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lawyer [7]
3 years ago
10

1.

Chemistry
1 answer:
Diano4ka-milaya [45]3 years ago
3 0

Answer:

The advantage of net ionic equations is that they show only those species that are directly involved in the reaction

Explanation:

You might be interested in
Solve for ‘x’: 12(x-7)+3(2x+2)=50x-2
Alex Ar [27]

Answer:

Explanation:

Hi there!

12(x-7) = 12x -84

3(2x+2) = 6x+6

Total 18x-78

18x-78 = 50x-2

-78= 32x-2

-76= 32x

19/8 =x

Hope you understand

Have a nice day,

Wolfiex.

5 0
4 years ago
Use the following terms to fill in the blanks:
Alenkinab [10]

Answer:

1. Molality.

2. Concentration.

3. Percent by mass.

4.  Mole fraction.

Explanation:

Molarity is the measurement by which the number of moles is measured by a solute. The solution here is one liter respectively.

Concentration is the product which is emerged when the amount of the solute is divided by the amount o the solvent or solution.

Mole fraction refers to the number of component moles divided by the number of the solution moles.

7 0
3 years ago
Calculate the molarity of a solution containing 5.6 mol of sodium hydroxide dissolved in 2.1 L of water
yKpoI14uk [10]

2.0 MOL is the correct answer I believe


5 0
4 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
A gas in a 6.2 mL cylinder has a pressure of 1.4 atmospheres. A piston is pushed in
Ronch [10]

Answer:

A) 2.8 atm

Explanation:

3 0
3 years ago
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