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3 years ago

How many significant figures does .150, 20.20, and 25.00 have?

1 answer:

6 0

.150 = 3
20.20 = 4
25.00 = 4
If a 0 does not have a number or a period after it, it is not significant.
If the 0 is behind the decimal point, it is always significant.

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The answer is package c because each cd would cost $1.87, which is the lowest price

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The average lung capacity of a human is 6.0L.

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Answer:

(a) 0.25 mol

(b) 0.11 mol

Explanation:

(a)

We use the equation given by ideal gas which follows:

$PV=nRT$

where,

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Putting values in above equation, we get:

$1.00 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1.00\times 6.0}{0.0821\times 298}=0.25mol$

(b)

We use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 0.296 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 200 K

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n = number of moles = ?

Putting values in above equation, we get:

$0.296 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 200K\\\\n=\frac{0.296\times 6.0}{0.0821\times 200}=0.11mol$

(c)

We use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 30 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 250 K

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

n = number of moles = ?

Putting values in above equation, we get:

$30 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 250K\\\\n=\frac{30\times 6.0}{0.0821\times 250}=8.77mol$

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Answer:

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