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3 years ago

What is the mass of oxygen in 25.0 grams of potassium permanganate , KMnO4?

Chemistry

2 answers:

MArishka [77]3 years ago

4 0

Answer: 10.1 g

I just did it and got i right

Katen [24]3 years ago

3 0

Answer:

10.76 grams

Explanation:

Given that the amount of $KMnO_4$ is 25.0 grams.

Mass of 1 mole of $KMnO_4$ = 158 grams

The number of atoms in 1 mole of $KMnO_4$ is 4.

Mass of oxygen in 1 mole of $KMnO_4$ = 16\times 4 = 68 grams.

Here, 158 grams of $KMnO_4$ has 68 grams of oxygen

So, the amount of oxygen in 1 gram of $KMnO_4$ = 68/158 grams

Therefore, the amount of oxygen in 1 gram of $KMnO_4$ = $\frac {68}{158}\times 25$ grams

=10.76 grams

Hence, 25 grams of $KMnO_4$ has 10.76 grams of oxygen.

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Explanation:

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3 years ago

Which of the following is an example of a phase?

Zielflug [23.3K]

Answer:

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Explanation:

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3 years ago

A scientist was asked to test the effect of a new vitamin for rats. His hypothesis was that young rats that had vitamins added t

Olin [163]

A: incorrect because there will be no effect on keeping the rats in separate cage in the same room at the time of the experiment.

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3 years ago

Read 2 more answers

Hi can someone help me quickly please its for chemistry

lara [203]

Answer:

Explanation:

The atom X has 1 1 protons (atomic number) and thus would have 11 electrons. This means the atom X would have 1 valence electron in it's outermost shell. While the atom Y has 17 protons (atomic number) and thus would have 17 electrons. This means the atom Y would have 7 electrons in it's outermost shell.

The type of bond this atoms (X and Y) would form is an Ionic (or electrovalent) bond. <u>This bond/combination involves the transfer of electron(s) and a molecule is not formed</u> as seen in the image below.

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3 years ago

Given the following formula for calculating the ionization energy of one-electron species such as Li2+, He+, and H, calculate th

Yuki888 [10]

Answer:

The answer is " $32819.9 \ \frac{J}{mol}\\\\$ "

Explanation:

$Boron: 5^{B}\to 1s^2 2s^2 2p^1$

$\Delta E=-2.18\times 10^{-18}\ \frac{J}{atom}\ (\frac{1}{\infity^2}-\frac{1}{n^2_{initial}})(z^2) (6.022\times 10^{23}\ \frac{atom}{mol})\\\\$

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4 0

3 years ago