Given :
Mass of oxygen containing carbon monoxide (CO) is 2.666 gram .
To Find :
How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O) .
Solution :
By law of constant composition , a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.
So , volume of solution does not matter .
Moles of oxygen ,
.
Now , molecule of CO contains 1 mole of C .
So , moles of C is also 0.167 mole .
Mass of carbon ,
.
Therefore , mass of carbon is 2 grams .
Hence , this is the required solution .
<u>Answer:</u>
<em>The basic theme behind the movie interstellar on which it is based is the theory of relativity that was suggested by Einstein.</em>
<u>Explanation:</u>
As per <em>Einstein theory</em> the twin sister theory explains, how time is relative in nature that if two sisters war twins one of them remains behind. And other travels in space after a certain <em>ears cap the one thing on earth would age. </em>
While the other would not the same theory was followed in the movie. And the main cast of the movie who was the hero did not age while his daughter ages with respect to time so when the meet her daughter is much older than him.
<em>The movie also shows the theory behind black holes and how it feels inside the black hole.
</em>
Answer : The enthalpy change or heat required is, 139.28775 KJ
Solution :
The conversions involved in this process are :
![(1):H_2O(s)(-15.5^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(124^oC)](https://tex.z-dn.net/?f=%281%29%3AH_2O%28s%29%28-15.5%5EoC%29%5Crightarrow%20H_2O%28s%29%280%5EoC%29%5C%5C%5C%5C%282%29%3AH_2O%28s%29%280%5EoC%29%5Crightarrow%20H_2O%28l%29%280%5EoC%29%5C%5C%5C%5C%283%29%3AH_2O%28l%29%280%5EoC%29%5Crightarrow%20H_2O%28l%29%28100%5EoC%29%5C%5C%5C%5C%284%29%3AH_2O%28l%29%28100%5EoC%29%5Crightarrow%20H_2O%28g%29%28100%5EoC%29%5C%5C%5C%5C%285%29%3AH_2O%28g%29%28100%5EoC%29%5Crightarrow%20H_2O%28g%29%28124%5EoC%29)
Now we have to calculate the enthalpy change.
![\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bvap%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cg%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= enthalpy change or heat required = ?
m = mass of water = 45 g
= specific heat of solid water = ![2.09J/g^oC](https://tex.z-dn.net/?f=2.09J%2Fg%5EoC)
= specific heat of liquid water = ![4.18J/g^oC](https://tex.z-dn.net/?f=4.18J%2Fg%5EoC)
= specific heat of liquid water = ![1.84J/g^oC](https://tex.z-dn.net/?f=1.84J%2Fg%5EoC)
n = number of moles of water = ![\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{45g}{18g/mole}=2.5mole](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BMass%20of%20water%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20water%7D%7D%3D%5Cfrac%7B45g%7D%7B18g%2Fmole%7D%3D2.5mole)
= enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole
= enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole
Now put all the given values in the above expression, we get
![\Delta H=[45g\times 4.18J/gK\times (0-(-15.5))^oC]+2.5mole\times 6010J/mole+[45g\times 2.09J/gK\times (100-0)^oC]+2.5mole\times 40670J/mole+[45g\times 1.84J/gK\times (124-100)^oC]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B45g%5Ctimes%204.18J%2FgK%5Ctimes%20%280-%28-15.5%29%29%5EoC%5D%2B2.5mole%5Ctimes%206010J%2Fmole%2B%5B45g%5Ctimes%202.09J%2FgK%5Ctimes%20%28100-0%29%5EoC%5D%2B2.5mole%5Ctimes%2040670J%2Fmole%2B%5B45g%5Ctimes%201.84J%2FgK%5Ctimes%20%28124-100%29%5EoC%5D)
(1 KJ = 1000 J)
Therefore, the enthalpy change is, 139.28775 KJ
Qualitative I believe is right
This problem is asking to predict the pressure in the container at a temperature of 1,135 K with no apparent background; however, in similar problems we can be given a graph having the pressure on the y-axis and the temperature on the x-axis and a trendline such as on the attached file, which leads to a pressure of 21.2 atm by using the given equation and considering the following:
<h3>Graph analysis.</h3>
In chemistry, experiments can be studied, modelled and quantified by using graphs in which we have both a dependent and independent variable; the former on the y-axis and the latter on the x-axis.
In addition, when data is recorded and graphed, one can use different computational tools to obtain a trendline and thus, attempt to find either the dependent or independent value depending on the requirement.
In this case, since the provided trendline by the graph and the program it was put in is y = 0.017x+1.940, we understand y stands for pressure and x for temperature so that we can extrapolate this equation even beyond the plotted points, which is this case.
In such a way, we can plug in the given temperature to obtain the required pressure as shown below:
y = 0.017 ( 1,135 ) + 1.940
y = 21.2
Answer that is in atm according to the units on the y-axis:
Learn more about trendlines: brainly.com/question/13298479