The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Answer:
Plants are found in the domain of eukarya
Answer: Solution A : ![[H_3O^+]=0.300\times 10^{-7}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D0.300%5Ctimes%2010%5E%7B-7%7DM)
Solution B : ![[OH^-]=0.107\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.107%5Ctimes%2010%5E%7B-5%7DM)
Solution C : ![[OH^-]=0.177\times 10^{-10}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.177%5Ctimes%2010%5E%7B-10%7DM)
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.
![pH=-\log[H_3O^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH_3O%5E%2B%5D)
![[H_3O^+][OH^-]=10^{-14}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%5BOH%5E-%5D%3D10%5E%7B-14%7D)
a. Solution A: ![[OH^-]=3.33\times 10^{-7}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D3.33%5Ctimes%2010%5E%7B-7%7DM)
![[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B3.33%5Ctimes%2010%5E%7B-7%7D%7D%3D0.300%5Ctimes%2010%5E%7B-7%7DM)
b. Solution B : ![[H_3O^+]=9.33\times 10^{-9}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D9.33%5Ctimes%2010%5E%7B-9%7DM)
![[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B9.33%5Ctimes%2010%5E%7B-9%7D%7D%3D0.107%5Ctimes%2010%5E%7B-5%7DM)
c. Solution C : ![[H_3O^+]=5.65\times 10^{-4}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D5.65%5Ctimes%2010%5E%7B-4%7DM)
![[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B5.65%5Ctimes%2010%5E%7B-4%7D%7D%3D0.177%5Ctimes%2010%5E%7B-10%7DM)
We know that:
Molar Mass H2O: 18 g/mol
<span>Molar Mass of Eugenol: 164 g/mol </span>
<span>Boiling point of H2O: 100 degrees C </span>
<span>Boiling point of Eugenol: 254 degrees C </span>
<span>Density of water: 1.0 g/mL </span>
<span>Density of Eugenol: 1.05 g/mL </span>
<span>Using formula:
V= [mole fraction x molar mass] / density </span>
<span>mH20: 0.9947 * 18
= 17.9046 / 1 g/mL
= 17.9046 </span>
<span>morg: 0.0053 * 164
= 0.8692/ 1.05 g/mL
= 0.8278 </span>
<span>V% = Vorg/(Vorg + VH2O) * 100 </span>
<span>(0.8278/18.7324) * 100 = 4.419% </span>
Yotal volume = 30 mL; therefore,
<span>0.0442 = (volume eugenol/30) </span>
<span>(m eug/mH2O) = (peug*164/pH2O*18) </span>
<span>(m eug/30) = (4*164/760*18) </span>
<span>m eug = about 1.44g and </span>
<span>
volume = mass/density
= 1.44/1.05
= about 1.37 mL </span>
Scientists should control most possible variables in experiments to get the most valid and correct data. If many variables are included in experiments it is more difficult to interpret what is causing a different outcome.
