Answer:
E. a small yellow ball that represents the Sun
Explanation:
key word on the question, 'physical'
Answer:
The right choice is c. Water molecules have a weakly positive hydrogen end.
Explanation:
The unequal sharing of electrons in water molecule gives a slight negative charge near its oxygen atom ( see image below) and a slight positive charge near its hydrogen atoms. A neutral molecule that has a partial positive charge at one end and a partial negative charge at the other, it is a polar molecule.
so
a. Water molecules have a nonpolar bond.
It is wrong choice because water has polar bond .
b. Water molecules have a weakly positive oxygen end.
Also, a wrong choice due to water molecule gives a slight negative charge near its oxygen atom.
c. Water molecules have a weakly positive hydrogen end.
This is the right choice.
d. Water molecules have two oxygen and two hydrogen atoms
It is wrong choice because water has one oxygen and two hydrogen atoms
So, the right choice is
c. Water molecules have a weakly positive hydrogen end.
PH + pOH = 14.
So, subtract 3.73 from 14.
The pOH is 10.27 :)
Answer:
Explanation:
Places near the Equator experience little seasonal variation. They have about the same amount of daylight and darkness throughout the year. These places remain warm year-round. Near the Equator, regions typically have alternating rainy and dry seasons.
Answer:
118.06 mL
Explanation:
The neutralization reaction between HBr (acid) and Ba(OH)₂ (base) is the following:
2HBr + Ba(OH)₂ → BaBr₂ + 2H₂O
According to the equation, 2 moles of HBr react with 1 mol Ba(OH)₂. Thus, at the equivalence point the moles of acid and base react completely:
2 moles HBr = 1 mol Ba(OH)₂
We can replace the moles by the product of the molar concentration (M) and volume (V):
2 x (M HBr) x (V HBr) = M Ba(OH)₂ x V Ba(OH)₂
Now, we introduce the data in the equation to calculate the volume in mL of Ba(OH)₂:
V Ba(OH)₂ = (2 x (M HBr) x (V HBr))/M Ba(OH)₂
= (2 x 0.311 M x 57.7 mL)/(0.304 M)
= 118.06 mL
Therefore, 118 mL of Ba(OH)₂ are needed.
