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:
ans is (2) 2,4- hexadiene
To balance this equation, first we should consider balancing C because it only presents in one reactant and one product. Assuming the coefficient of C6H6 is 1, there are 6 C's in the reactant, so it generates 6CO2. Then consider balancing H for the same reason. If the coefficient of C6H6 is 1, there are 6 H's in the reactant, so it generates 3H2O.
Now that the coefficient of the products are determined, we can balance O. There are 6*2=12 O's in CO2 and 3*1=3 O's in H2O. So the total number of O in the products is 12+3 = 15. O2 is the only reactant that contains O, so to balance the equation, the coefficient of O2 should be 15/2.
Now the equation looks like:
C6H6 + 15/2O2 ⇒ 6CO2 + 3H2O.
Times both sides of the equation by 2 results the final answer:
2C6H6 + 15O2 ⇒ 12CO2 + 6H2O
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thanx
<u>h</u><u><em>ope it helps you mark as brainlest</em></u>
Answer:
Q = 30284.88 j
Explanation:
Given data:
Mass of ethanol = 257 g
Cp = 2.4 j/g.°C
Chnage in temperature = ΔT = 49.1°C
Heat required = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Now we will put the values in formula.
Q = 257 g× 2.4 j/g.°C × 49.1 °C
Q = 30284.88 j
