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4 years ago

What is the correct way to represent 5600 L using scientific notation?

Chemistry

1 answer:

Law Incorporation [45]4 years ago

4 0

Answer:

5.6 * 10^3

this is because when thinking about scientific notation you know that you have to include a decimial and how many times you are moving it and whether it is a negative of a positive.

this means that you would take 5600 and you would put the decimal inbetween 5 and 6 because they are significant numbers and then you would have to figure out how many places to the right you need to go inorder to make it 5600

meaning you have to move it three times therefore it becomes

5.6 * 10^3

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Thepotemich [5.8K]

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2 years ago

The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av

nlexa [21]

<u>Answer:</u> The percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope be 'x'. So, fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope will be '1 - x'

<u>For $_{77}^{191}\textrm{Ir}$ isotope:</u>

Mass of $_{77}^{191}\textrm{Ir}$ isotope = 190.9606 amu

Fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope = x

<u>For $_{77}^{193}\textrm{Ir}$ isotope:</u>

Mass of $_{77}^{193}\textrm{Ir}$ isotope = 192.9629 amu

Fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

$192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710$

Percentage abundance of $_{77}^{191}\textrm{Ir}$ isotope = $0.3710\times 100=37.10\%$

Percentage abundance of $_{77}^{193}\textrm{Ir}$ isotope = $(1-0.3710)=0.6290\times 100=62.90\%$

Hence, the percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

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3 years ago

Which describes an atom that has fewer neutrons than protons and more electrons than protons?

Eva8 [605]

Answer:

Option d is correct = Negative ion

Explanation:

We know that an atom consist of electrons, protons and neutrons. Neutrons and protons are present inside the nucleus while electrons are present out side the nucleus. Electron has a negative charge and is written as e⁻. The mass of electron is 9.10938356×10⁻³¹ Kg . While mass of proton and neutron is 1.672623×10⁻²⁷Kg and 1.674929×10⁻²⁷ Kg respectively.

Symbol of proton= P⁺

Symbol of neutron= n⁰

The number of electron or number of protons are called atomic number while mass number of an atom is sum of protons and neutrons. The umber of protons and electrons are always equal to make the atom electrically neutral and when an atom loses its valance electron the number of protons increases and thus positive charge increased and atom form cation.

When an atom gain electron negative charge increase because of more number of electron thus atom form negative ion or anion. For example,

Anion formation:

X + e⁻ → X⁻

Cation formation:

X → X⁺ + e⁻

Thus option d is correct option.

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4 years ago

The carbon cycle involves an exchange of carbon between ___

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<h2>Answer:</h2>

D) The geosphere, hydrosphere, and atmosphere

<h2>Explanation:</h2>

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