A sample containing 43.98g of metal pellets is poured into a graduated cylinder initially containing 13.8 mL of water, causing t
he water level in the cylinder to rise to 21.3 mL. Calculate the density of the metal.
1 answer:
Answer:
5.9 g/cm³
Step-by-step explanation:
Data:
Mass of metal = 43.98 g
Initial level of water = 13.8 mL
Final level of water = 21.3 mL
Calculations:
The picture below illustrates what's happening in this experiment.
The water level was originally at 13.8 mL.
Then you added the metal. It fell to the bottom and displaced its volume of water.
The water level rose to 21.3 mL, so
Volume of metal = volume of displaced water
= 21.3 mL – 13.8 mL
= 7.5 mL
= 7.5 cm³
Density = mass/volume
= 43.98/7.5
= 5.9 g/cm³
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Answer:
1.327 g Ag₂CrO₄
Explanation:
The reaction that takes place is:
- 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)
First we need to <em>identify the limiting reactant</em>:
We have:
- 0.20 M * 50.0 mL = 10 mmol of AgNO₃
- 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄
If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.
Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:
- 4 mmol K₂CrO₄ *
= 1326.92 mg Ag₂CrO₄
- 1326.92 mg / 1000 = 1.327 g Ag₂CrO₄