1. 3.0% ----> 3.0 kg fat= 100 kg body weigh
also remember that 1 kg= 2.20 lbs

2. 0.94 g/mL----> 0.94 grams= 1 mL
1 Liters= 1000 mL
1kg= 1000 grams
C—stems and leaves growing upward?
(a) We know that work is the product of Force and Distance so: (in this
case Distance is negative since going down so –d)
work = force * distance
work = M * (g - g/4) * -d
work = -3Mgd/4 <span>
(b) The work by the weight of the block is simply:</span>
work = Mgd <span>
(c) The kinetic energy is simply equivalent to the
net work, therefore:</span>
KE = net work
KE = Mgd/4 <span>
(d) The velocity is:</span>
v = √(2*KE/M)
Plugging in the value of KE from c:
v = √(2*Mgd / 4M)
<span>v = √(gd / 2) </span>
The correct answer is industrial smog. This type of smog exists in coal power plants which creates smoke and sulfur dioxide which may mix with fog creating a thick blanket of haze. Sulfur dioxide is one primary component of an industrial smog.
Answer:
- 13.56 g of sodium chloride are theoretically yielded.
- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.
- 0.50 g of sodium nitrate remain when the reaction stops.
- 92.9 % is the percent yield.
Explanation:
Hello!
In this case, according to the question, it is possible to set up the following chemical reaction:

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

Therefore, the leftover of sodium nitrate is:

Finally, the percent yield is computed via:

Best regards!