The kinetic energy of emitted electrons when cesium is exposed to UV rays of frequency 1.80 * 10 ^15 is 3.054 x 10 ^- 19 J.
Explanation:
To calculate the kinetic energy of emitted electrons,
It is given that the frequency is 1.80* 10^15 Hz
We have,
KE = E – Eo = hv –hvo
Where, h = 6.626 x 10^ - 34 Js
Given frequency = 1.4 x 10 ^ 15 Hz
vo (Threshold frequency) for cesium = 9.39 x 10 ^ 14 Hz
Applying in equation,
we get
KE = 6.626 x 10^34 (1.4 x 10^15 - 9.39 x 10^14)
KE= 3.054 x 10^-19 J
[Note: Here, threshold frequency of Cesium is not provided. Apply the correct threshold frequency from the part A]
Answer : Option D) Hot and dry with low growing shrubs.
Explanation : According to the attached image of the Biome, this particular biome displays the hot and dry climatic region along with low growing shrubs. A biomes are the way to divide the largely occurring flora and fauna in a particular place on the Earth's surface. These divisions are made on the basis of climatic patterns, soil types, and the animals and plants that are found to inhabit in that area.
Answer:
0.4 moles of KOH is required to neutralize 0.4 moles of HNO3.
Explanation:
The equation of the reaction is
KOH(aq) + HNO3(aq) ------> KNO3(aq) + H2O(l)
This is a neutralization reaction. A neutralization reaction is a reaction between an acid and a base to form salt and water only.
Having written the balanced chemical reaction equation, we can now solve the prob!em stoichiometrically.
From the balanced reaction equation;
1 mole of KOH is required to neutralize 1 mole of HNO3
Therefore x moles of KOH is required to neutralize 0.4 moles of HNO3
x= 1×0.4/1 = 0.4 moles
Therefore, 0.4 moles of KOH is required to neutralize 0.4 moles of HNO3.
Answer: combination reaction
