Answer:
salt bridge balances the charge when electrons move from one half cell to another half cell.
Explanation:
Explanation: A salt bridge balances the charge when electrons move from one half cell to another half cell. During this process the salt bridge uses its electrolyte solution which further helps in balancing charges in both the half cells. ... Therefore, for each electrochemical cell a new salt bridge is used.
Answer: pH = 2,897 , basic![[H+][OH-] = 10^{-14} ==> [H+] = \frac{10^{-14}}{7,89*10^{-12} } =\frac{1}{789} \\pH= -lg([H+]) = 2,897 \\pH basic](https://tex.z-dn.net/?f=%5BH%2B%5D%5BOH-%5D%20%3D%2010%5E%7B-14%7D%20%3D%3D%3E%20%5BH%2B%5D%20%3D%20%5Cfrac%7B10%5E%7B-14%7D%7D%7B7%2C89%2A10%5E%7B-12%7D%20%7D%20%3D%5Cfrac%7B1%7D%7B789%7D%20%5C%5CpH%3D%20-lg%28%5BH%2B%5D%29%20%3D%202%2C897%20%5C%5CpH%3C7%20%3D%3D%3E%20basic)
Explanation:
Answer:
The molar mass of the organic solid is 120.16 g/mol.
The molecular formula of an organic solid is 
Explanation:
Let the molecular mass of an organic solid be 


where,
=Elevation in boiling point = 
Mass of organic solid= 0.561 g
Mass of diphenyl = 24.9 g = 0.0249 kg (1 kg = 1000 g)
= boiling point constant = 8.00 °C/m
m = molality
Now put all the given values in this formula, we get



Percentage of carbon in an organic solid = 40.0%

x = 4.0
Percentage of hydrogen in an organic solid = 6.7%

y = 8.0
Percentage of hydrogen in an organic solid = 6.7%

y = 4.0
The molecular formula of an organic solid is 
Answer is: the percent by mass of NaHCO₃ is 2,43%.
m(NaHCO₃) = 10 g.
V(H₂O) = 400 ml.
d(H₂O) = 1 g/ml.
m(H₂O) = V(H₂O) · d(H₂O).
m(H₂O) = 400 ml · 1 g/ml.
m(H₂O) = 400 g.
m(solution) = m(H₂O) + m(NaHCO₃).
m(solution) = 400 g + 10 g.
m(solution) = 410 g.
ω(NaHCO₃) = 10 g ÷ 410 g · 100%.
ω(NaHCO₃) = 2,43 %
Given:
A compound with:
Number of carbon atoms = 9
Number of double bonds = 1
A double bond between 5th and 6th carbon
A propyl group (CH2CH2CH3) branching off the 3rd carbon from the left
Try to illustrate the given and observe the formation of the atoms. Now, follow the correct IUPAC naming system. The name of the compound is
4-propyl-1-hexene
Count from the right to the left, the double bond is between the 1st and 2nd carbon, thus, 1-hexene. The propyl branches out the 4th carbon from the right, thus 4-propyl.