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polet [3.4K]
3 years ago
7

PLEASE HELP 16. (4 points) Sodium reacts with oxygen to produce sodium oxide as described by the balanced equation below. If 54.

1g of sodium reacts with excess oxygen gas to produce 61.8g of sodium oxide, what is the percent yield? Show all work. (hint: be sure to calculate theoretical yield first)
4Na + O2 --> 2Na2O
Chemistry
1 answer:
never [62]3 years ago
4 0
Percent yield of the equation at above condidtions will be 42.4%
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When .080 moles of propane burn at STP, what volume of carbon dioxide is produced?
rodikova [14]

The volume of carbon dioxide produced is : 5.4 L

<u>Given data:</u>

moles of propane = 0.080 moles

<h3>Determine the volume of Carbon dioxide produced </h3>

The chemical reaction

C₃H₈  + 5O₂  ---- > 3CO₂ + 4H₂O

From the reaction

I mole of C₃H₈ = 3CO₂

0.080 moles of C₃H₈ =  3 * 0.080 = 0.24

Applying the equation below to determine the volume of CO₂

Pv = nRT

    = 0.24 * R * T

v = ( 0.24 * 8.314 * 273 ) / 1 atm

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Hence we can conclude that The volume of carbon dioxide produced is : 5.4 L

Learn more about Propane burning at STP : brainly.com/question/11903456

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1 year ago
A formula for a cough syrup contains 60 mg of codeine per fluid ounce. How many mg are contained in one teaspoonful?
Kruka [31]

Answer : The amount of codeine in one teaspoonful is, 30 mg

Explanation : Given,

Amount of codeine per fluid ounce = 60 mg

Now we have to determine the amount of codeine present in one teaspoonful.

As we know that:

1 teaspoonful = 0.5 fluid ounce

As, the amount of codeine per fluid ounce = 60 mg

So, the amount of codeine 0.5 fluid ounce = \frac{0.5}{1}\times 60mg=30mg

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3 years ago
For the cells in a human body, an isotonic solution is 0.9% NaCl. If a red blood cell is placed in a 1% NaCl solution, what will
igomit [66]

Answer: The red blood cell will shrink due to water loss

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Witch object has most potential energy.
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3 years ago
Use the Henderson-Hasselbalch equation, eq. (3), to calculate the pH expected for a buffer solution prepared from this acid and
notka56 [123]

Answer:

pH=4.56

Explanation:

Hello there!

In this case, given the Henderson-Hasselbach equation, it is possible for us to compute the pH by firstly computing the concentration of the acid and the conjugate base; for this purpose we assume that the volume of the total solution is 0.025 L and the molar mass of the sodium base is 234 - 1 + 23 = 256 g/mol as one H is replaced by the Na:

n_{acid}=\frac{0.2g}{234g/mol}=0.000855mol\\\\n_{base}= \frac{0.2g}{256g/mol}=0.000781mol

And the concentrations are:

[acid]=0.000855mol/0.025L=0.0342M

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Then, considering that the Ka of this acid is 2.5x10⁻⁵, we obtain for the pH:

pH=-log(2.5x10^{-5})+log(\frac{0.0312M}{0.0342M} )\\\\pH=4.60-0.04\\\\pH=4.56

Best regards!

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