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Dafna1 [17]
3 years ago
13

A 93-L sample of dry air is cooled from 145 oC to -22 oC at a constant pressure of 2.85 atmospheres). What is the final volume?

Chemistry
1 answer:
Aloiza [94]3 years ago
3 0

Answer: 55.84L

Explanation: Please see attachment for explanation.

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Can you give the specific problem?
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Which compound can act as both a Bronsted-Lowry acid and a Bronsted-Lowry base?
Dvinal [7]
The answer is A. Water

Bronsted-Lowry base compounds are those that can accept protons

Bronsted-Lowry Acid Compounds are those that can recieve one

Water / H2O is an Amphoteric compund which mean that its molecul can act as a Base and Acid compound, so the answer is A.
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3 years ago
PLEASEEEE HELP MEEEEE
fiasKO [112]

Answer:

3.82 x 10²¹ molecules As₂O₃

Explanation:

To find the amount of molecules arsenic (III) oxide (As₂O₃), you need to (1) convert kg to lbs, then (2) convert g As₂O₃ to moles As₂O₃ (via molar mass), and then (3) convert moles to molecules (via Avogadro's number).

1 kilogram = 2.2 lb

Molar Mass (As₂O₃): 2(74.992 g/mol) + 3(15.998 g/mol)

Molar Mass (As₂O₃): 197.978 g/mol

Avogadro's Number:

6.022 x 10²³ molecules = 1 mole

0.0146 g As₂O₃            1 kg                  189 lb
------------------------  x  ---------------  x  ------------------  x   ................
         1 kg                     2.2 lb          

       1 mole                6.022 x 10²³ molecules
x  ------------------  x  ---------------------------------------  = 3.82 x 10²¹ molecules As₂O₃
      197.978 g                        1 mole

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2 years ago
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PLS HELP What is one of the benefits that scientists have discovered from organisms' daily exposure to radioisotopes? 1, being a
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Calculate the standard molar enthalpy for the complete combustion of liquid ethanol (C2H5OH) using the standard enthalpies of fo
PSYCHO15rus [73]

Answer:

Explanation:

For the reaction

C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O

We can calculate the  standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:

ΔHºc =  2ΔHºf CO2 (g) + 3ΔHºfH2O(l)  - ( ΔHºf C2H5OH (l) - 3ΔHºfO2 (g) )

( we were not give the water state but we know we are at standard conditions so it is in its liquid state )

The ΔHºfs can be found in appropiate reference or texts.

ΔHºc =  2ΔHºf CO2 (g)+ 3ΔHºfH2O(l)  - (  ΔHºf C2H5OH (l) -+3ΔHºfO2 (g) )

= [ 2 ( -393.52 ) + 3 ( -285.83 ) ] - [(  -276.2 + 0 ) ] kJ

ΔHºc =  -1368.33 kJ

5 0
4 years ago
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