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Anna35 [415]
3 years ago
13

I don't really understand this worksheet question.​

Chemistry
1 answer:
Citrus2011 [14]3 years ago
3 0
25/2 and 96/X
CROSS MULTIPLY.

2x=2,400.
divide by 2.
x=1,200.

you take the GIVEN MASS of an element, and you put it on top, the coefficient is what it’s over. i believe this is right
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What is the mass of 61.9 L of oxygen gas collected at STP?
Tcecarenko [31]

Answer:

D. 44.2 g O₂

General Formulas and Concepts:
<u>Gas Laws</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at <em>1 atm, 273 K</em>

<u>Stoichiometry</u>

  • Dimensional Analysis
  • Mole Ratio

Explanation:

<u>Step 1: Define</u>

<em>Identify given.</em>

61.9 L O₂ at STP

<u>Step 2: Convert</u>

We know that the oxygen gas is at STP. Therefore, we can set up and solve for how many <em>moles</em> of O₂ is present:

\displaystyle 61.9 \ \text{L} \ \text{O}_2 \bigg( \frac{1 \ \text{mol} \ \text{O}_2}{22.4 \ \text{L} \ \text{O}_2} \bigg) = 2.76339 \ \text{mol} \ \text{O}_2

Recall the Periodic Table (Refer to attachments). Oxygen's atomic mass is roughly 16.00 grams per mole (g/mol). We can use a mole ratio to convert from <em>moles</em> to <em>grams</em>:

\displaystyle 2.76339 \ \text{mol} \ \text{O}_2 \bigg( \frac{16.00 \ \text{g} \ \text{O}_2}{1 \ \text{mol} \ \text{O}_2} \bigg) = 44.2143 \ \text{g} \ \text{O}_2

Now we deal with sig figs. From the original problem, we are given 3 significant figures. Round your answer to the <u>exact</u> same number of sig figs:

\displaystyle 44.2143 \ \text{g} \ \text{O} \approx \boxed{ 44.2 \ \text{g} \ \text{O}_2 }

∴ our answer is letter choice D.

---

Topic: AP Chemistry

Unit: Stoichiometry

6 0
2 years ago
In Part A, we saw that the theoretical yield of aluminum oxide is 0.700 mol . Calculate the percent yield if the actual yield of
s2008m [1.1K]

Considering the definition of percent yield, the percent yield is 76%.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

percent yield=\frac{actual yield}{theorical yield}x100

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Percent yield in this case</h3>

In this case, you know:

  • actual yield= 0.532 moles
  • theorical yield= 0.700 moles

Replacing in the definition of percent yield:

percent yield=\frac{0.532 moles}{0.700 moles}x100

Solving:

<u><em>percent yield= 76%</em></u>

Finally, the percent yield is 76%.

Learn more about percent yield:

brainly.com/question/14408642

#SPJ1

3 0
2 years ago
Name the following Compound: CrF2
pav-90 [236]
Chromium(II) fluoride
8 0
3 years ago
Suppose you are investigating how the amount of exercise a hamster gets affects how long the hamster lives. In order to control
ozzi
Hello, to answer your question fully.

<span>D. Make all the hamsters run on a hamster wheel for the same amount of time.

 Because your doing the experiment on exercise not anything else 
 so keeping charts in a notebook and set a time on how long they were on it and how long you want them to be on it. 
   
                Signed by, Virtuoso Sargedog</span>

4 0
3 years ago
Read 2 more answers
Need help please! Theres three parts to this that I don't understand at all
butalik [34]

Answer:

Chemical reaction B governs the process

Explanation:

The first part of the question asks to convert the mass of the calcium carbonate given to number of moles.

Mathematically;

Number of moles = mass/molar mass

Molar mass of CaCO3 = 100 g/mol

So the number of moles of CaCO3 will be 2.49/100 = 0.0249 moles

The second part of the question asks to convert the mass of carbon iv oxide to moles of carbon iv oxide

Mathematically;

That is same as ;

Number of moles = mass/molar mass

molar mass of CO2 is 44 g/mol

Number of moles of CO2 = 1.13/44 = 0.0256 moles

Now, if we compare the values of these number of moles, we can see that there are almost equal.

What this means is that the number of moles of calcium carbonate reacted is equal to the number of moles of carbon iv oxide produced.

So what we conclude here is that we have an equal mole ratio between the two compounds.

So the reaction that would be the correct answer will present equal number of moles of carbon iv oxide and calcium carbonate

Thus, we can see that reaction B is the one that governs this process as it is the only reaction out of the three options that present the two compounds with equal number of moles.

4 0
4 years ago
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