Hydrogen H 1.00794
Carbon C 12.0107
Oxygen O 15.9994
Potassium K 39.0983
Answer:
- 0.1852
- 0.0947
- 0.7201
- 3.0345 kg CO
/ Kg C
H
- 15.3848 Kg air / kg C H
Explanation:
Molar masses of each product are :
Butane = 58 kg /kmol
Oxygen = 32 kg/kmol
Nitrogen = 28 kg/kmol
water = 18 kg/kmol
<u><em>1) Calculate the mass fraction of carbon dioxide </em></u>
= ( 4 * 44 ) / ( (5 * 18) + (4 *44 )+ (24.44 * 28) )
= 176 / 950.32
= 0.1852
<em><u>2) Calculate the mass fraction of water </u></em>
= ( 5 * 18 ) / (( 5* 18 ) + ( 4*44) + ( 24.44 * 28 ))
= 90 / 950.32
= 0.0947
<em><u>3) Calculate the mass fraction of Nitrogen </u></em>
= (24.44 * 28 ) / ((4 * 44 ) + ( 24.44 * 28 ) + ( 5 * 18 ))
= 684.32 / 950.32
= 0.7201
<em><u>4) Calculate the mass of Carbon dioxide in the products</u></em>
Mco2 = ( 4 * 44 ) / 58 = 3.0345 kg CO / Kg C H
<u>5) Mass of Air required per unit of fuel mass burned </u>
Mair = ( 6.5 * 32 + 24.44 *28 ) / 58 = 15.3848 Kg air / kg C H
Answer: The sample absorbed 3061.565J
Explanation:
Given that
Mass of water = 335g
Initial temperature = 24.5°C
Final temperature = 26.4°C
Heat absorbed by the sample is given as = mass x specific heat of water x temperature change
Heat absorbed,q=mCΔt
The specific heat of water,C = 4.81J/g°C
Therefore, Heat absorbed,q = 335 x 4.81h x (26.4 - 24.5) = 3061.565J
Answer:
See explanation
Explanation:
Copper II oxide is a base but not an alkali. An alkali is a soluble base. Since Copper II oxide is not soluble in water then it is not an alkali.
Let us recall that the change of colour of litmus with an alkali requires the presence of water. In the absence of water, solid Copper II oxide does not turn red litmus paper blue.
The ability to turn red litmus paper blue is commonly observed with alkalis and Copper II oxide is not an alkali.
Also recall that since Copper II oxide is not soluble, hydroxide ions are absent hence Copper II oxide does not turn red litmus paper blue.
Answer:
Forged parts are often tougher than cast parts. This can be determined by performing tensile tests on various areas on the parts. Additionally, the microstructures of forged and cast parts can be used to determine if a part was forged or cast. The microstructure of a cast part will have a more uniform grain structure.
Explanation:
