Answer:
13 km
Explanation:
Distance travelled = 5 km + 3 km + 2 km + 3 km = 13 km
Nonmetals which are located in the second row form pi bonds
more easily than the elements situated in the third row and below. Actually there
are no compounds or molecules known that forms covalent bonds to the noble gas
Ne and Ar. Hence the other second row element which is Carbon, is the element that
forms
pi bonds most readily.
Answer:
<span>C</span>
Dihydrogen oxide is the right answer. Dihydrogen oxide is just 2 hydrogen and 1 oxygen which is H2O or water.
Volume of the tank is 5.5 litres.
Explanation:
mass of the CO2 is given 8.6 grams
Pressure of the gas is 89 Kilopascal which is 0.8762 atm
Temperature of the gas is 29 degrees ( 0 degrees +273.5= K) so (29+273)
R = gas constant 0.0821 liter atmosphere per kelvin)
FROM THE IDEAL GAS LAW
PV=nRT ( P Pressure, V Volume, n is number of moles of gas, R gas constant, Temperature in Kelvin)
no of moles = mass/atomic mass
= 8.6/44
= 0.195 moles
now putting the values in equation
V=nRT/P
= 0.195*0.0821*302/ 0.8762
= 5.5 litres.
As the carbon dioxide gas occupies the volume os the tank hence volume of tank is 5.5 litres.
Answer:
390.85mL
Explanation:
Step 1:
Data obtained from the question.
Initial pressure (P1) = 780 torr
Initial volume (V1) = 400mL
Initial temperature (T1) = 40°C = 40°C + 273 = 313K
Final temperature (T2) = 25°C = 25°C + 273 = 298K
Final pressure (P2) = 1 atm = 760torr
Final volume (V2) =?
Step 2:
Determination of the final volume i.e the volume of the gas outside Matt's body.
The volume of the gas outside Matt's body can be obtained by using the general gas equation as shown below:
P1V1/T1 = P2V2/T2
780 x 400/313 = 760 x V2 /298
Cross multiply to express in linear form
313 x 760 x V2 = 780 x 400 x 298
Divide both side by 313 x 760
V2 = (780 x 400 x 298) /(313 x 760)
V2 = 390.85mL
Therefore, the volume of the gas outside Matt's body is 390.85mL
