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Ber [7]
2 years ago
5

0.750mol of gas are stored in a tank at a pressure of 275kpa and a temperature of 0.00 degrees * C What is the volume of the tan

k?
Chemistry
1 answer:
nordsb [41]2 years ago
4 0

Answer:

The volume on the tank is 6, 20 L

Explanation:

We use the formula PV=nRT. We convert the units of pressure in kPa into atm and temperature in Celsius into Kelvin:

0°C=273K

101,325kPa---1 atm

275kPa --------x=(275kPax 1 atm)/101,325kPa= 2,71 atm

PV=nRT --> V=nRT/P

V= 0,750 mol x 0,082 l atm /K mol x 273 K/ 2, 71 atm= <em>6, 20 L</em>

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Media X contains 2mM glucose, 5% NaCl and 10mM beef extract. This is an example of a (COMPLEX/DEFINED) media.
konstantin123 [22]

Answer:

Complex media.

Explanation:

In a defined medium the exact composition of each component is known and it contains pure ingredients. In a complex medium, the exact composition of each component is unknown and it often contains extracts of animal or plant tissues (whose exact composition is unknown). In this case, media X contains beef extract, which is an undefined ingredient so it is a complex medium.  

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How many moles of each element are in one mole of (NH4)2S? (3 points)
cricket20 [7]

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b 1 mole of nitrogen 6 hydro 1 sulfur

3 0
3 years ago
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Why can't polyatomic ions ever stand alone?
natta225 [31]
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8 0
3 years ago
2L of hydrogen has an initial pressure of 750 mmHg, what is the final pressure in mmHg if the volume increases to 20 L with a co
Verdich [7]

Answer: The final pressure is 75 mm Hg.

Explanation:

According to Boyle's law, at constant temperature the pressure of a gas in inversely proportional to volume.

Since, it is given that the temperature is constant. Hence, formula used is as follows.

P_{1}V_{1} = P_{2}V_{2}

Substitute the values into above formula as follows.

P_{1}V_{1} = P_{2}V_{2}\\750 mm Hg \times 2 L = P_{2} \times 20 L\\P_{2} = \frac{750 mm Hg \times 2 L}{20 L}\\= 75 mm Hg

Thus, we can conclude that the final pressure is 75 mm Hg.

3 0
3 years ago
How much energy is evolved during the formation of 197 g of Fe, according to the reaction below?
hoa [83]

Answer:

ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'

Explanation:

Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj

197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)

From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...

3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).

______

NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g.  Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.  

8 0
2 years ago
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