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Maru [420]
3 years ago
5

A sample of a compound contains 1.65 × 10²³ atoms of C, 0.552g of H, and 4.39g of O. What is the molecular formula and molar mas

s of the compound?​
Chemistry
1 answer:
Rudiy273 years ago
5 0

Answer:

C1 H2 O16

Explanation:

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What is the surface composition of planets in the solar system?
Stels [109]

Answer:

A: Helium or Hydrogen

Explanation:

Terrestrial planets are the 4 inner most planets of the solar system which are mercury, venus, earth, Mars, while the giant planets are the 4 outer most which are Jupiter, Saturn, Uranus and Neptune.

Now, these outer most ones are the surface ones and are surrounded primarily by layers of hydrogen and helium gases.

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3 years ago
The alkali metals all have
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D-1 electron in their outer energy level.
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3 years ago
What is the identity of the element which had the following electron configuration? 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 ​
kati45 [8]

Answer:

The element with electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 ​ is manganese (25Mn).

Explanation:

Step 1: Data given

The element with electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 ​

has 25 electrons.

This element has 2 electrons on the first shell, 8 electrons on the second shell, 13 electrons on the third shell and 2 electrons on the outer shell (valence electrons).

This means this element is part of group VII.

The element with 25 electrons, we can find on the periodic table, with atomic number 25.

The element with electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 ​ is manganese (25Mn).

5 0
3 years ago
0.0200 M Fe3+ is initially mixed with 1.00 M oxalate ion, C2O42-, and they react according to the equation: Fe3+(aq) + 3 C2O42-(
Studentka2010 [4]

Answer : The concentration of Fe^{3+} at equilibrium is 0 M.

Solution :  Given,

Concentration of Fe^{3+} = 0.0200 M

Concentration of C_2O_4^{2-} = 1.00 M

The given equilibrium reaction is,

                            Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)

Initially conc.       0.02         1.00                   0

At eqm.             (0.02-x)    (1.00-3x)                x

The expression of K_c will be,

K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}

1.67\times 10^{20}=\frac{(x)^2}{(1.00-3x)^3\times (0.02-x)}

By solving the term, we get:

x=0.02M

Concentration of Fe^{3+} at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M

Therefore, the concentration of Fe^{3+} at equilibrium is 0 M.

4 0
3 years ago
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