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3 years ago

explain why an aqueous solution of Na2SO4 is neutral while an aqueous solution of Na2CO3 is basic in nature

Chemistry

1 answer:

exis [7]3 years ago

5 0

The latter contains hydroxyl ions when dissolved in water making it basic.

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silver has an atomic mass of 107.868 amu. silver has two common isotopes. one of the isotopes has a mass of 106.906 amu and a re

Sunny_sXe [5.5K]

A. The abundance of the 2nd isotope is 48.119%

B. The mass of the 2nd isotope is 108.905 amu

Let the 1st isotope be A

Let the 2nd isotope be B

A. Determination of the abundance of the 2nd isotope

Abundance of isotope A = 51.881%.

<h3>Abundance of isotope B =? </h3>

Abundance of B = 100 – A

Abundance of B = 100 – 51.881

<h3>Abundance of B = 48.119%</h3>

B. Determination of the mass of the 2nd isotope

Atomic mass of silver = 107.868 amu.

Mass of 1st isotope (A) = 106.906 amu

Abundance of isotope A (A%) = 51.881%.

Abundance of isotope B (B%) = 48.119%

<h3>Mass of 2nd isotope (B) =? </h3>

$atomic \: mass = \frac{mass \: of \:A \times \:A\%}{100} + \frac{mass \: of \:B \times \:B\%}{100} \\ \\ 107.868 = \frac{106.906\times \ \: 51.881}{100} + \frac{mass \: of \:B \times \:48.119}{100} \\ \\ 107.868 = \: 55.464 + 0.48119 \times mass \: of \:B \\ \\ collect \: like \: terms \\ \\ 0.48119 \times mass \: of \:B = 107.868 - 55.464 \\ \\ divide \: both \: side \: by \: 0.48119 \\ \\ mass \: of \:B = \frac{107.868 - 55.464 }{0.48119} \\ \\ mass \: of \:B =108.905 \: amu \\ \\$

Therefore, the mass of the 2nd isotope is 108.905 amu

Learn more: brainly.com/question/7955048

6 0

2 years ago

Which two statements about redox reactions are true?

Mars2501 [29]

Answer:

Explanation:

Reduction is a gain of electrons, oxidation is a loss of electrons, and electron transfer reactions are also called redox reactions.

this link might be helpful

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Oxidation-Reduction_Reactions

4 0

3 years ago

Read 2 more answers

Silver metal can be prepared by reducing its nitrate, AgNO3 with copper according to the following equation:

Lerok [7]

%yield = 88.5%

<h3>Further explanation</h3>

Given

Reaction

Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)

Required

The percent yield

Solution

mol AgNO₃(MW=169,87 g/mol) :

= mass : MW

= 127 : 169.87

= 0.748

mol Ag from equation :

= 2/2 x mol AgNO₃

= 2/2 x 0.748

= 0.748

Mass Ag (theoretical) :

= mol x Ar Ag

= 0.748 x 108

= 80.784

% yield = (actual/theoretical) x 100%

%yield = 71.5/80.784 x 100%

<em>%yield = 88.5%</em>

7 0

3 years ago

What is the concentration, in mass percent (m/m), of a solution prepared from 50.0 g nacl and 150.0 g of water?

Nataly [62]

Mass of solute ( m1 ) = 50.0 g

mass of solvent ( m2 ) = 150.0 g

Therefore:

m/m = ( m1 / m1 + m2 )

m/m = ( 50.0 / 50.0 + 150.0 )

m/m = ( 50.0 / 200 )

m/m = 0.25

8 0

3 years ago

Read 2 more answers

What is the volume of 120mg of water

iVinArrow [24]

Answer:

560mg

Explanation:

120mg-560mg trust dude

7 0

3 years ago