Answer: Volume of the 1M EtOH and water should be 0.75 ml and 9.25 ml respectively to obtain the working concentration.
Explanation:
According to the dilution law,

where,
= molarity of stock solution = 1M
= volume of stock solution = ?
= molarity of diluted solution = 0.075 M (1mM=0.001M)
= volume of diluted solution = 10 ml
Putting in the values we get:


Thus 0.75 ml of 1M EtOH is taken and (10-0.75)ml = 9.25 ml of water is added to make the volume 10ml.
Therefore, volume of the 1M EtOH and water should be 0.75 ml and 9.25 ml respectively to obtain the working concentration
<h3>Answer:</h3><h3>1865.5g</h3><h3>Explanation:</h3><h3 /><h2> first the chemical formular for ammonium hydroxide is NH4OH</h2><h3>its molarmass is given as N=14H=1O=16 </h3><h3> so we have 14 +1(2) +16+1 =35</h3><h2>also no of moles = mass / molarmass</h2><h3> we have 5.33×10 = mass/35 </h3><h2>therefore mass = 35 ×5.33×10 = 1865.5g</h2>
To get a result with the best degree of precision, the
number of significant figures should be equal to the smallest number of
significant figures of the given numbers. In this case, the smallest is 3 as
given by the number 9.03 mL.
Therefore density is:
<span>11.50 g / 9.03 mL = 1.27 g/mL</span>
Mass/volume cause you divide to get density