Ammonia undergoes combustion with oxygen to produce nitric oxide and water. The volume of the oxygen required to react with 720 ml of ammonia is 900 ml.
<h3>What is volume?</h3>
Volume is the area occupied by the substance and is the ratio of the mass to the density.
At STP, 1 mole of gas occupies 22.4 L of volume
Given,
Volume of ammonia reacted = 0.720 L
The combustion reaction is shown as,
From the stoichiometry of the reaction, it can be said that,
L of ammonia reacts with 
L of oxygen gas.
So, 0.720 L of ammonia will react with:
Therefore, the volume of oxygen required is 900 mL.
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Answer: The independent variable is the type of metal being used.
{Note: The "dependent variable" is the "measured density" that corresponds to each of the metals."}.
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Explanation:
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The "independent variable", which is plotted on the "x-axis" (horizontal axis), is the variable that can be "controlled/manipulated". In this case, this would be the type of metal chosen.
The "dependent variable" , which is plotted on the "y-axis" (vertical axis) is the "obtained value/measurement/result" (that "cannot be controlled/manipulated").
In this case, the "density", which is the "measured value" that corresponds to the selected "meal", is the "dependent variable".
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Hope this helpful to you!
Wishing you well!
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Answer:
I do not bake but my mom is a professional baker.
Explanation:
A roof over my head and a warm home.
Plenty of drinkable water.
I don't have to go hungry.
I can enjoy the small and free pleasures of life.
Access to the internet.
My friends and family.
My health.
The kindness of people I have never met before.
Answer: hundred cubic meters of carbon dioxide initially at 150◦C and 50 bar is to beisothermally compressed in a frictionless piston-and-cylinder device to a final pressure of 300 bar.Calculatei The volume of the compressed gas.ii The work done to compress the gas.iii The heat flow on compression.assuming carbon dioxide(a) Is an ideal gas.(b) Obeys the principle of corresponding states of Sec. 6.6(c) Obeys the Peng-Robinson equation of state.SolutionWe haveT1= 150◦C,P1= 50 bar,T2= 150◦C,P2= 300 bar. (1 and 2 denote the initialand final conditions in this ’snapshot’ problem, respectively - we have sometimes called themt1andt2)(a) If CO2is an idea gas, we havePV=NRT.The number of moles can be calculated from theinitial conditions:N1=P 1 V 1 RT 1 = (6 × 10 6 Pa)(100 m 3 ) (8 . 314 J/(mol K))(150 + 273 K) = 142127 mol = 142 . 1 kmol i. Since we know N 1 = N 2 , T 2 , P 2 V 2 = N 2 RT 2 P 2 = (142127 mol)(8 . 314 J/(mol K))(150 + 273 K ) 30 × 10 6 Pa = 16 . 66 m 3 ii. Since there is no shaft work, and since the gas is isothermally compressed , we only have pressure-volume work: W = - Z V 2 V 1 PdV = - Z V 2 V 1 NRT V dV = - NRT ln V 2 V 1 W = - (142127 mol)(8 . 314 J/(mol K))(150 + 273 K ) ln 16 . 66 100 = 8 . 958 × 10 8 J iii. Energy balance (integral form) for the closed system is: U 2 - U 1 = Q + W Back from homework 2, for an ideal gas, stating from equation 6.2-21, dU = C V dT + " T ∂P ∂T V - P # dV reduces to: dU = C V dT However, the process is isothermal, so dT = 0 Which gives: dU = Δ U = 1 N Δ U Therefore 0 = Q + W → Q = - W Q = - 8 . 958 × 10 8 J 3
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Explanation:As revealed above, the stimuli connections are clearly stated
