AHHHHH HELP ME WITH THESE QUESTIONS FOR MY QUIZ

Which evidence cannot be found in spoiled foods?

eimsori [14]

Answer:

a. Change of state

Explanation:

Because you will see that the state has changed

8 0

2 years ago

What part of the atom is represented by the letter Z?

pashok25 [27]

Answer:

The proton of the atom

4 0

3 years ago

Juan Carlos placed 35 grams of ke into a dry, 200-gram container. The top of the container was attached tightly. When the ice wa

lianna [129]

Correct answer is D

8 0

3 years ago

HELP ME WITH ONE OR BOTH OF THESE QUEATIONS PLEASEEEE

marta [7]

Answer:

2. 181.25 K.

3. 0.04 atm.

Explanation:

2. Determination of the temperature.

Number of mole (n) = 2.1 moles

Pressure (P) = 1.25 atm

Volume (V) = 25 L

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) =?

The temperature can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

1.25 × 25 = 2.1 × 0.0821 × T

31.25 = 0.17241 × T

Divide both side by 0.17241

T = 31.25 / 0.17241

T = 181.25 K

Thus, the temperature is 181.25 K.

3. Determination of the pressure.

Number of mole (n) = 10 moles

Volume (V) = 5000 L

Temperature (T) = –10 °C = –10 °C + 273 = 263 K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure (P) =?

The pressure can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

P × 5000 = 10 × 0.0821 × 263

P × 5000 = 215.923

Divide both side by 5000

P = 215.923 / 5000

P = 0.04 atm

Thus, the pressure is 0.04 atm

6 0

3 years ago

Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

3 years ago