There are a lot of empty space between the particles
<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:
55.75g
Explanation:
From
m/M = CV
Where
m= required mass of solute
M= molar mass of solute
C= concentration of solution
V= volume of solution=675ml
Molar mass of solute= 3(23) + 31 + 4(16)= 69+31+64=164gmol-1
Number of moles of sodium ions present= 1.5× 675/1000= 1.01 moles
Since 1 mole of Na3PO4 contains 3 moles of Na+
It implies that 1.01/3 moles of Na3PO4 are present in solution= 0.34moles
mass of Na3PO4= number of moles × molar mass= 0.34 × 164 =55.75g
The limiting reagent will be Al
<h3>What are limiting reagents?</h3>
They are reagents that limit the quantity of products that are formed in reactions.
From the equation of the reaction:
The mole ratio of Al to O2 is 4:3.
With 2 moles of Al and 2 moles of O2, Al becomes limiting while O2 is in excess.
With 2 moles of O2, the amount of Al required should be:
2 x 4/3 = 2.67 moles.
With 2 moles of Al, the amount of O2 required should be:
2 x 3/4 = 1.5 moles
Thus, O2 is in excess by 0.5 moles.
More on limiting reagents can be found here: brainly.com/question/11848702
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Answer:
moles Fe₂O₃ = 0.938 mole
Explanation:
Convert given data to moles, solve in terms of moles by reaction ratios in balanced equation. After obtaining answer in moles, convert to needed dimension. In this case, no conversions are needed.
4Fe + 3O₂ => 2Fe₂O₃
105g/56g·mol⁻¹
= 1.875 mol Fe => => => => => => 2/4(1.875 mol Fe₂O₃) = 0.938 mol Fe₂O₃
