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3 years ago

7

Answer (example of a bar graph)

1 answer:

aniked [119]3 years ago

4 0

Basically it's a chart that shows how much something has for instance number of candy bars: let's say for instance bob has thirty you have to make a bar that is up to the 30 line. Lol hope that helps

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If you remove enough heat from a liquid, it will become a?

Yanka [14]

It will become a solid or a gas

8 0

3 years ago

2. Which of the following would produce a precipitate when mixed water (H20)?

Colt1911 [192]

Answer: #2 : s, solid & #3 : PbBr2

Explanation:

4 0

3 years ago

A 2.0 mL sample of air in a syringe exerts a pressure of 1.02 atm at 22 C. If that syringe is placed into boiling water at 100 C

emmainna [20.7K]

Answer: The new volume of the air in the syringe is 7.3 mL.

Explanation:

Given: $V_{1}$ = 2.0 mL, $P_{1}$ = 1.02 atm, $T_{1} = 22^{o}C$

$V_{2}$ = ?, $P_{2}$ = 1.27 atm, $T_{2} = 100^{o}C$

Formula used to calculate the new volume in syringe is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{1.02 atm \times 2.0 mL}{22^{o}C} = \frac{1.27 atm \times V_{1}}{100^{o}C}\\V_{1} = 7.3 mL$

Thus, we can conclude that the new volume of the air in the syringe is 7.3 mL.

6 0

3 years ago

Phosphoric acid is a triprotic acid ( K a1 = 6.9 × 10 − 3 Ka1=6.9×10−3, K a2 = 6.2 × 10 − 8 Ka2=6.2×10−8, and K a3 = 4.8 × 10 −

PilotLPTM [1.2K]

<u>Answer:</u> To calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

<u>Explanation:</u>

Phosphoric acid is a triprotic acid and it will undergo three dissociation reaction each having their respective dissociation constants.

The chemical equation for the first dissociation reaction follows:

$H_3PO_4\rightleftharpoons H_2PO_4^-+H^+;K_a1=6.9\times 10^{-3}$

The chemical equation for the second dissociation reaction follows:

$H_2PO_4^-\rightleftharpoons HPO_4^{2-}+H^+;K_a2=6.2\times 10^{-8}$

The chemical equation for the third dissociation reaction follows:

$HPO_4^{2-}\rightleftharpoons PO_4^{3-}+H^+;K_a3=4.8\times 10^{-13}$

To form a buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a$ of second dissociation process

To calculate the $pK_a$ , we use the equation:

$pK_a=-\log (K_a)\\\\pK_a=-\log(6.2\times 10^{-8})\\\\pK_a=7.21$

To calculate the pH of buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a2+\log(\frac{[\text{conjugate base}]}{[\text{weak acid}]})$

$pH=pK_a2+\log(\frac{[HPO_4^{2-}]}{[H_2PO_4^-]})$

We are given:

$pK_a2$ = negative logarithm of second acid dissociation constant of phosphoric acid = 7.21

$[HPO_4^{2-}]$ = concentration of conjugate base

$[H_2PO_4^{-}]$ = concentration of weak acid

Hence, to calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

8 0

3 years ago

5.05 g<br> Express your answer as an integer.

Jet001 [13]

Answer:5.5

Explanation:

8 0

4 years ago