Answer:
A
Explanation:
Hydrocarbons with short chain lengths are more volatile than those with longer chains. A practical example of this can be seen in the first few members of the alkane series. They are mostly gaseous in nature and this is quite a contrast to the next few members which are solid in nature.
As we move down the group, we can see that there is an increase in the number of solids. Hence, as we go down the group we can see a relative increase in order and thus we expect more stability at room temperature compared to the volatility of the shorter chain
Answer:
Answer is B.
Explanation:
Hydrogen bonds forms when hydrogen atom is attracted towards oxygen atom of other water. A proton is shared by two ion electrons pair in which oxygen atom is partially negatively charged while hydrogen atom is partially positively charged.
C. Unbalanced describes the forces acting on the car
Enthalpy of formation is calculated by subtracting the total enthalpy of formation of the reactants from those of the products. This is called the HESS' LAW.
ΔHrxn = ΔH(products) - ΔH(reactants)
Since the enthalpies are not listed in this item, from reliable sources, the obtained enthalpies of formation are written below.
ΔH(C2H5OH) = -276 kJ/mol
ΔH(O2) = 0 (because O2 is a pure substance)
ΔH(CO2) = -393.5 kJ/mol
ΔH(H2O) = -285.5 kJ/mol
Using the equation above,
ΔHrxn = (2)(-393.5 kJ/mol) + (3)(-285.5 kJ/mol) - (-276 kJ/mol)
ΔHrxn = -1367.5 kJ/mol
<em>Answer: -1367.5 kJ/mol</em>
The value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
Aa we know that, 125mL of 0.06M Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl.
Given, T = 25°C.
<h3>Chemical equation:</h3>
Pb(NO3)2 + NaCl ---- NaNO3 + PbCl2
PbCl2 in aqueous solution split into following ions
PbCl2 ------ Pb(+2) + 2Cl-
Q = [Pb(+2)] [Cl-]^2
The Concentration of Pb(+2) ions and Cl- ions can be calculated as
[Pb(+2)] = 0.06 × 125/200
= 0.0375
[Cl-] = 0.02 × 75/200
= 0.0075
By substituting all the values, we get
[0.0375] [0.0075]^2
= 2.11 × 10^(-6).
Thus, we calculated that the value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
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