To prepare 350 mL of 0.100 M solution from a 1.50 M
solution, we simply have to use the formula:
M1 V1 = M2 V2
So from the formula, we will know how much volume of the
1.50 M we actually need.
1.50 M * V1 = 0.100 M * 350 mL
V1 = 23.33 mL
So we need 23.33 mL of the 1.50 M solution. We dilute it
with water to a volume of 350 mL. So water needed is:
350 mL – 23.33 mL = 326.67 mL water
Steps:
1. Take 23.33 mL of 1.50 M solution
<span>2. Add 326.67 mL of water to make 350 mL of 0.100 M
solution</span>
When there’s water in a department such as a lake, pound or/and ocean, the water tends to evaporated into the air and that’s how/when clouds are created. The same water from those locations are the same water that comes from the rain water, it’s a cycle that creates and takes water
<span>In a titration, the substance that is unknown and being identified is called analyte. A titration is where a known solution or concentration called the titrant is used to identify and measure an unknown substance which is the analyte.</span>
Answer:
Option D. 30 g
Explanation:
The balanced equation for the reaction is given below:
2Na + S —> Na₂S
Next, we shall determine the masses of Na and S that reacted from the balanced equation. This is can be obtained as:
Molar mass of Na = 23 g/mol
Mass of Na from the balanced equation = 2 × 23 = 46 g
Molar mass of S = 32 g/mol
Mass of S from the balanced equation = 1 × 32 = 32 g
SUMMARY:
From the balanced equation above,
46 g of Na reacted with 32 g of S.
Finally, we shall determine the mass sulphur, S needed to react with 43 g of sodium, Na. This can be obtained as follow:
From the balanced equation above,
46 g of Na reacted with 32 g of S.
Therefore, 43 g of Na will react with = (43 × 32)/46 = 30 g of S.
Thus, 30 g of S is needed for the reaction.
Answer:
<em>1 mole is equal to 1 moles NaOH, or 39.99711 grams.</em>
Explanation:
<em>Hope this helps have a nice day :)</em>
