) 1.248 x 10⁴ ton
Explanation:
a) Since 1000 g = 1 kg we can convert grams to kg by multiplyig any given quantity in grams by the conversion factor ( 1 kg / 1000 g):
1.248 x 10¹⁰ g * (1 kg / 1000 g) = 1.248 x 10⁷ kg
b) Since 1 Mg = 1 x 10⁶ g, the conversion factor will be ( 1 Mg / 1 x 10⁶ g):
1.248 x 10¹⁰ g * ( 1 Mg / 1 x 10⁶ g) = 1.248 x 10⁴ Mg
c) Since 1 mg = 1 x 10⁻³ g, the conversion factor will be ( 1 mg / 1 x 10⁻³ g):
1.248 x 10¹⁰ g ( 1 mg / 1 x 10⁻³ g) = 1.248 x 10¹³ mg
d) Since 1 metric ton = 1000 kg and 1000 g = 1 kg, we can use these conversions factors: ( 1 kg / 1000 g) and (1 ton / 1000 kg):
1.248 x 10¹⁰ g * ( 1 kg / 1000 g) * ( 1 ton / 1000 kg) = 1.248 x 10⁴ ton
Answer:
His speed was 2 m/s, and his velocity was 0. So it is B
Explanation:
took the test
