Answer:
The halogens located in group 17
Explanation:
The change in energy of a neutral atom when an electron is added is known as electron affinity. <em>Fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At) are known as halogens and conform the group 17 of the periodic table, the group that presents higher electron affinity.</em> This characteristic happens because of their atomic structures, it is easier for them to gain electrons because they are closer to form octets.
Here I present you halogens' electron affinities values:
- Fluorine (F) -328 kJ mol-1
- Chlorine (Cl) -349 kJ mol-1
- Bromine (Br) -324 kJ mol-1
- Iodine (I) -295 kJ mol-1
<em>The use of negative signs is to indicate the RELEASE of energy.</em>
I hope you find this information useful and interesting! Good luck!
The standard temperature and pressure is 273 K and 1 atm. Since, pressure is not changed we can use Charle's law for the calculations.
<span>Charle's law says "at a constant pressure, the volume of a fixed amount
of gas is directly proportional to its absolute temperature".
V α T
Where V is the volume and T is the temperature
in Kelvin of the gas. We can use this for two situations as,
V</span>₁/T₁ = V₂/T₂<span>
</span>V₁ =
806 mL<span>
T</span>₁ =
26 ⁰C = 299 K
V₂ <span>=
? </span><span>
T</span>₂ =
273 K<span>
<span>
By applying the
formula,
</span></span>(806 mL / 299 K) = (V₂ / 273 K)
V₂ = 735. 91 mL
<span>
Hence, the answer is "a".</span>
<span>The vaporization of br2 from liquid to gas state requires 7.4 k/cal /mol.</span>
Answer:
E) All three samples have the same number of hydrogen atoms
Explanation:
The statements are:
A) Sample A has more hydrogen atoms than sample B or sample C.
B) Sample B has more hydrogen atoms than sample A or sample C.
C) Sample C has more hydrogen atoms than sample A or sample B
D) Both samples A and C have the same number of hydrogen atoms, but more than in sample B.
E) All three samples have the same number of hydrogen atoms
Hydrogens in sample A are:
180g × (1mol Glucose / 180.156g) × (12mol H / 1 mol Glucose) =<em> 12 moles of H</em>
Hydrogens in sample B are:
90g Glucose × (1mol Glucose / 180.156g) × (12mol H / 1 mol Glucose) =<em> 6 moles of H</em>
90g mannose× (1mol Mannose / 180.156g) × (12mol H / 1 mol Glucose) =<em> 6 moles of H</em>
<em>Total moles: 12</em>
Hydrogens in sample C are:
180g × (1mol Mannose / 180.156g) × (12mol H / 1 mol Glucose) =<em> 12 moles of H</em>
<em />
Thus, right answer is:
<em>E) All three samples have the same number of hydrogen atoms</em>
<em></em>
I hope it helps!
