Answer:
(A) The work done by the system is -101.325J
(B) The workdone by the system is -90.75J
Explanation:
(A) Workdone = -PΔV
Given that A = 100cm2 = 0.01m2
distance d = 10cm = 0.1m
ΔV= Area × distance
ΔV= 0.01 ×0.1
ΔV = 0.001m3
P= external pressure = 1atm = 101325Pa
Workdone = -0.001 × 101325
W= - 101.325Pa m3
1Pam3 = 1J
Therefore W = - 101.325J
The work done on the system is -101.325J
(B) Workdone = -PΔV
Given that A = 50cm2 = 0.005m2
distance d = 15cm = 0.15m
ΔV= Area × distance
ΔV= 0.005×0.15
ΔV = 0.00075m3
P=121kPa = 121000Pa
W= - 121000 × 0.00075
W= -90.75Pa m3
1Pam3 = 1J
W = - 90.75J
The woekdone by the system is -90.75J
500 water molecules and the remaining 500 O2 molecules. Remember the ratio of H to O in H2O.
B is correct. As you move down group 1, the elements become more reactive with other elements because the electrons have a weaker attraction to their own atoms nucleus which means attraction with other elements is much stronger, making the atom more reactive.
<span>decomposition of SrCO3 to SrO and CO2 =change in mass
moles of CO2 =(1.850 g - 1.445 g).
</span>Mass of <span>C<span>O2</span></span><span> in mixture: 1.850-1.445 = 0.405g
</span>0.405g/44.01 g/mol <span>C<span>O2</span></span><span> = 0.0092 moles </span><span>C<span>O2</span></span><span>.
</span>ratio of <span>C<span>O2</span></span><span> to SrO in Sr</span><span>C<span>O3</span></span><span> is 1:1
</span><span> mass ratio = 1.358/1.850 = 0.7341, </span>
or 73.41% Sr<span>C<span>O3</span></span><span>.
</span>hope this helps
1,000 grams = 1 kilogram
20 grams = 0.02 kilogram
Kinetic energy = (1/2) (mass) x (speed)²
(1/2) (0.02) x (15)² =
(0.01) x (225) = 2.25 joules
