<span><span>Argon,</span><span>Carbon dioxide,</span><span>Neon,</span><span>Helium, and </span><span>Methane</span></span>
Answer:
energy required=qnet=87.75kJ
Explanation:
we will do it in three seperate step and then add up those value.
first step is to heat the sample of water upto 100C i.e upto boiling pont. because just after this sample of water started vaporization.
q 1= m c (T2-T1)
q1 = 36.0 g (4.18 J/gC) (100 - 65 C)
q1 = 5267 J
=5.267kJ
next is to vaporize the sample at 100C
q2 = 36.0 g / 18.0 g/mol X 40.7 kJ/mol
q2= 81.4 kJ
Finally, heat the steam upto 115C
q3 = m c (T2-T1)
q 3= 36.0 g (2.01 J/gC)(115-100C)
q3 = 1085 J
=1.085kJ
qnet=q1 +q2 +q3
energy required=qnet=87.75kJ
Answer:
yes 150.0grams = 150grams
Answer:
Explanation:
a ) false.
NH₃ is more polar molecule than PH₃ so inter-molucular attraction is greater in NH₃ ( hydrogen bond ) . Hence vapour pressure is low for NH₃ .
b ) false .
The average kinetic energy of boiling water molecules is lower on a mountaintop than it is at sea level. It is so because water boils at lower temperture on mountain and kinetic energy of molecules depends upon temperature .
c ) false
vapour pressure depends upon temperature .
d ) True
CCl4 is more volatile than CBr4
e ) false
vapour pressure increases as temperature increases.
Answer:
2370.0 contains 4 significant digits and Option (c) is correct .
1.20\times 10^{-3}\ contains\ three\ significant\ digit.
Option (b) is correct .
Step-by-step explanation:
Rules for finding significant digit .
1 : Non-zero digits are always significant.
2: Any zeros between two significant digits are significant .
3: Trailing zeros in the decimal number is also significant.
As the number given be 2,370.0.
= \frac{23700}{10}
Simplify the above
= 2370
Thus by using the rule given above.
2370.0 contains 4 significant digits.
Option (c) is correct .
As the number given be 0.00120 .
= \frac{120}{100000}
Simplify the above
= \frac{1.20}{1000}
= 1.20\times 10^{-3}
Thus by using the rule given above.
1.20\times 10^{-3}\ contains\ three\ significant\ digit.
Option (b) is correct .
