Mass in kilograms of liquid air required = 0.78 kg
<u>Given that </u>
1 Litre of liquid air contains 1.3 grams of oxygen ( air )
<u />
<u> Determine the a</u><u>mount of liquid air</u><u> in Kg</u>
volume of air given = 600 L
mass of liquid air required = x
1 litre = 1.3 grams
600 L = x
∴ x ( mass of liquid air ) = 1.3 * 600
= 780 g = 0.78 kg
Hence we can conclude that Mass in kilograms of liquid air required = 0.78 kg
Learn more about liquid air : brainly.com/question/636295
Answer:
particles of gold
Explanation:
To convert the number of moles of any substance, in this case gold, you need Avogadro's number.
Avogadro's number is always 
× 
moles Au × 
= particles of gold
The new pressure, P₂ is 6000 atm.
<h3>Calculation:</h3>
Given,
P₁ = 1.5 atm
V₁ = 40 L = 40,000 mL
V₂ = 10 mL
To calculate,
P₂ =?
Boyle's law is applied here.
According to Boyle's law, at constant temperature, a gas's volume changes inversely with applied pressure.
PV = constant
Therefore,
P₁V₁ = P₂V₂
Put the above values in the equation,
1.5 × 40,000 = P₂ × 10
P₂ = 1.5 × 4000
P₂ = 6000 atm
Therefore, the new pressure, P₂ is 6000 atm.
Learn more about Boyle's law here:
brainly.com/question/23715689
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Explanation:
k so basically u gotta do 59/1000000 then multiply that by 972 which gives u 0.057348
