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malfutka [58]
2 years ago
12

Determine the number of moles of 300g or Molybdenum (VI) oxide.

Chemistry
1 answer:
True [87]2 years ago
8 0

Answer:

How many moles Molybdenum in 1 grams? The answer is 0.010423181154888.

We assume you are converting between moles Molybdenum and gram.

You can view more details on each measurement unit:

molecular weight of Molybdenum or grams

The molecular formula for Molybdenum is Mo.

The SI base unit for amount of substance is the mole.

1 mole is equal to 1 moles Molybdenum, or 95.94 grams.

Note that rounding errors may occur, so always check the results.

Use this page to learn how to convert between moles Molybdenum and gram.

Type in your own numbers in the form to convert the units!

Explanation:

Does this help you any i hope it does.If it does not i am sorry.

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A study of the decomposition reaction 3RS2  3R + 6S yields the following initial rate dat
shutvik [7]
Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150                0.0394
0.250                0.109
0.350                0.214
0.500                0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.
3 0
2 years ago
Given the speed of light as 3.0 × 108 m/s, calculate the wavelength of the electromagnetic radiation whose frequency is 7.5 × 10
arlik [135]
Wavelength= velocity/frequency
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you can do the math
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3 years ago
PLZZZZZZZZZZZZZZZZZZ15. As much as 90 percent of the oxygen in our atmosphere is the result of
vladimir2022 [97]

Answer:

The correct answer is - option D. photosynthesis.

Explanation:

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Please help me, I really don't want to fail but I don't know how to do this
Brrunno [24]

Answer:

A)

<u>4, 7, 4, 6</u>

B)

<u>12 moles</u>

Explanation:

NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)

__↑______↑

8.00 mol | 14.00 mol

________________

NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)

You can turn this into a system of variables which are solvable.

To do this, create variables for the coefficients of each compound in the reaction respectively.

a(NH_{3}(g)) + b(O_{2}(g)) → \\c(NO_{2}) + d(H_{2}O(g))

Because to be balanced, the count of atoms in each element of the compound correspond to the coefficient of the variable in that compound so that the count of the left (reactant) side is set equal to the right (product) side.

a corresponds to the coefficient of the first compound, b corresponds to the coefficient of the second compound, c corresponds to the coefficient of the third compound, and d corresponds to the coefficient of the fourth compound.

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Reactant: 2b [O] Product: 2c + 1d.

Thus the system is:

1a = 1c

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Then just use the substitution methods to solve.

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