Answer:
36.09% is the mass percent composition of calcium in calcium chloride
Explanation:
Mass percent is defined as one hundred times the ratio between the mass of a compound (In this case, Calcium), and the total mass of the sample:
<em>Mass Percent = Mass compound / Total mass of the sample × 100</em>
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Computing the values of the problem:
Mass Percent = 0.690g / 1.912g * 100
Mass percent =
<h3>36.09% is the mass percent composition of calcium in calcium chloride</h3>
Answer:
Boron
Explanation:
An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.
All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other. For example if neutral atom has 6 protons than it must have 6 electrons. The sum of neutrons and protons is the mass number of an atom while the number of protons are number of electrons is the atomic number of an atom.
For example
In given atomic model we can see that there are five electrons out side the nucleus it means this atomic model is of that element which have atomic number 5 and it is boron.
It has 5 electron 5 proton and 6 neutrons.
It means its atomic number is five and mass number is 5+6 = 11
Most scientific questions are developed from Observations.
Whats the question? Im not sure what your asking
Answer:
0.42 M
Explanation:
The reaction that takes place is:
- Cu(CH₃COO)₂ + Na₂CrO₄ → Cu(CrO₄) + 2Na(CH₃COO)
First we <u>calculate the moles of Na₂CrO₄</u>, using the <em>given volume and concentration</em>:
(200 mL = 0.200L)
- 0.70 M * 0.200 L = 0.14 moles Na₂CrO₄
Now we <u>calculate the moles of Cu(CH₃COO)₂</u>, using its <em>molar mass</em>:
- 40.8 g ÷ 181.63 g/mol = 0.224 mol Cu(CH₃COO)₂
Because the molar ratio of Cu(CH₃COO)₂ and Na₂CrO₄ is 1:1, we can directly <u>substract the reacting moles of Na₂CrO₄ from the added moles of Cu(CH₃COO)₂</u>:
- 0.224 mol - 0.14 mol = 0.085 mol
Finally we <u>calculate the resulting molarity</u> of Cu⁺², from the <em>excess </em>cations remaining:
- 0.085 mol / 0.200 L = 0.42 M
