All categories

3 years ago

Helppppppppppppppppppppppppp

Chemistry

2 answers:

Zigmanuir [339]3 years ago

8 0

Answer is c ok
Bye...

bezimeni [28]3 years ago

7 0

The awnser would be c i got you fam

You might be interested in

How would you know if a precipitate was formed without performing the chemical reaction

Goshia [24]

Hey there

Thats easy

Precipitation occurs when cations and anions in aqueous solution combine to form an insoluble ionic compound called precipitation. So, you can refer to a solubility chart or draw polar/nonpolar lewis structures, that might be helpful.
Remember water is polar so polar molecules will dissolve in water

4 0

3 years ago

What law is known as the law of action-reaction

Marianna [84]

Answer:

It would Newton's third law.

Explanation:

4 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

In a titration of hno3, you add a few drops of phenolphthalein indicator to 50.00 ml of acid in a flask. You quickly add 20.00 m

bekas [8.4K]

Answer:

The HNO3 solution has a concentration of 0.07 M

Explanation:

Step 1: find a balanced equation

HNO3 (aq) + NaOH (aq) → NaNO3 (aq) + H2O (l)

⇒ for 1 mole of HNO3 reacted, there will also react 1 mole of NaOH, and be produced 1 mole of NaNO3 and 1 mole of H2O, since the ratio is 1:1

Step 2: Calculating moles

Since we know that for 1 mole of HNO3 there will react 1 mole of NaOH, we can calculate the number of moleNaOH

⇒ Concentration = mole / volume

⇒ 0.210 = mole / ((20 + 7.23 ml) *10^-3)

mole = 0.005733 mole NaOH = 0.005733 mole HNO3

Step 3: Calculating the concentration of HNO3

Concentration = mole / volume

C(HNO3) = 0.005733 mole / ((50 + 30 ml) *10^-3)

C(HNO3) = 0.07 M

The HNO3 solution has a concentration of 0.07 M

To control this we can calculate through the following formule:

0.02723L x 0.21 M x ( 1mol HNO3 / 1 mol NaOH) x (1/ 0.08L) = 0.07M

8 0

3 years ago

The heat energy needed to raise the temperature of vapor above its saturation point is called _____________________. A. sensible

lana [24]

Answer:

Option (D)

Explanation:

The super-heating is usually defined as a phenomenon where a certain amount of energy is needed to raise the temperature of the water vapor beyond its normal saturation point. This is also known as the boiling delay.

The super-heat can be mathematically written as:

Super-heat = Current temperature - Boiling point of the liquid.

Thus, super-heat refers to the amount of energy that is required to increase the temperature of vapor beyond its point of saturation.

This super-heat is essential as it helps in preventing the damages of machines like air conditioner, fridge and also helps in their soft running.

Hence, the correct answer is option (D).

7 0

3 years ago