Which of the following is NOT a qualitative observation?

First,
Moles of HBu = Volume per liter * moles of HBu/ liter
= 25ml/1000 * 0.15 = 0.00375 moles
moles of NaOH = volume per liter * moles of NaOH/liter
= 1 ml/1000 * 0.15 = 0.00015 moles
according to this equation:
HBu + NaOH → H2O + NaBu
when 1 mol of NaOH gives 1mol of HBu
So 0.00015 of NaOH will give 0.00015 mol of HBu
∴moles of HBu which remains = 0.00375- 0.00015 = 0.0036 moles

∴moles of Bu- produced = 0.00015 moles
when the total volume = 0.025 + 0.026 =0.051 L
[HBu] = 0.0036moles / 0.051 L = 0.071 moles
[Bu] = 0.00015 / 0.051L = 0.0029 moles
when Ka = [H+] [Bu] / [HBu]
1.5x10^-5 = [H+] (0.0029) /(0.071)
∴[H+] =1x10^-6 / 0.076 = 1.5 x 10^-5
∴PH = -㏒[H+]
= - ㏒ 1.5x10^-5
= 4.81

3 0

2 years ago

How much energy has your body used, in joules, if your health indicates the 450 Calories were burned during your workout? How ma

11Alexandr11 [23.1K]

Answer:

107.55 J, 88.67 cal

Explanation:

1 Joule = 0.239 Calories

450 cal x 0.239 = 107.55 J

3.5 x 106 = 371

371 J = 371 x 0.239 cal

371 J = 88.67 cal

3 0

3 years ago

Calculate the amount of heat energy in KJ required to convert 45.0 g of ice at -15.5'C to steam at 124.0°C. (Cwater 118 Jig'c, G

dangina [55]

Answer : The enthalpy change or heat required is, 139.28775 KJ

Solution :

The conversions involved in this process are :

$(1):H_2O(s)(-15.5^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(124^oC)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change or heat required = ?

m = mass of water = 45 g

$c_{p,s}$ = specific heat of solid water = $2.09J/g^oC$

$c_{p,l}$ = specific heat of liquid water = $4.18J/g^oC$

$c_{p,g}$ = specific heat of liquid water = $1.84J/g^oC$

n = number of moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{45g}{18g/mole}=2.5mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

$\Delta H_{vap}$ = enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole

Now put all the given values in the above expression, we get

$\Delta H=[45g\times 4.18J/gK\times (0-(-15.5))^oC]+2.5mole\times 6010J/mole+[45g\times 2.09J/gK\times (100-0)^oC]+2.5mole\times 40670J/mole+[45g\times 1.84J/gK\times (124-100)^oC]$

$\Delta H=139287.75J=139.28775KJ$ (1 KJ = 1000 J)

Therefore, the enthalpy change is, 139.28775 KJ

3 0

2 years ago