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3 years ago

A certain element has 52 protons, 52 electrons, and 76 neutrons. its atomic mass is _____. 52 76 128 104

Chemistry

2 answers:

kifflom [539]3 years ago

5 0

Pretty sure it is 128 hope this helps you :D

Nikitich [7]3 years ago

4 0

Answer:

Atomic mass is 128

Explanation:

Atoms are composed of fundamental subatomic particles: protons, neutrons and electrons.

They are represented by a standard nuclear notation which includes the symbol of the element, its atomic number and atomic mass.

i.e. $_{Z}^{A}\textrm{X}$

Z = atomic mass = number of protons + number of neutrons -----(1)

A = atomic number = number of protons = number of electrons----(2)

<u>Given data:</u>

Protons = 52

Electrons = 52

Neutrons = 76

Based on equation (1)

Atomic mass = 52 + 76 = 128

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When a 3.00 grams sample of a compound containing only c, h, and o was completely burned, 1.17 grams of h2o and 2.87 grams of co

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molar mass of water = 18.02 gram per mol

From given info, combustion of compound gives 1.17 grams of water and 2.87 grams of carbon dioxide. Let's calculate the moles of these:

$1.17gH_2O(\frac{1mol}{18.02g})$

= $0.0649molH_2O$

Similarly, $2.87gCO_2(\frac{1mol}{44g})$

= $0.0652molCO_2$

One mol of water has two moles of H. So, the moles of H would be two times the moles of water as calculated above.

So, moles of H = 2* 0.0649 = 0.1298 mol

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moles of C = 0.0652 mol

Let's convert the moles of H and C to grams so that we could calculate the amount of oxygen present in the sample as:

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If we subtract the sum of the masses of C and H from sample mass then it would give as the mass of oxygen since the sample has only C, H and O.

mass of O in sample = 3.00g - (0.1308 g + 0.7824 g)

= 3.00 g - 0.9132 g

= 2.0868 g

Let's convert these grams of oxygen to moles on dividing by it's atomic mass as:

$2.0868gO(\frac{1mol}{15.999g})$

= 0.130 mol O

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C = $\frac{0.0652}{0.0652}$ = 1

H = $\frac{0.1298}{0.0652}$ = 2

O = $\frac{0.130}{0.0652}$ = 2

The ratio of C, H and O is 1:2:2. So, the simplest formula of the compound is $CH_2O_2$ .

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