Answer:
As
Explanation:
For any element to exhibit the pattern of ionization energy shown in the question, it must possess five electrons in its outermost shell. These five electrons are not lost at once. They are lost progressively until the valence shell becomes empty. The ionization energy increases steadily as more electrons are lost from the valence shell.
The only pentavalent element among the options in arsenic, hence the answer.
Answer:
223.08 K
Explanation:
First we <u>convert 173.0 °C to K</u>:
- 173.0 °C + 273.16 = 446.16 K
With the absolute temperature we can use <em>Charles' law</em> to solve this problem:
Where in this case:
We <u>input the data</u>:
- 446.16 K * 50 L = T₂ * 100 L
And <u>solve for T₂</u>:
Answer:
24e⁻ are transferred by the reaction of respiration.
Explanation:
C₆H₁₂O₆ + 6O₂ → 6 H₂O + 6CO₂
This is the reaction for the respiration process.
In this redox, oxygen acts with 0 in the oxidation state on the reactant side, and -2 in the product side - REDUCTION
Carbon acts with 0 in the glucose (cause it is neutral), on the reactant side and it has +4, on the product side - OXIDATION
6C → 6C⁴⁺ + 24e⁻
In reactant side we have a neutral carbon, so as in the product side we have a carbon with +4, it had to lose 4e⁻ to get oxidized, but we have 6 carbons, so finally carbon has lost 24 e⁻
6O⁻² + 6O₂ + 24e⁻ → 6O₂²⁻ + 6O⁻²
In reactant side, we have 6 oxygen from the glucose (oxidation state of -2) and the diatomic molecule, with no charge (ground state), so in the product side, we have the oxygen from the dioxide with -2 and the oxygen from the water, also with -2 at the oxidation state. Finally the global charge for the product side is -36, and in reactant side is -12, so it has to win 24 e⁻ (those that were released by the C) to be reduced.
Answer:
The boiling point of water at 550 torr will be 91 °C or 364 Kelvin
Explanation:
Step 1: Data given
Pressure = 550 torr
The heat of vaporization of water is 40.7 kJ/mol.
Step 2: Calculate boiling point
⇒ We'll use the Clausius-Clapeyron equation
ln(P2/P1) = (ΔHvap/R)*(1/T1-1/T2)
ln(P2/P1) = (40.7*10^3 / 8.314)*(1/T1 - 1/T2)
⇒ with P1 = 760 torr = 1 atm
⇒ with P2 = 550 torr
⇒ with T1 = the boiling point of water at 760 torr = 373.15 Kelvin
⇒ with T2 = the boiling point of water at 550 torr = TO BE DETERMINED
ln(550/760) = 4895.4*(1/373.15 - 1/T2)
-0.3234 = 13.119 - 4895.4/T2
-13.4424= -4895.4/T2
T2 = 364.2 Kelvin = 91 °C
The boiling point of water at 550 torr will be 91 °C or 364 Kelvin
<span>Molar mass(C)= 12.0 g/mol
Molar mass (O2)=2*16.0=32.0 g/mol
Molar mass (CO2)=44.0 g/mol
18g C*1mol C/12 g C = 1.5 mol C
C + O2 → CO2
from reaction 1 mol 1 mol 1 mol
from problem 1.5 mol 1.5 mol 1.5 mol
1.5 mol O2*32 g O2/1 mol O2 = 48 g O2
In reality this reaction requires only 48 g O2 for 18 g carbon.
And from 18 g carbon you can get only
1.5 mol CO2*44 g CO2/1 mol CO2=66 g CO2
But these problem has 72g CO2. The best that we can think, it is a mix of CO2 and O2.
So to find all amount of O2 that was added for the reaction (probably people who wrote this problem wanted this)
we need (the mix of 72g - mass of carbon 18 g)= 54 g.
So the only answer that is possible is </span><span>2.) 54 g.</span>
