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oksano4ka [1.4K]
3 years ago
14

Number of electrons present in 3.6mg of NH₄⁺?​

Chemistry
1 answer:
nata0808 [166]3 years ago
4 0
0.0036g of NH4+ x 1mol of NH4+/18.04g of NH4+ x 6.02*10^23 electrons of NH4+/1mol of NH4+

= 1.20 x 10^20 electrons of NH4+
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When energy is transferred to ocean water from moving air above it what happens to most of the energy
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1- Which process are fast in carbon cycle?
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Answer:

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3 years ago
Find the no. Of electron involved in the electro deposition of 63.5g of cu from a solution of cuso4
sp2606 [1]
This is a redox reaction, meaning reduction-oxidation reaction. This represents the reaction in one side of the electrode in an electrolysis set-up. First, we find the oxidation number of Cu in CuSO4:

(ox. # of Cu)+ ox.# of S + 4(ox.# of oxygen) = 0
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The stoichiometric equation would be 2 electrons per mole Copper. Copper has a molar mass of <span>63.5 g/mol. Then, it would only need 2 electrons.


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What is the maximum number of electrons in p&amp;q shell​
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Read 2 more answers
An element has an average atomic mass of 1.008 amu. It consists of two isotopes , one having a mass of 1.007 amu, and one having
dimulka [17.4K]

Answer:

The most abundant isotope is 1.007 amu.

Explanation:

Given data:

Average atomic mass = 1.008 amu

Mass of first isotope = 1.007 amu

Mass of 2nd isotope = 2.014 amu

Most abundant isotope = ?

Solution:

First of all we will set the fraction for both isotopes

X for the isotopes having mass  2.014 amu

1-x for isotopes having mass 1.007 amu

The average atomic mass is 1.008 amu

we will use the following equation,

2.014x + 1.007  (1-x) = 1.008  

2.014x + 1.007  - 1.007 x = 1.008  

2.014x - 1.007x  =  1.008  -  1.007

1.007 x = 0.001

x= 0.001/ 1.007

x= 0.0009

0.0009 × 100 = 0.09 %

0.09 % is abundance of isotope having mass  2.014 amu because we solve the fraction x.

now we will calculate the abundance of second isotope.

(1-x)

1-0.0009 = 0.9991

0.9991 × 100= 99.91%

6 0
3 years ago
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