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Bingel [31]
3 years ago
8

A piece of dry ice (solid carbon dioxide) with a mass of 28.8 g sublimes (converts from solid to gas) into a large balloon. Assu

ming that all of the carbon dioxide ends up in the balloon, what is the volume of the balloon at 22 °C and a pressure of 742 mmHg?
Chemistry
1 answer:
Ulleksa [173]3 years ago
6 0

Answer:16.22L

Explanation:

Mass of CO2=28.8g

Molar mass of CO2=44.00964g/mol

No of moles (n) = mass(g)/molar mass(g/mol)

= 28.8g/44.00964g = 0.6544mol

From the ideal gas formula

PV=nRT

V=nRT/P

Where n=0.6544mol

R=62.363LmmHg/Kmol

T=22+273=295K

V=0.6544x62.3636x295/742

=16.22L

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Draw a complete Lewis structure for the conjugate acid of nitromethane that shows all bonds, unshared electron pairs, and minimi
Marysya12 [62]
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4 years ago
Which quantity is equivalent to 50 kilocalories
Tanya [424]

I believe this problem have these following choices:

5,000 cal

0.05 cal

5 x 10^3 cal

5 x 10 ^4 cal

 

<span>Now 1 kilocalorie  = 1000 calories, therefore:</span>

 

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5 x 10 ^4 cal         (ANSWER)

7 0
4 years ago
One liter of ocean water contains 35.06 g of salt. What volume of ocean water would contain 1.00 kg of salt? Express your answer
sweet [91]

Answer:

28.52 L

Explanation:

First, let's calculate the density of the ocean, which is the mass divided by the volume:

d = m/V

d = 35.06/1

d = 35.06 g/L

So, for a mass of 1.00 kg = 1000.00 g

d = m/V

35.06 = 1000.00/V

V = 1000.00/35.06

V = 28.52 L

How all the data are expressed with two significant figures, the volume must also be expressed with two.

7 0
4 years ago
the ph of a naoh solution is 12.05. what is the concentration of [oh-] of the naoh solution? express your answer in standard not
Alborosie

Answer:

0.01 M

Explanation:

NaOH (sodium hydroxide) is a strong base, so it is completely dissociated into ions in aqueous solution:

NaOH ⇄ Na⁺ + OH⁻

So, we have OH⁻ ions in solution.

From the problem, we have the pH value. The pH is calculated as:

pH = -log [H⁺]

As NaOH is a base, we can also calculate the pOH:

pOH = -log [OH⁻]

Thus. we can use the relation between pH and pOH to calculate the pOH and then use it to calculate [OH⁻]:

pH + pOH = 14 ⇒ pOH = 14 - pH = 14 - 12.05 = 1.95

pOH = -log [OH⁻] ⇒ [OH⁻] = 10^{-pOH}= 10^{1.95}= 0.0112 M ≅ 0.01 M

3 0
3 years ago
Read 2 more answers
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