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boyakko [2]
3 years ago
12

A chemist adds of a zinc nitrate solution to a reaction flask. Calculate the mass in kilograms of zinc nitrate the chemist has a

dded to the flask. Round your answer to significant digits.
Chemistry
1 answer:
Ray Of Light [21]3 years ago
8 0

Answer:

5.3 × 10⁻³ kg

Explanation:

There is some info missing. I think this is the original question.

<em>A chemist adds 135.0 mL of a 0.21 M zinc nitrate (Zn(NO₃)₂) solution to a reaction flask. Calculate the mass in kilograms of zinc nitrate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.</em>

<em />

We have 135.0 mL of a 0.21 M zinc nitrate (Zn(NO₃)₂) solution. The moles of zinc nitrate are:

0.1350 L × 0.21 mol/L = 2.8 × 10⁻² mol

The molar mass of zinc nitrate is 189.36 g/mol. The mass corresponding to 2.8 × 10⁻² moles is:

2.8 × 10⁻² mol × 189.36 g/mol = 5.3 g

1 kilogram is equal to 1000 grams. Then,

5.3 g × (1 kg/1000 g) = 5.3 × 10⁻³ kg

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2.3 Zinc has five naturally occurring isotopes: 48.63% of 64 Zn with an atomic weight of 63.929 amu; 27.90% of 66Zn with an atom
lakkis [162]

<u>Answer:</u> The average atomic mass of element Zinc is 65.40 amu.  

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i .....(1)

  • <u>For _{30}^{64}\textrm{Zn} isotope:</u>

Mass of _{30}^{64}\textrm{Zn} isotope = 63.929 amu

Percentage abundance of _{30}^{64}\textrm{Zn} isotope = 48.63 %

Fractional abundance of _{30}^{64}\textrm{Zn} isotope = 0.4863

  • <u>For _{30}^{66}\textrm{Zn} isotope:</u>

Mass of _{30}^{66}\textrm{Zn} isotope = 65.926 amu

Percentage abundance of _{30}^{66}\textrm{Zn} isotope = 27.90 %

Fractional abundance of _{30}^{66}\textrm{Zn} isotope = 0.2790

  • <u>For _{30}^{67}\textrm{Zn} isotope:</u>

Mass of _{30}^{67}\textrm{Zn} isotope = 66.927 amu

Percentage abundance of _{30}^{67}\textrm{Zn} isotope = 4.10 %

Fractional abundance of _{30}^{67}\textrm{Zn} isotope = 0.0410

  • <u>For _{30}^{68}\textrm{Zn} isotope:</u>

Mass of _{30}^{68}\textrm{Zn} isotope = 67.925 amu

Percentage abundance of _{30}^{68}\textrm{Zn} isotope = 18.75 %

Fractional abundance of _{30}^{68}\textrm{Zn} isotope = 0.1875

  • <u>For _{30}^{70}\textrm{Zn} isotope:</u>

Mass of _{30}^{70}\textrm{Zn} isotope = 69.925 amu

Percentage abundance of _{30}^{70}\textrm{Zn} isotope = 0.62 %

Fractional abundance of _{30}^{70}\textrm{Zn} isotope = 0.0062

Putting values in equation 1, we get:

\text{Average atomic mass of Zinc}=[(63.929\times 0.4863)+(65.926\times 0.2790)+(66.927\times 0.0410)+(67.925\times 0.1875)+(69.925\times 0.0062)]

\text{Average atomic mass of Zinc}=65.40amu

Hence, the average atomic mass of element Zinc is 65.40 amu.

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3 years ago
What is tins atomic symbol
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Tin
Chemical Element
Tin is a chemical element with the symbol Sn and atomic number 50. It is a main group metal in group 14 of the periodic table. Wikipedia
Symbol: Sn
Electron configuration: [Kr] 4d105s25p2
Atomic number: 50
Melting point: 449.5°F (231.9°C)
Atomic mass: 118.71 u
Boiling point: 4,717°F (2,603°C)
Electrons per shell: 2, 8, 18, 18, 4
4 0
4 years ago
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(26) All atoms area...<span><span>with the number of protons equaling the number of electrons

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4 years ago
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A mineral consisted of 29.4% calcium, 23.5% sulfur and 47.1% oxygen. What is the empirical formula?
inessss [21]
Step  one  calculate  the  moles  of  each  element
that  is  moles= %  composition/molar  mass
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moles  of      Ca = 29.4 /40g/mol=0.735 moles,      S= 23.5/32  =0.734 moles,  O= 47.1/16= 2.94   moles

calculate  the mole  ratio  by  dividing   each  mole with  smallest  mole   that   is  0.734
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3 years ago
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