60.7 ml is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa.
Explanation:
Data given:
V1 = 75 ml
T1 = 30 Degrees or 273.15 + 30 = 303.15 K
P1 = 91 KPa
V2 =?
P2 = 1 atm or 101.3 KPa
T2 = 273.15 K
At STP the pressure is 1 atm and the temperature is 273.15 K
applying Gas Law:
= 
putting the values in the equation of Gas Law:
V2 = 
V2 = 
V2 = 60.7 ml
at STP the volume of carbon dioxide gas is 60.7 ml.
According to Boyle's law, if the temperature were tripled as the number of moles and the volume were held constant, the pressure would triple (option C).
<h3>What is Boyle's law?</h3>
Boyle's law is the observation that the pressure of an ideal gas is inversely proportional to its volume at constant temperature.
However, when the temperature of a gas is increased, the pressure of the gas also increases provided the volume is constant.
According to this question, the temperature of a gas tripled as the number of moles and the volume were held constant.
Therefore, according to Boyle's law, if the temperature were tripled as the number of moles and the volume were held constant, the pressure would triple.
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Answer:
Explanation:
E = hc/λ
h = planck's constant = 6.66 x 10 ^ -34
c = speed of light = 2.98 *10^8 m/s
E = (6.66 x 10 ^ -34 )(2.98 *10^8)/32
= 19.85 * 10 ^-26
=0.62 x 10^-26
= 6.2 x 10^-27 J
The volume of 6.40 grams of O₂ gas at STP is 4.48L (option A). Details about volume can be found below.
<h3>How to calculate volume?</h3>
The volume of a gas can be calculated using the following formula:
p = m/v
Where;
- p = density
- m = mass
- v = volume
According to this question, the mass of O₂ gas at STP is 6.40 grams. The density of the gas at STP is 1.43 g/L.
1.43g/L = 6.4g/V
Volume of O2 = 6.4 ÷ 1.43 = 4.48L
Therefore, the volume of 6.40 grams of O₂ gas at STP is 4.48L.
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The answer would be a mixture. A mixture is when two or more substances are combined to create heterogeneous mix, where the substances do not combine with each other in an equal or even state, but just sit together.
