Answer:
84.4g of AgCl
Explanation:
Based on the reaction:
2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃)₂
<em>2 moles of AgNO₃ and 1 mole of CaCl₂ priduce 2 moles of AgCl and 1 mole of Ca(NO₃)₂</em>
<em />
100g of each reactant are:
AgNO₃: 100g × (1mol / 169.87g) = 0.589 moles
CaCl₂: 100g × (1mol / 110.98g) = 0.901 moles
For a complete reaction of 0.901 moles of CaCl₂ are necessaries 0.901×2 = <em>1.802 moles of AgNO₃. </em>As there are just 0.589moles, <em>AgNO₃ is limitng reactant</em>
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0.589 moles of AgNO₃ produce:
0.589 moles × ( 2 moles AgCl / 2 moles AgNO₃) =
<em>0.589 moles of AgC</em>l. In mass:
0.589 moles of AgCl × (143.32g / mol) =<em> 84.4g of AgCl</em>
Answer:
0.13 cm^3
Explanation:
Volume = Mass/ density
1.8 g/ml / 14g = 0.128571429
Both numbers have 2 significant digits.
With that, the answer rounds to 0.13
The final answer is 0.13 cm(don't forget the units for volume)
Hope it helped!
<span>The instructor should be questioned to see if the filtrate is able to be recycled. This precipitate can contaminate the filtrate, rendering it useless for repeated experiments. If it is able to be recycled, a second pass through the filter might be required to remove the precipitate.</span>
Answer:
1.54 atm
Explanation:
By Dalton's Law Of partial pressure,
Total Pressure = Sum of all partial pressures
So,P= P1 + P2 + P3
Therefore, P=0.23+0.42+0.89
=1.54 atm
