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3 years ago

Indicate the number of significant figures in the measured number 5.80

Chemistry

1 answer:

levacccp [35]3 years ago

4 0

<u>Answer:</u>

The correct answer option is C. 3.

<u>Explanation:</u>

We are given the number '5.80' and we are to indicate the number of significant figures in the given measured number.

According to the rules of significant figures, numbers that are non-zero, zeros between any two significant numbers and the ending zeros in the decimal position are categorized as significant figures.

Therefore, the number 5.80 has 3 significant figures in it.

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3 years ago

What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91

Vikentia [17]

Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

$1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L$

Temperature = $91^{o}C = (91 + 273) K = 364 K$

As molar mass of $O_2$ is 32 g/mol. Hence, the number of moles of $O_2$ are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol$

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

$PV = nRT\\P \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm$

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

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3 years ago

Which is the strongest acid listed in the table?

andrew11 [14]

Answer:

Hydrofluoric acid.

Explanation:

To know which of the acid is the strongest, let us determine the pka of each acid. This is illustrated below:

1. Acetic acid

Ka = 1.8x10^-5

pKa =..?

pKa = –logKa

pKa = –Log 1.8x10^-5

pKa = 4.74

2. Benzoic acid

Ka = 6.5x10^-5

pKa =..?

pKa = –logKa

pKa = –Log 6.5x10^-5

pKa = 4.18

3. Hydrofluoric acid.

Ka = 6.8x10^-4

pKa =..?

pKa = –logKa

pKa = –Log 6.8x10^-4

pKa = 3.17

4. Hypochlorous acid

Ka = 3.0x10^-8

pKa =..?

pKa = –logKa

pKa = –Log 3.0x10^-8

pKa = 7.52

Note: the smaller the pKa value, the stronger the acid.

The pka of the various acids as calculated above is given below:

Acid >>>>>>>>>>>>>>>>>> pKa

1. Acetic acid >>>>>>>>>> 4.74

2. Benzoic acid >>>>>>>> 4.18

3. Hydrofluoric acid >>>> 3.17

4. Hypochlorous acid >> 7.52

From the above illustration, we can see that hydrofluoric acid has the lowest pKa value. Therefore, hydrofluoric acid is the strongest among them.

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3 years ago