Question

98 POINTS FOR WHO CAN GET THIS FOR ME PLEASE

Answer 1

Answer: For Number #1)

The solution above is an acidic solution  

Kw= {H+}*{OH^-}

[OH^-] =Kw/[H+]

OH^-= 1.0*10^14/ 1.0*10^-4  

=1.0*10^4

Answer 2

Answer:

1) [OH-] = 1*10⁻¹⁰ mol/L

Solution is Acidic

2) V = 14.4 L

3) q = -26125 J

4) Concentration of NaCl = 2.59 M

Explanation:

1) Given:

[H+] = 1*10⁻⁴ mol/L

Formula:

$[H+][OH-] = 10^{-14}$

$[OH-] = \frac{10^{-14} }{10^{-4} } \\\\[OH-] = 1*10^{-10} mol/L$

$p[H] = -log[H+] = -log[10^{-4} ] = 4$

Since pH < 7, the solution is acidic

2) Given:

Initial conditions:

Pressure, P1 = 1 atm

Temperature, T1 = 273 K

Volume, V1 = 10.0 L

Final conditions:

Pressure, P2 = 2.0 atm

Temperature, T2 = 512+273 = 785 K

Volume, V2 = ?

Formula:

$\frac{P1V1}{T1} = \frac{P2V2}{T2} \\\\V2 = \frac{P1V1}{T1} * \frac{T2}{P2} \\\\V2 = \frac{1*10.0}{273} * \frac{785}{2} = 14.4 L$

3) Given:

Volume of tea = 250 ml

Initial temp T1 = 375 K

Final temp, T2 = 350 K

Formula:

Energy transferred, q = mcΔT = mc(T2-T1)

m = mass of tea (water) = density * volume = 1 g/ml * 250 ml = 250 g

c = specific heat of tea (water) = 4.18 J/ gK

ΔT = T2-T1 = 350-375 = -25 K

q = 250*4.18*(-25) = -26125 J

4) Given:

Volume of sea water = 0.500 L

Mass of NaCl = 75 g

Molar mass of NaCl = 58 g/mol

Formula:

$Molarity = \frac{Moles \ NaCl}{Volume\ of\ solution} \\\\Moles\ NaCl = \frac{mass}{molar\ mass} = \frac{75}{58} =1.293\\\\Molarity = \frac{1.293}{0.500} =2.59 M$