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Stels [109]
3 years ago
10

determine the average volume from the following volume reading: 3.00ml, 2.0ml, 2.987 x 10^-3 l, and 3.4856 ml

Chemistry
1 answer:
lesya [120]3 years ago
8 0

Answer:

The average of given values is 2.1221 ml

Explanation:

Given data:

Given measurements = 3.00 ml , 2.0 ml, 2.987 × 10⁻³ml , 3.4856 ml

Average value of given measurements = ?

Solution:

Formula:

Average value = sum of all measurement  / total number of measurements

2.987 × 10⁻³ml = 0.002987 ml

Now we will put the  values.

Average value = 3.00 ml + 2.0 ml + 0.002987 ml+ 3.4856 ml / 4

Average value = 8.488587 ml / 4

Average value =  2.1221 ml

The average of given values is 2.1221 ml

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For the following reaction, 4.34 grams of sulfur dioxide are mixed with excess oxygen gas . The reaction yields 3.89 grams of su
Degger [83]

Answer:

5.42g, 71.77%

Explanation:

First, we have to write out the balanced chemical equation. The unbalanced equation can be written as “SO2+O2 -> SO3” and to balance it, we can see that having two mols of SO2 and two mols of SO3 will make each side have the same amount of mols per element on each side. So the balanced chemical equation is “2SO2 + O2 -> 2SO3”

Now, we want to solve for the theoretical yield in grams of SO3. To do this, we have to use dimensional analysis. We convert g SO2 into mols SO2 using the molar mass of the elements. Then we convert mols of SO2 into mols of SO3 using the balanced equation. Once we’ve done that, we can convert mols of SO3 into grams of SO3.

You should know how to look up the molar mass of elements on the periodic table by now. Find the masses and set up the terms so they cancel like so:

4.34g \times  \frac{1mol \: so2}{64.07g \: so2}  \times  \frac{2 \: mol \: so3}{2 \: mol \: so2}  \times  \frac{80.07gso3}{1 \: mol \: so3}

Doing the math, we get 5.42g so3 as the theoretical yield. This is the most amount that you could ever get if the world was a perfect place. But alas, it isn’t and mistakes are gonna happen, so the number is going to be less than that. So the best we can do, is to figure out the percent yield that we got.

In a lab scenario, this was calculated to be 3.89 g as stated by the problem. The percent composition formula is

| \frac{result}{expected \: result} |  \times 100

and plugging the numbers into it, we get:

| \frac{3.89}{5.42} |  \times 100 = 71.77\%

make sure to follow the decimal/significant figure rules of your instructor, but only round at the end. My professor didn't care too much thankfully, but some professors do

6 0
2 years ago
The ratio of the lengths of two sides of a right triangle forms a
Delicious77 [7]

Answer:

perimeter

Explanation:

6 0
3 years ago
What is 44.25 l to ml
Oliga [24]

44250

hope this helps

6 0
3 years ago
What is the molality of a solution in which 0.42 moles aluminum chloride has been dissolved in 4200 water ?
madreJ [45]

Answer:

0.1 M

<h3>Explanation:</h3>
  • Molarity refers to the concentration of a solution in moles per liter.
  • It is calculated by dividing the number of moles of solute by the volume of solvent;
  • Molarity = Moles of the solute ÷ Volume of the solvent

<u>In this case, we are given;</u>

  • Number of moles of the solute, NH₄Cl as 0.42 moles
  • Volume of the solvent, water as 4200 mL or 4.2 L

Therefore;

Molarity = 0.42 moles ÷ 4.2 L

            = 0.1 mol/L or 0.1 M

Thus, the molarity of the solution will be 0.1 M

7 0
3 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
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