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Lelu [443]
3 years ago
9

A student dissolved 3.50g of copper (II) nitrate in water and mixed it with a solution of sodium carbonate. The student recovere

d 1.89g of copper carbonate precipitate. If the theoretical yield is 2.35g, what is the percent yield
Chemistry
1 answer:
Arisa [49]3 years ago
3 0

Answer: If the theoretical yield is 2.35g then the percent yield is 80.4%

Explanation:

Given: Actual yield = 1.89 g

Theoretical yield = 2.35 g

Formula used to calculate the percentage yield is as follows.

Percent yield = \frac{actual yield}{theoretical yield} \times 100

Substitute the values into above formula as follows.

Percent yield = \frac{actual yield}{theoretical yield} \times 100\\= \frac{1.89}{2.35} \times 100\\= 80.4 percent

Thus, we can conclude that if the theoretical yield is 2.35g then the percent yield is 80.4%

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Which of the following always contains
erica [24]

Answer:

1 mole of atom is correct.

8 0
3 years ago
Read 2 more answers
a swimming pool is 50 meters long. each day a swimmer swam total of 75 lengths of the pool for practice. how many kilometers did
Brilliant_brown [7]
50x75=3750 -- metres per day
3750x15=56250 -- metres in 15 days

Divide by 1000 to convert to km
56.250km
<u>56.25km</u>
8 0
3 years ago
If the distance between London and New York is 3600 miles, how long would you
Tju [1.3M]

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3 hours

Explanation:

s=d/t

1200=3600/t

3600/1200=t

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3 0
3 years ago
How many liters of carbon dioxide will be produced at STP if 3.56 g calcium carbonate reacts completely with carbon dioxide? CaC
sp2606 [1]

Answer:

V = 0.798 L

Explanation:

Hello there!

In this case, for this gas stoichiometry problem, we first need to compute the moles of carbon dioxide via stoichiometry and the molar mass of starting calcium carbonate:

3.56gCaCO_3*\frac{1molCaCO_3}{100gCaCO_3} *\frac{1molCO_2}{1molCaCO_3} =0.0356molCO_2

Next, we use the ideal gas equation for computing the volume, by bearing to mind that the STP conditions stand for a pressure of 1 atm and a temperature of 273.15 K:

PV=nRT\\\\V=\frac{nRT}{P}\\\\V=\frac{0.0356mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm} \\\\V=0.798L

Best regards!

4 0
3 years ago
A solution is prepared by dissolving 15.0 g of nh3 in 250.0 g of water. the density of the resulting solution is 0.974 g/ml. the
sergeinik [125]
The formula for molality---> m = moles solute/ Kg of solvent

the solute here is NH₃ because it's the one with less amount. which makes water the solvent.

1) let's convert the grams of NH₃ to moles using the molar mass

molar mass of NH₃= 14.0 + (3 x 1.01)= 17.03 g/ mol

15.0 g (1 mol/ 17.03 g)= 0.881 mol NH₃

2) let's convert the grams of water into kilograms (just divide by 1000)

250.0 g= 0.2500 kg

3) let's plug in the values into the molality formula

molality= mol/ Kg---> 0.881 mol/ 0.2500 kg= 3.52 m
6 0
3 years ago
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