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3 years ago

How many significant figures are in the quantity 12.0 mL? O A. 1 B. 2 C. 3 O D. 12

Chemistry

1 answer:

insens350 [35]3 years ago

7 0

Answer:

C. 3

Explanation:

12.0 - 3 significant figures 1,2 and 0, because 0 is after decimal point.

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Earthquakes change the surface of the Earth. Write a response analyzing how the author of "Earthquakes" used cause and effect to

Alex_Xolod [135]

Earthquakes make the surface all cracky and hard where it’s hard to dig up and the soil is like hard to step on because one false move and you could fall down

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3 years ago

Which environmental conditions are MOST likely to increase the speed of chemical weathering? A) cold and wet B) warm and dry C)

luda_lava [24]

D) warm and wet.
Water can be a catalyst, and heat speeds up reactions.

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3 years ago

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A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

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:

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

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

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3 years ago

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Which element decreases its oxidation number in this reaction? bicl2 + na2so4 → 2nacl + biso4?

kifflom [539]

The balanced reaction is as follows;
BiCl₂ + Na₂SO₄ --> 2NaCl + BiSO₄
this is a double displacement reaction
the oxidation number of Bi is +2 in both BiCl₂ and BiSO₄
oxidation number of Cl is -1 in both BiCl₂ and NaCl
oxidation number of Na is +1 in both Na₂SO₄ and NaCl
oxidation numbers of elements in SO₄²⁻ remains the same in both compounds.Therefore the oxidation state in any of the elements in the reaction doesn't change. Neither of the elements show an increase or decrease in the oxidation numbers .
Answer for this question is no element decreases its oxidation number.

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3 years ago

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Does changing the number of neutrons change the identity of the element you have built ?

worty [1.4K]

Answer:

If you change the number of neutrons somehow, nothing will happen because it carry's no charge at all.

Explanation:

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4 years ago