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OLga [1]
2 years ago
12

Will ag2so4 precipitate when 100ml of .050M agno3 is mixed with 10ml of 5x10-2m na2so4 solution

Chemistry
1 answer:
Lynna [10]2 years ago
4 0

No, Ag₂SO₄ will not precipitate when 100ml of .050M AgNO3 is mixed with 10ml of 5x10-2 m Na₂SO₄ solution because the precipitate is made only in an aqueous solution.

<h3>What is precipitation?</h3>

Precipitation is the solid extract that is collect in a place. Precipitate is the concentration of the substance in a solution in a specific place.

Thus, No, Ag₂SO₄ will not precipitate when 100ml of .050M AgNO3 is mixed with 10ml of 5x10-2 m Na₂SO₄ solution because the precipitate is made only in an aqueous solution.

Learn more about precipitation

brainly.com/question/18109776

#SPJ1

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Convert 11.03 moles of calcium nitrate to grams
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6 0
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The nucleus of unstable _____ of an element will decay leading to emission of radiation.question 1 options:moleculescationsanion
Marta_Voda [28]

Isotopes of same element has different number of neutrons with different masses and having same number of protons and electrons.

Radioactive isotopes are those isotopes which are radioactive in nature. The unstable nucleus results in the radioactivity process and this process will go on until the stable isotope (element) forms.

Thus, the nucleus of unstable isotopes of an element will decay leading to emission of radiation.

6 0
3 years ago
If you dispense 40 ml of hexane, but it turns out you only need 5 ml, what should you do with the remainder?
Natasha2012 [34]

After subtracting the volume needed from the volume dispensed, we got a remainder of 35ml

<h3>Subtraction of Numbers</h3>

Given Data

  • Volume of Hexane dispensed = 40ml
  • Volume needed = 5 ml

Let us compute the amount of excess hexane/ the volume that will remain

Remainder = The difference in volume dispensed and the volume needed

Remainder = 40-5

Remainder = 35 ml

The remainder is 35ml

Learn more about subtraction of numbers here:

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7 0
2 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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