Answer:
D (The last answer)
Explanation:
In a transverse wave, particles oscillate perpendicular to the direction of wave motion.
In a longitudinal wave, the oscillations of particles are parallel to the direction of propagation.
C.) Newton. & it's S.I. Unit of Force.
Hope this helps!
The heat cause 300g water temperature increase from 20 to 26 celcius. The heat transferred would be: 300g * (26 °C -20 °C) *4.2 joule/gram °C= 7560J
The unknown substance is added to the water, so its final temperature should be the same as the water. The calculation would be:
7560J= 124g * (100-26)* specific heat
specific heat= 7560J / 124g / 74 °C= 0.8238 J/gram °C
<u>Answer:</u>
<em>4.5 L water we have in litres (L).</em>
<em><u></u></em>
<u>Explanation:</u>

where
= Final T - Initial T
Q is the heat energy in calories
c is the specific heat capacity (for water 1.0 cal/(g℃))
m is the mass of water
Plugging in the values

So,
Volume of water = mass/density

=4.5 L (Answer)
<u>Answer:</u> The concentration of hydrogen ion and bromoacetate ion is 0.0132 M and 0.0132 M resepectively and that of bromoacetic acid is 0.0868 M
<u>Explanation:</u>
We are given:
Molarity of bromoacetic acid = 0.100 M
Percent of ionization = 13.2 %
The chemical equation for the ionization of bromoacetic acid follows:

1 mole of bromoacetic acid produces 1 mole of bromoacetate ion and 1 mole of hydrogen ion
Molarity of hydrogen ion = 13.2 % of 0.100 = 
Molarity of bromoacetate ion = molarity of hydrogen ion = 0.0132 M
Molarity of bromoacetic acid = Molarity of solution - Molarity of ionized substance
Molarity of bromoacetic acid = 0.100 - 0.0132 = 0.0868 M
Hence, the concentration of hydrogen ion and bromoacetate ion is 0.0132 M and 0.0132 M resepectively and that of bromoacetic acid is 0.0868 M