Select the true statement about the diagram.

A plane takes 230 passengers from New York to Chicago in 3 hours.

devlian [24]

Answer:263mph

Explanation:

3 0

2 years ago

What type of animal might become preserved in amber

goldfiish [28.3K]

Insect's or Bugs preserved in Amber

6 0

3 years ago

What new evidence led scientists to change their minds about the geocentric model?

IRISSAK [1]

Galileo disproved the Ptolemaic theory, sanctioned for centuries by the Church, which held the Earth to be the central and principal object in the universe, about which all celestial objects orbited.

3 0

3 years ago

A sample of gas exerts a pressure of 16 atm at a temperature of 340 K. What is the pressure of the gas in atm if the temperature

gavmur [86]

Answer:

P2 = 19.2atm

Explanation:

Initial pressure (P1) = 16atm

Initial temperature (T1) = 340K

Final temperature (T2) = 408K

Final pressure (P2) = ?

This question involves the use of pressure law

Pressure law states that the pressure of a fixed mass of gas is directly proportional to it's temperature provided that volume is kept constant.

Mathematically,

P = kT, k = P / T

Therefore,

P1 / T1 = P2 / T2 = P3 / T3 = ......=Pn / Tn

P1 / T1 = P2 / T2

We need to solve for P2

P2 = (P1 × T2) / T1

Now we can plug in the values and solve for P2

P2 = (16 × 408) / 340

P2 = 6528 / 340

P2 = 19.2atm

The final pressure (P2) of the gas is 19.2atm

5 0

3 years ago

50.0 mL solution of 0.160 M potassium alaninate ( H 2 NC 2 H 5 CO 2 K ) is titrated with 0.160 M HCl . The p K a values for the

scZoUnD [109]

Answer:

a) 6.12

b) 1.87

Explanation:

At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ ------> $H_3}^+NC_2H_5CO^-_2$

1 mole of alaninate react with 1 mole of acid to give 1 mole of alanine;

therefore 50.0 mL of 0.160 M alaninate required 50.0 mL of 0.160M HCl to reach the first equivalence point.

The concentration of alanine can be gotten via the following process as shown below;

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{initial moles of alaninate}{total volume}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+50.0mL)}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{8}{100mL}$

$[H_3}^+NC_2H_5CO^-_2]$ = 0.08 M

Alanine serves as an intermediary form, however the concentration of $H^+$ and the pH can be determined as follows;

$[H^+]$ = $\sqrt{\frac{K_{a1}K_{a2}{[H_3}^+NC_2H_5CO^-_2]+K_{a1}K_w}{ K_{a1}{[H_3}^+NC_2H_5CO^-_2] } }$

$[H^+]$ = $\sqrt{\frac{ (10^{-pK_{a1})}(10^{-pK_{a2})}(0.08)+(10^{-pK_{a1})}(1.0*10^{-14})} {(10^{-pK_{a1}})+(0.08)} }$

$[H^+]$ = $\sqrt{\frac{ (10^{-2.344})(10^{-9.868})(0.08)+(10^{-2.344})(1.0*10^{-14})} {(10^{-2.344})+(0.08)} }$

$[H^+]$ = $7.63*10^{-7}M$

pH = - log $[H^+]$

pH = $-log[7.63*10^{-7}]$

pH= 6.12

Therefore, the pH of the first equivalent point = 6.12

b) At the second equivalence point; all alaninate is converted into protonated alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO^-_2$

$H^+_3NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO_2H$

Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;

Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.

Thus, the concentration of protonated alanine can be determined as:

$[H^+_3NC_2H_5CO_2H]$ = $\frac{initial moles of alaninate}{total volume}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+100.0mL)}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{8}{150}$

$[H^+_3NC_2H_5CO_2H]$ = 0.053 M

The pH at the second equivalence point can be calculated via the dissociation of protonated alanine at equilibrium which is represented as:

$H^+_3NC_2H_5CO_2H$ ⇄ $H^+_3NC_2H_5CO^-_2$ $+$ $H^+$

(0.053 - x) x x

$K_{a1}$ = $\frac{[H^+] [H^+_3NC_2H_5CO^-_2]}{[H^+_3NC_2H_5CO_2H]}$

$10^{-PK_{a1}}$ = $\frac{x*x}{(0.053-x)}$

$10^{-2.344} =\frac{x^2}{(0.053-x)}$

$0.00453 = \frac{x^2}{(0.053-x)}$

$0.00453(0.053-x) =x^2$

$x^2+0.00453x-(2.4009*10^{-4})$

Using quadratic equation formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

we have:

$\frac{-0.00453+\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$ OR $\frac{-0.00453-\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$

= 0.0134 OR -0.0179

So; we go by the positive integer which says

x = 0.0134

So $[H^+]=[H_3^+NC_2H_5CO^-_2]= 0.0134 M$

pH = $-log[H^+]$

pH = $-log[0.0134]$

pH = 1.87

Thus, the pH of the second equivalent point = 1.87

3 0

3 years ago