1 kg/L -------------- 0.001 kg/mL
22.4 kg/L --------- ??
22.4 x 0.001 / 1 => 0.0224 kg/mL
Answer:
active because he built up antibodies due to exposure to the flu
Explanation:
The volume in liters of 3245 aluminum cans is 24.8 L.
A recycling center collects 3245 aluminum cans and there are 22 aluminum cans in 1 lb. The mass of 3245 aluminum cans is:

To convert mass to volume, we need the density of aluminum (5.95 lb/L). The volume corresponding to 147.5 lb of aluminum is:

The volume in liters of 3245 aluminum cans is 24.8 L.
You can learn more about density here: brainly.com/question/1841285
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Plants and animals are multi-cellular organisms composed of eukaryotic cells, while bacteria are single-cell prokaryotic organisms. Each eukaryotic cell of a plant or animal includes a central nucleus containing DNA and membrane-bound organelles, such as endoplasmic reticulum and mitochondria. A bacterial cell has no nucleus or membrane-bound organelles.
hope it helps! :)
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Answer:
- The limiting reactant is lead(II) nitrate.
- 7.20 g of precipitate are formed.
- 1.9 g of the excess reactant remain.
Explanation:
The reaction that takes place is:
- Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)
With a percent yield of 87.5%.
To determine the limiting reactant, first we <u>convert the masses of each reactant to moles</u>, using their molar mass:
- 9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂
- 5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl
Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.
To calculate the mass of precipitate (PbCl₂) formed, we <u>use the moles of the limiting reactant</u>:
- 0.0296 mol Pb(NO₃)₂
*
* 87.5/100 = 7.20 g PbCl₂
- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:
- 0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂
Now we <u>convert that amount to moles of KCl and finally into grams of KCl</u>:
- 0.0259 mol Pb(NO₃)₂
*
= 3.86 g KCl
3.86 g of KCl would react, so the amount remaining would be: