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kondaur [170]
4 years ago
14

When one atom loses an electron and another atom accepts that electron a(n) bond between the two atoms results?

Chemistry
1 answer:
morpeh [17]4 years ago
8 0
When an element losses its electron its called a cation. When an element accepted that electron it called anion. This is called an ionic bond.
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A sample of gaseous arsine (AsH3) in a 460 mL flask at 332 Torr and 223 K, is heated to 437 K, at which temperature arsine decom
Taya2010 [7]

Answer:

28/95 = 29.,5 % of Arsine decomposed

Explanation:A sample of gaseous arsine (AsH3) in a 460 mL flask at 332 Torr and 223 K, is heated to 437 K, at which temperature arsine decom- poses to solid arsenic and hydrogen gas. The flask is then cooled to 273 K, at which tem- perature the pressure in the flask is 488 Torr. What percentage of arsine molecules have de- composed?

Answer in units of %.

initial pressure 332 Torr initial volume 0.46 L initial temperature 223K

final pressure 488 Torr final volume 0.46 L final 273 K

Torr is 1/760 atm 332 torr = 0.437 atm 488 Torr =0.642 atm

PV = nRT so n=RT/PV

INITIAL n= 0.082 X 223/(0.437)(0.46) = 91 moles

final n= 0.082 X 273 / (.437)(488) = 105 moles

2AsH3----------> 2As + 3H2

x moles of Arsine decomposed to make 1.5 moles of H2

the final number of moles was

(91 -X)+ 1.5 X = 105 moles

91 + 0.5 X = 105

0.5 X = 14

X =28

CHECK

if 28 moles of Arsine , then the container would have

91 --28 + 1.5(28) = 91 +14 =105 check

so 28/95 = 29.,5 % of Arsine decomposed

Your answer

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polyalchemVirtuoso

Answer:

Explanation:

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A 1.40-L gaseous system absorbs 75 J of heat and expands its volume to 2.00 L against an external pressure of 1.02 atm. What is
kramer

Answer:

15.24 J is the change in internal energy for this process.

Explanation:

According to the first law of thermodynamics:-

\Delta U = q + w

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U is the internal energy

q is the heat

w is the work done

From the question,

q = + 75 J

(+ sign as the heat is being absorbed)

The expression for the calculation of work done is shown below as:

w=-P\times \Delta V

Where, P is the pressure

\Delta V is the change in volume

From the question,  

\Delta V = 2.00 - 1.40 L = 0.60 L

P = 1.02 atm

w=-1.02\times0.60\ atmL

Also, 1 atmL = 101.3 J

So,  

w=-1.02\times0.60\times 101.3\ J=-59.76\ J (work is done by the system)

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\Delta U = +75\ J-59.76\ J = 15.24\ J

<u>15.24 J is the change in internal energy for this process.</u>

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