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Anna11 [10]
4 years ago
9

Which of the following best describes elements?

Chemistry
1 answer:
sukhopar [10]4 years ago
6 0

Explanation:

Elements are distinct substances that cannot be split-up into simpler substances. Such substances are made up of only one kind of atom.

  • Elements are classified as pure substances.
  • They are homogeneous  in all parts.
  • They have a definite composition.
  • They cannot be split or broken into simpler substances by physical means.
  • There are over a hundred known elements.
  • Elements are usually symbolized by a capital letter or capital letter followed by a smaller letter derived from the English or Latin or Greek name of the element concerned.

learn more:

Element brainly.com/question/2572495

#learnwithBrainly

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Consider the following isotopic symbol: 137Ba2+
aliya0001 [1]

Answer:

a) 56 protons

b) 54 electrons

c) 81 neutrons

d) The sum of protons and neutrons is shown. The number of protons is always the same. So we can calculate the number of neutrons ( and also the isotopes)

e)137Ba (with 56 protons and 81 neutrons)

f) atomic mass is 136.9 u ; the mass number is the sum of protons and neutrons and is 137

Explanation:

Step 1: Data given

137 Ba2+ is an isotope of barium. The atomic number of barium( and its isotopes) is 56. This shows the number of protons.

For a neutral atom, the number of protons is equal to the number of electrons.

The different isotopes of an element have the same number of protons but a different number of neutrons.

137Ba2+ has 56 protons (this is the same as the atomic number)

137Ba2+ has 54 electrons ( since it's Ba2+, this means it has 2 electrons less than protons, that's why it's charged +2)

137Ba2+ has 81 neutrons ( 137 - 56 = 81)

In the symbol, the atomic number is not shown. The sum of the protons and neutrons is shown. (Since the number of protons is the same for every isotope, we can calculate the number of neutrons that way. By knowing the neutrons, we also know the isotope.

This isotope is 137Ba

Atomic mass is also known as atomic weight. The atomic mass is the weighted average mass of an atom of an element based on the relative natural abundance of that element's isotopes.

The atomic mass of 137Ba2+ is 136.9 u

The mass number is a count of the total number of protons and neutrons in an atom's nucleus.

The mass number of 137Ba2+ is 137

3 0
3 years ago
What biome has an extremely warm, dry climate and supports plants that are adapted to living without a lot of water?
____ [38]
I think thats a tundra
6 0
3 years ago
a gas has an intial volume of 2.75L at a temperature of 285K. if the temperature changes to 380K, what is the new volume of the
olganol [36]
Suppose the gas is ideal gas. According to the ideal gas equation of state, pV = nRT, the pressure is unchanged. So Vi/Ti = nR/p = Vn/Tn. (i stands for initial and n stands for new) So the new volume Vn = 3.67 L
5 0
4 years ago
How do you prepare copper sulphate​
Elena-2011 [213]
Prepare a 1% copper sulfate solution. To make this solution, weigh 1 gram of copper sulfate (CuSO4 ·5H2O), dissolve in a small amount of distilled water in a 100 ml volumetric flask and bring to volume. Label this as 1% copper sulfate solution.
4 0
3 years ago
If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________. If the initial concen
Nikolay [14]

This is an incomplete question, here is a complete question.

If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________.

The reaction

2NOBr(g)\rightarrow 2NO(g)+Br_2(g)

It is a second-order reaction with a rate constant of 0.80 M⁻¹s⁻¹ at 11 °C.

Answer : The concentration of after 9.0 seconds is, 0.00734 M

Explanation :

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.80 M⁻¹s⁻¹

t = time taken  = 142 second

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 0.0440 M

Putting values in above equation, we get:

0.80M^{-1}s^{-1}=\frac{1}{142s}\left (\frac{1}{[A]}-\frac{1}{(0.0440M)}\right)

[A]=0.00734M

Hence, the concentration of after 9.0 seconds is, 0.00734 M

5 0
3 years ago
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