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emmainna [20.7K]
3 years ago
5

Can someone help its not hard. What are the cons of using a cup made of plastic?

Chemistry
1 answer:
slavikrds [6]3 years ago
7 0

Answer:

it could burn if a hot liquid is poured into it and it could break easily

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When you dissolve an antacid tablet in water and it produces bubbles, what kind of change is this?
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Explanation:

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Sodium carbonate is commonly known as washing soda and is used in detergents as a water softener. What is the molarity of 5.30 g
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0.1 m

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After reading the list of physical properties above, you realize that they are describing a substance that is a
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What conversion factor does a balanced chemical equation give you?
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From google but i can explain further if needed. <span> The </span>balanced<span> equation for the reaction of interest contains the stoichiometric ratios of the reactants and products; these ratios </span>can<span> be used as </span>conversion factors<span> for mole-to-mole </span>conversions<span>.</span>
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3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
3 years ago
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