Answer:
- <em>Hydration number:</em> 4
Explanation:
<u>1) Mass of water in the hydrated compound</u>
Mass of water = Mass of the hydrated sample - mass of the dehydrated compound
Mass of water = 30.7 g - 22.9 g = 7.8 g
<u>2) Number of moles of water</u>
- Number of moles = mass in grams / molar mass
- molar mass of H₂O = 2×1.008 g/mol + 15.999 g*mol = 18.015 g/mol
- Number of moles of H₂O = 7.9 g / 18.015 g/mol = 0.439 mol
<u>3) Number of moles of Strontium nitrate dehydrated, Sr (NO₃)₂</u>
- The mass of strontium nitrate dehydrated is the constant mass obtained after heating = 22.9 g
- Molar mass of Sr (NO₃)₂ : 211.63 g/mol (you can obtain it from a internet or calculate using the atomic masses of each element from a periodic table).
- Number of moles of Sr (NO₃)₂ = 22.9 g / 211.63 g/mol = 0.108 mol
<u>4) Ratio</u>
- 0.439 mol H₂O / 0.108 mol Sr(NO₃)₂ ≈ 4 mol H₂O : 1 mol Sr (NO₃)₂
Which means that the hydration number is 4.
Explanation:
how are you confused there just tell me the problem
It's a mixture of H2O + a soluble substance like, salt.
Answer:
Volume of liquid = 28.7 mL
Explanation:
Given data;
Density of solid = 3.57 g/ml
Mass of solid = 19.5 g
Volume of water = 23.2 mL
Total volume when solid is dropped into graduated cylinder= ?
Solution:
Density = mass/ volume
v = m/d
v = 19.5 g/ 3.57 g/ml
v = 5.5 mL
Volume of liquid = volume of water + volume of solid
Volume of liquid = 23.2 mL + 5.5 mL
Volume of liquid = 28.7 mL
<h3>
Answer:</h3>
200 mL
<h3>
Explanation:</h3>
Concept tested: Dilution formula
We are given;
- Concentration of stock solution as 1.00 M
- Volume of the stock solution as 50 mL
- Molarity of the dilute solution as 0.25 M
We are required to calculate the volume of diluted solution;
- The stock solution is the original solution before dilution while diluted solution is the solution after dilution.
- Using the dilution formula we can determine the volume of diluted solution;
M1V1 = M2V2
Rearranging the formula;
V2 = M1V1 ÷ M2
= (1.00 M × 50 mL) ÷ 0.25 M
= 200 mL
Therefore, a volume of 200mL of 0.25 M solution could be made from the stock solution.
