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2 years ago

9

Help Light reflecting from the aluminum coating on a CD ____________ and then interferes.

1 answer:

adelina 88 [10]2 years ago

8 0

Answer:

Reflection

Hope it will help you

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What volume of 0.152 M KMnO4 solution would completely react with 20.0 mL of 0.381 M FeSO4 solution according to the following n

ANEK [815]

<u>Answer:</u> The volume of permanganate ion (potassium permanganate) is 10.0 mL

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of ferrous sulfate solution = 0.381 M

Volume of solution = 20.0 mL = 0.020 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.381M=\frac{\text{Moles of ferrous sulfate}}{0.020L}\\\\\text{Moles of ferrous sulfate}=(0.381mol/L\times 0.020L)=0.00762mol$

For the given chemical equation:

$5Fe^{2+}+8H^++MnO_4^-\rightarrow 5Fe^{3+}+Mn^{2+}+4H_2O$

By Stoichiometry of the reaction:

5 moles of iron (II) ions (ferrous sulfate) reacts with 1 mole of permanganate ion (potassium permanganate)

So, 0.00762 moles of iron (II) ions (ferrous sulfate) will react with = $\frac{1}{5}\times 0.00762=0.00152mol$ of permanganate ion (potassium permanganate)

Now, calculating the volume of permanganate ion (potassium permanganate) by using equation 1, we get:

Molarity of permanganate ion (potassium permanganate) = 0.152 M

Moles of permanganate ion (potassium permanganate) = 0.00152 mol

Putting values in equation 1, we get:

$0.152mol/L=\frac{0.00152mol}{\text{Volume of permanganate ion (potassium permanganate)}}\\\\\text{Volume of permanganate ion (potassium permanganate)}=\frac{0.00152mol}{0.152mol/L}=0.01L=10.0mL$

Hence, the volume of permanganate ion (potassium permanganate) is 10.0 mL

3 0

3 years ago

Un lote de 250 pijamas tiene un de 20.000, ¿cuál es el costo de cada pijama?<br>

zysi [14]

Answer in file below

Jsjsjsijsbsjsodjdnsjdnxj

6 0

2 years ago

Calculate the number of moles contained in 550ml of carbon dionide at stp

g100num [7]

The answer is 20 I think

3 0

3 years ago

(A tablet containing calcium carbonate and fillers with a mass of 1.631 g was dissolved in HCl. After the fillers were filtered

FrozenT [24]

Answer:

1) 0.825 grams is the mass of pure calcium carbonate product collected at the end of the experiment.

2) The mass % calcium carbonate in the tablet is 50.58%.

3) The mass % of calcium in calcium carbonate is 40.00 %.

4) Mass of calcium in a tablet is 0.33 g.

Explanation:

Mass of tablet = 1.631 g

Mass of the watch glass = 46.719 g

Mass of the watch glass + mass of calcium carbonate = 47.544 g.

Mass of calcium carbonate = 47.544 g - 46.719 g = 0.825 g

0.825 grams is the mass of pure calcium carbonate product collected at the end of the experiment.

Mass of calcium carbonate = 0.825 g

Percentage of calcium in tablet:

$\%=\frac{\text{Mass of calcium carbonate }}{\text{Mass of tablet}}\times 100$

$\%=\frac{0.825 g}{1.631 g}\times 100=50.58\%$

The mass % calcium carbonate in the tablet is 50.58%.

Percentage of calcium in calcium carbonate :

Molar mass of calcium carbonate = 100 g/mol

Atomic mass of calcium = 40 g/mol

Percentage of calcium in calcium carbonate :

$\%=\frac{\text{atomic mass of calcium}}{\text{Molar mass of}CaCO_3}\times 100$

$\%=\frac{40 g/mol}{100 g/mol}\times 100=40.00\%$

The mass % of calcium in calcium carbonate is 40.00 %.

Mass of calcium carbonate = 0.825 g

The mass % of calcium in calcium carbonate is 40.00 %.

Mass of calcium in a tablet : x

$40.00\%=\frac{\text{Mass of calcium in a tablet}}{0.825 g}\times 100$

$x=40.00\times \frac{0.825 g}{100}=0.33 g$

Mass of calcium in a tablet is 0.33 g.

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3 years ago

Hello tell me how to find this <br><br>k+o2>>k2o its o ok

umka2103 [35]

Explanation:

whay is the wietwion lol

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2 years ago