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Novay_Z [31]
3 years ago
8

A 17.4 L sample of oxygen gas (O2) was collected at a temperature of 23.0°C and a pressure of 2.18 atmospheres. What volume woul

d the gas occupy at STP?
Chemistry
1 answer:
timofeeve [1]3 years ago
6 0

The volume of the gas at STP = 35.01 L

<h3>Further explanation</h3>

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure).

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm  

V = volume, liter  

n = number of moles  

R = gas constant = 0.08206 L.atm / mol K  

T = temperature, Kelvin  

V=17.4 L

T = 23 + 273 = 296 K

P = 2.18 atm

\tt mol=n=\dfrac{PV}{RT}\\\\n=\dfrac{2.18\times 17.4}{0.082\times 296}\\\\n=1.563

The volume of the gas occupy at STP :

\tt 1.563\times 22.4=35.01~L

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According to the graph, if you are in a biome where the annual precipitation is about 300 cm and the average temperature is arou
alina1380 [7]

Answer:

We are in Tropical Rainforests.

Explanation:

Biome is basically an ecological region of earth that has specific type of flora and fauna (plants and animals). Every biome has specific abiotic factors like specific climate, temperature, geology, vegetation, soils and relief.

There are total 8 biomes in the world

  • Desert
  • Grassland
  • Temperate Boreal Forest  
  • Chaparral
  • Tropical Savanna
  • Tropical Rainforest
  • Tundra
  • Temperate Deciduous Forest

Now in our question, we have been given some conditions and are asked to identify the biome. The annual precipitation is about 300 cm and the average temperature is around 25 degrees Celsius.

Now, if we study the characteristics of all these biomes, it is very simple to identify that the biome is Tropical Rainforest.

Important characteristics of Tropical Rainforest:

  • They receive more than 200 cm of rain each year.
  • The temperature of the biome is between 20 degree Celsius and 25 degree Celsius during the whole year.
  • Around 50 percent of the animal specie of the world are found here.

<em>Note: </em><em>You can study the characteristics of world’s other biomes in the link : http://www.physicalgeography.net/fundamentals/9k.html </em>


Hope it helps!  :)


8 0
3 years ago
The molecular weight of NaCl is 58.44 grams/moles. How many grams of NaCl are found in a beaket with 100 ml of a 0.0050 M soluti
Viktor [21]

Answer:

The mass of NaCl is 0.029 grams

Explanation:

Step 1: Data given

Molecular weight of NaCl = 58.44 g/mol

Volume of solution = 100 mL = 0.100 L

Molarity = 0.0050 M

Step 2: Calculate moles NaCl

Moles NaCl = molarity * volume

Moles NaCl = 0.0050 M * 0.100 L

Moles NaCl = 0.00050 moles

Step 3: Calculate mass NaCl

Mass NaCl = moles NaCl * molar mass NaCl

Mass NaCl = 0.00050 moles * 58.44 g/mol

Mass NaCl = 0.029 grams

The mass of NaCl is 0.029 grams

4 0
3 years ago
Which consequence of new fishing methods had the most immediate impact on the environment?
maxonik [38]
I believe the answer is D
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3 years ago
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dexar [7]

Is the atoms of electric

6 0
3 years ago
Which of the following has the strongest buffering capacity? A. H2O B. 0.1 M HCl C. 0.1 M carbonic/bicarbonate (H2CO3/HCO3-) at
enyata [817]

Explanation:

(A)   As we know that carbonic acid (H_{2}CO_{3}) and Sodium bicarbonate (NaHCO_{3}) forms an acidic buffer.

Therefore, pH of an acidic buffer is given by Hendeerson-Hasselbalch equation as follows.

               pH = pK_{a} + log(\frac{[Salt]}{[Acid]}) ........... (1)

So mathematically,  if [Salt] = [Acid]  then \frac{[Salt]}{[Acid]} = 1 .

And,  log (\frac{[Salt]}{[Acid]}) = 0

Therefore, equation (1) gives us the following.

         pH = pK_{a} (when acid and salt are equal in concentration)

Hence, pK_{a} of H_{2}CO_{3} (carbonic acid) is 6.35.

And, with this we have following results.

In (A) and (D) we have the case \frac{[NaHCO_{3}]}{[H_{2}CO_{3}]}[/tex] i.e. [Salt] = [Acid].

Hence, for the cases pH = pK_{a} = 6.35.

(B)    [NaHCO_{3}] = 0.045 M and,  [H_{2}CO_{3}] = 0.45 M

Hence,   pH = 6.35 + log([NaHCO_{3}][[H_{2}CO_{3}])

                     = 6.35 + log(\frac{0.045}{0.45})

                     = 6.35 + (-1)

                     = 5.35

Therefore, it means that this buffer will be most suitable buffer as it has pH on acidic side and addition of slight excess base will not affect much of its pH value.

(C)    [NaHCO_{3}] = 0.45 M [H_{2}CO_{3}]

                          = 0.045 M

So,       pH = 6.35 + log(\frac{[NaHCO_{3}]}{[H_{2}CO_{3}]})

                  = 6.35 + log(\frac{0.45}{0.045})

                  = 6.35 + (+1)

                 = 7.35

This means that pH on Basic side makes it no more acidic buffer.

5 0
3 years ago
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