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Harlamova29_29 [7]
2 years ago
9

16.

Chemistry
1 answer:
mixas84 [53]2 years ago
8 0

Answer:

Explanation:

When going from top to bottom in any group of the periodic table, the atomic radius always tends to increase

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Balance the following chemical equation, then answer the following question.
julsineya [31]
Molar mass:

O2 = 31.99 g/mol
C8H18 = 144.22 g/mol

<span>2 C8H18(g) + 25 O2(g) = 16 CO2(g) + 18 H2O(g)

2 x 144.22 g --------------- 25 x 31.99 g
10.0 g ----------------------?? ( mass of O2)

10.0 x 25 x 31.99 / 2 x 144.22 =

7997.5 / 288.44 => 27.72 g of O2

hope this helps!


</span>

3 0
3 years ago
What is the total number of atoms in a 75 g sample of zinc?
velikii [3]
Jinx kxknxbxbxbjxj dbdbcj
8 0
2 years ago
Write a hypothesis:<br> What is the mole ratio of the cation and the anion in a precipitate?
tankabanditka [31]

Answer:

The mole ratio of the cation and the anion in a precipitate is a simple fraction. ( im sorry if this dosent help a lot.)

Explanation:

8 0
2 years ago
What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances
belka [17]

The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope                    mass amu        Relative abundance

1                                77.9                     14.4

2                               81.9                     14.3

3                               85.9                      71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9  

% abundance of isotope 1 = 14.4% = \frac{14.4}{100}=0.144

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = \frac{14.3}{100}=0.143

Mass of isotope 3 = 85.9

% abundance of isotope 2 = 71.3% = \frac{71.3}{100}=0.713

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]

A=84.2amu

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu

4 0
3 years ago
UU
BigorU [14]

Answer:

Samarium

Explanation:

The element Sm describe is called Samarium. This element has unique sets of properties that makes it very unique and distinct.

The lanthanides are found in the f-block on the periodic table of elements.

This element is a moderately hard silvery metal that readily oxidizes in air. It assumes an oxidation state of +3. The element has an atomic number of 62

7 0
3 years ago
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