The moles of potassium dichromate , K₂Cr₂O₇ are required to prepare a 250 mL solution of with a concentration of 2.16 M is 0.54 mol.
given that :
molarity = 2.16 M
volume = 250 mL = 0.25 L
the molarity is given as :
molarity = number of moles / volumes in L
from this we can calculate the number of moles, we get :
number of moles of K₂Cr₂O₇ = molarity × volume
number of moles of K₂Cr₂O₇ = 2.16 × 0.25
number of moles of K₂Cr₂O₇ = 0.54 mol
Thus, The moles of potassium dichromate , K₂Cr₂O₇ are required to prepare a 250 mL solution of with a concentration of 2.16 M is 0.54 mol.
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Answer: Bubbles, burning, odor, color change, and rusting.
Explanation: Chemical changes are changes to something that is irreversible.
<u>Answer:</u> The red litmus paper turns blue on dipping in NaOH solution.
<u>Explanation:</u>
Litmus paper is the indicator that detects the nature of the solution, whether it is acidic or basic.
There are 2 types of litmus paper:
- <u>Red litmus paper:</u> This paper will turn blue if it is dipped in basic solution and will remain as such if it is dipped in acidic solution.
- <u>Blue litmus paper:</u> This paper will turn red if it is dipped in acidic solution and will remain as such if it is dipped in basic solution.
NaOH is a strong base, so when a red litmus paper is dipped in the beaker having necessary amount of NaOH, the red litmus paper turns into blue.
Answer:
The average atomic weight = 121.7598 amu
Explanation:
The average atomic weight of natural occurring antimony can be calculated as follows :
To calculate the average atomic mass the percentage abundance must be converted to decimal.
121 Sb has a percentage abundance of 57.21%, the decimal format will be
57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .
123 Sb has a percentage abundance of 42.79%, the decimal format will be
42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .
Next step is multiplying the fractional abundance to it masses
121 Sb = 0.5721 × 120.904 = 69.169178400
123 Sb = 0.4279 × 122.904 = 52.590621600
The final step is adding the value to get the average atomic weight.
69.169178400 + 52.590621600 = 121.7598 amu
The element with 8 electrons in its 3d sublevel is Ni,
Nickel. The answer is letter D. The
rest of the choices do not answer the question above.
