Salt lowers the freezing/melting point of water, so in both cases the idea is to take advantage of the lower melting point. Ice forms when the temperature of water reaches 32 degrees Fahrenheit (0 degrees Celsius).
Answer:
1) pentane is of higher Molecular mass than propane hence heavier member
2) due to intermolecular hydrogen bonding methanol is liquid having lower molecular weight
3) I) in methanol C = sp3
ii) in methanoic acid C= sp2
5)I) i) The two carbon-oxygen bonds in the methanoate anion, HCO2-, have the same length due to resonance stabilization of the C-O bond
Explanation:
See attachment for question 4
Answer:
Hg2^2+(aq) + 2Cl^-(aq) —> Hg2Cl2(s)
Explanation:
The balanced equation for the reaction is given below:
2NaCl(aq) + Hg2(NO3)2(aq) —> 2NaNO3(aq) + Hg2Cl2(s)
Considering the states of each compound in the reaction, we can see that Hg2Cl2 is in solid form meaning it will precipitate out of the solution
In solution the following occurs:
NaCl —> Na+(aq) + Cl-(aq)
Hg2(NO3)2 —> Hg2^2+(aq) + 2NO3^-(aq)
Combining the two equation together, a balanced double displacement reaction occurs as shown below:
2Na+(aq) + 2Cl-(aq) + Hg2^2+(aq) + 2NO3^-(aq) —> 2Na+2NO3^-(aq) + Hg2^2+2Cl-(s)
From the above we can thus right the insoluble precipitate as
Hg2^2+(aq) + 2Cl^-(aq) —> Hg2Cl2(s)
Answer:
The answer is 0.01365.
Explanation:
C(t) = 0.0225te^?0.0467t
We can derivate,
C'(t) = d/dt ( 0.0225 te^-0.0467t)
after solving,
C'(t) = 0.0225 (e^-0.0467t - (0.0467e^-0.0467t . t)
Plug in t =5,
Answer is C'(t) = 0.01365.
