Answer: The final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml is 0.017 M
Explanation:
According to the dilution law,
where,
= molarity of stock 
solution = 2.5 M
= volume of stock solution = 5 ml
= molarity of diluted solution = ?
= volume of diluted solution = 750 ml
Putting in the values we get:
Therefore the final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml is 0.017 M
Answer : The molal freezing point depression constant of X is 
Explanation : Given,
Mass of urea (solute) = 5.90 g
Mass of X liquid (solvent) = 450.0 g
Molar mass of urea = 60 g/mole
Formula used :
where,
= change in freezing point
= freezing point of solution = 
= freezing point of liquid X= 
i = Van't Hoff factor = 1 (for non-electrolyte)
= molal freezing point depression constant of X = ?
m = molality
Now put all the given values in this formula, we get
![[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}](https://tex.z-dn.net/?f=%5B0.4-%28-0.5%29%5D%5EoC%3D1%5Ctimes%20k_f%5Ctimes%20%5Cfrac%7B5.90g%5Ctimes%201000%7D%7B60g%2Fmol%5Ctimes%20450.0g%7D)
Therefore, the molal freezing point depression constant of X is
Answer:
Oxygen is limiting reactant
Explanation:
Based on the chemical reaction:
2C2H6 + 7O2 → 6H2O + 4CO2
<em>2 mole of ethane reacts with 7 moles of oxygen</em>
<em />
For a complete reaction of 5.25 moles of ethane are required:
5.25 moles Ethane * (7mol Oxygen / 2mol Ethane) = 18.38 moles of oxygen
As there are just 15.0 moles of oxygen
<h3>Oxygen is limiting reactant</h3>
<span>The first method to determine the chemical composition of a substance in space was using light. By determining red shift in the observed spectrum of light they could determine the elements they were observing. Different elements change the way light behaves and from this scientists can determine the makeup of things such as stars and nebulas.</span>
The number of atoms present, on average, will be the natural abundance of the isotope times the number of atoms in the sample => number of C-13 atoms = C-13 abundance * number of atoms in the sample = 1.07% * 30,000 = 1.07 * 30,000 / 100 = 321 atoms.<span> Answer: 321 atoms.</span>
