The answer to your question is A <span>4.184 J</span>
Answer:
1.02 × 10⁶ g
Explanation:
Step 1: Given data
- Volume of the balloon (V): 5400 m³
- Absolute pressure (P): 1.10 × 10⁵ Pa
- Molar mass of He (M): 4.002 g/mol
Step 2: Convert "V" to L
We will use the conversion factor 1 m³ = 1000 L.
5400 m³ × 1000 L/1 m³ = 5.400 × 10⁶ L
Step 3: Convert "P" to atm
We will use the conversion factor 1 atm = 101325 Pa.
1.10 × 10⁵ Pa × 1 atm / 101325 Pa = 1.09 atm
Step 4: Calculate the moles of He (n)
We will use the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.09 atm × 5.400 × 10⁶ L / 0.08206 atm.L/mol.K × 280 K
n = 2.56 × 10⁵ mol
Step 5: Calculate the mass of He (m)
We will use the following expression.
m = n × M
m = 2.56 × 10⁵ mol × 4.002 g/mol
m = 1.02 × 10⁶ g
The element of the group 17 that is most active non metal is fluorine.
The group 17 of the periodic table contains bromine(Br), iodine(I), Chlorine(Cl) and fluorine(F).
Among all the elements of the group 17. Fluorine is the smallest in size.
Because of the small size of fluorine it has the highest electronegativity in group 17.
This high electronegativity makes it a very active non metal. It provides a very high oxidizing power and low dissociation energy to the fluorine atom.
Also because of the very small size the source of attraction between the nucleus and the electrons is very high in floor in atom.
It reacts readily to form oxides and hydroxides.
So, we can conclude here that fluorine is the most active non metal of group 17.
To know more about group 17, visit,
brainly.com/question/26440054
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The answer is B
If one circuit fails, it is most likely that all the components in the circuit will fail.
This set up of a conversion table should show you that if you multiply
the grams of BeI2 times .02 moles, it equals <span>5.256 g (your answer) </span>
