Answer: 20.7 kJ
Explanation:
According to first law of thermodynamics:
=Change in internal energy
q = heat absorbed or released
w = work done or by the system
w = work done on the system= {Work is done on the system is positive as the final volume is lesser than initial volume}
w = 4.51 kcal = (1kcal = 4.184kJ)
q = +1.79 kJ {Heat absorbed by the system is positive}
Thus for a system that receives 1.79 kJ of heat from surroundings and has 4.51 kcal of work done on it at the same time is 20.7 kJ
Hey there,
Answer:
4 valence electrons.
Hope this helps :D
<em>~Top☺</em>
Answer:
9.80 g
Explanation:
The molecular mass of the atoms mentioned in the question is as follows -
S = 32 g / mol
F = 19 g / mol
The molecular mass of the compound , SF₆ = 32 + ( 6 * 19 ) = 146 g / mol
The mass of 6 F = 6 * 19 = 114 g /mol .
The percentage of F in the compound =
mass of 6 F / total mass of the compound * 100
Hence ,
The percentage of F in the compound = 114 g /mol / 146 g / mol * 100
78.08 %
Hence , from the question ,
In 12.56 g of the compound ,
The grams of F = 0.7808 * 12.56 = 9.80 g
Answer:
All atoms heavier than barium
Explanation:
In the periodic table, elements are divided into blocks. We have the;
s- block elements
p- block elements
d- block elements
f- block elements
However, immediately after Barium, we now encounter elements that have f-orbitals. Barium possesses a fully filled d-orbital. Hence after it, we see elements with 4f and 5f orbitals called the Lanthanides and actinides. The elements following the lanthanide and actinide series possess completely filled f-orbitals as inner orbitals.
Hence elements heavier than barium all possess f-orbitals.
Answer:
2.387 mol/L
Explanation:
The reaction that takes place is:
- 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
First we <u>calculate how many moles of each reagent were added</u>:
- HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl
- Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂
460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.
Now we <u>calculate how many moles of Ba(OH)₂ reacted</u>, by c<em>onverting the total number of HCl moles to Ba(OH)₂ moles</em>:
- 203.85 mmol HCl * = 101.925 mmol Ba(OH)₂
This means the remaining Ba(OH)₂ is:
- 460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂
There are two OH⁻ moles per Ba(OH)₂ mol:
- OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻
Finally we <u>divide the number of OH⁻ moles by the </u><u><em>total</em></u><u> volume</u> (100 mL + 200 mL):
- 716.15 mmol OH⁻ / 300.0 mL = 2.387 M
So the answer is 2.387 mol/L