Answer:
ΔG°rxn = +50.8 kJ/mol
Explanation:
It is possible to obtain ΔG°rxn of a reaction at certain temperature from ΔH°rxn and S°rxn, thus:
<em>ΔG°rxn = ΔH°rxn - T×S°rxn (1)</em>
In the reaction:
2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)
ΔH°rxn = 3×ΔHfNO2 + ΔHfH2O - (2×ΔHfHNO3 + ΔHfNO)
ΔH°rxn = 3×33.2kJ/mol + (-285.8kJ/mol) - (2×-207.0kJ/mol + 91.3kJ/mol)}
ΔH°rxn = 136.5kJ/mol
And S°:
S°rxn = 3×S°NO2 + S°H2O - (2×S°HNO3 + S°NO)
ΔH°rxn = 3×0.2401kJ/molK + (0.0700kJ/molK) - (2×0.146kJ/molK + 0.2108kJ/molK)
ΔH°rxn = 0.2875kJ/molK
And replacing in (1) at 298K:
ΔG°rxn = 136.5kJ/mol - 298K×0.2875kJ/molK
<em>ΔG°rxn = +50.8 kJ/mol</em>
<em />
Answer:
633 grams of sugar can be dissolved in 300 g of H₂O
Explanation:
Solubility is the measure of the ability of a certain substance to dissolve in another and form a homogeneous system. Solubility is then the maximum amount of a solute that a solvent can receive and is expressed by concentration units.
The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.
If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:
a ⇒ b
c ⇒ x
Then:
![x=\frac{c*b}{a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bc%2Ab%7D%7Ba%7D)
You can apply the rule of three as follows: if by definition of solubility in 100 grams of H₂O there are 211 grams of sugar, in 300 g of H₂O how much sugar is there?
![sugar=\frac{300 grams of H_{2}O *211 grams of sugar}{100 grams of H_{2}O}](https://tex.z-dn.net/?f=sugar%3D%5Cfrac%7B300%20grams%20of%20H_%7B2%7DO%20%2A211%20grams%20of%20sugar%7D%7B100%20grams%20of%20H_%7B2%7DO%7D)
sugar= 633 grams
<u><em>633 grams of sugar can be dissolved in 300 g of H₂O</em></u>
Answer: 0.0 grams
Explanation:
To calculate the moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%20%7BMolar%20mass%7D%7D)
a) moles of butane
![\text{Number of moles}=\frac{5.2g}{58.12g/mol}=0.09moles](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B5.2g%7D%7B58.12g%2Fmol%7D%3D0.09moles)
b) moles of oxygen
![\text{Number of moles}=\frac{32.6g}{32g/mol}=1.02moles](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B32.6g%7D%7B32g%2Fmol%7D%3D1.02moles)
![2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O](https://tex.z-dn.net/?f=2C_4H_%7B10%7D%2B13O_2%5Crightarrow%208CO_2%2B10H_2O)
According to stoichiometry :
2 moles of butane require 13 moles of ![O_2](https://tex.z-dn.net/?f=O_2)
Thus 0.09 moles of butane will require =
of ![O_2](https://tex.z-dn.net/?f=O_2)
Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.
Thus all the butane will be consumed and 0.0 grams of butane will be left.
The ribosomes are the ones delivering the products of the endoplasmic reticulum
Partial pressure=mole fraction×Pt
x=0.044÷44(maolarmass of CO2)×Pt
x=0.044÷(44)2×Pt
x=5×10^-4×Pt
x=5×10^-4×Pt
where Pt:1atm=760mmHg
xatm=750mmHg
750×1÷760=0.99
now;5×10^-4×099=4.95×10^-4.
Pt=4.95×10^-4