Answer:
Molecular Mass:
1. 40g per mole
2. 19g per mole
3. 64g per mole
4. 32 g per mole
5. 17g per mole
6. 59.5g per mole
7.58.5 g per mole
8.220g per mole
9. 78g per mole
10. 200g per mole
Explanation:
Molecular Mass = Number of atom x atomic mass of an element
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Answer:
A dichotomous key is a tool that allows the user to determine the identity of items and organisms in the natural world. ... In each step, the user is presented with two statements based on characteristics of the organism. If the user makes the correct choice every time, the name of the organism will be revealed at the end
Explanation:
Answer: 1.99 x 10²² molecules H2
Explanation:First we will solve for the moles of H2 using Ideal gas law PV= nRT then derive for moles ( n ).
At STP, pressure is equal to 1 atm and Temperature is 273 K.
Convert volume in mL to L:
750 mL x 1 L / 1000 mL
= 0.75 mL
n = PV/ RT
= 1 atm ( 0.75 L ) / 0.0821 L.atm/ mole.K ( 273 K)
= 3.3x10-² moles H2
Convert moles of H2 to atoms using Avogadro's Number.
3.3x10-² moles H2/ 6.022x10²³ atoms H2 / 1 mole H2
= 1.99x10²² atoms H2
Answer: 0.0014 atm
Explanation:
Given that,
Original pressure of air (P1) = 1.08 atm
Original volume of air (T1) = 145mL
[Convert 145mL to liters
If 1000mL = 1l
145mL = 145/1000 = 0.145L]
New volume of air (V2) = 111L
New pressure of air (P2) = ?
Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law
P1V1 = P2V2
1.08 atm x 0.145L = P2 x 111L
0.1566 atm•L = 111L•P2
Divide both sides by 111L
0.1566 atm•L/111L = 111L•P2/111L
0.0014 atm = P2
Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm
The relation between vapour pressure , enthalpy of vapourisation and temperature is
ln (88/ 39) = DeltaH / 8.314 (1 / 318 - 1 / 298)
0.814 = DeltaH / 8.314 (2.11 X 10^-4 )
DeltaH = -32.07 kJ
