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katovenus [111]
2 years ago
13

What is the molar mass of CuCIO3

Chemistry
1 answer:
Genrish500 [490]2 years ago
5 0
<h3>Answer:</h3>

147.05 g/mol

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
<h3>Explanation:</h3>

<u>Step 1: Define</u>

CuClO₃

<u>Step 2: Find MM</u>

Molar Mass of Cu - 63.55 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CuClO₃ - 63.55 + 35.45 + 3(16.00) = 147.05 g/mol

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Answer: The mass in 3.75 \times 10^{21} atoms of zinc is 0.405 g.

Explanation:

Given: Atoms of zinc = 3.75 \times 10^{21}

It is known that 1 mole of every substance contains 6.022 \times 10^{23} atoms. So, the number of moles in given number of atoms is as follows.

Moles = \frac{3.75 \times 10^{21}}{6.022 \times 10^{23}}\\= 0.622 \times 10^{-2}\\= 0.0062 mol

As moles is the mass of a substance divided by its molar mass. So, mass of zinc (molar mass = 65.39 g/mol) is calculated as follows.

Moles = \frac{mass}{molar mass}\\0.0062 mol = \frac{mass}{65.39 g}\\mass = 0.405 g

Thus, we can conclude that the mass in 3.75 \times 10^{21} atoms of zinc is 0.405 g.

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2 years ago
Does anyone know why electron affinity increases as going upward in the periodic table ​
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Electron affinity increases upward for the groups and from left to right across periods of a periodic table because the electrons added to energy levels become closer to the nucleus, thus a stronger attraction between the nucleus and its electrons.

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"The molecular formula for hydrogen sulfide is H2S. What does the formula indicate about the composition of a 1.0-gram sample of
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Which of the following compounds is SOLUBLE? <br> A. SrSO4 <br> B. CaCO3 <br> C. BaS <br> D. CaCl2
andriy [413]

Answer:

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In May 2016, William Trubridge broke the world record in free diving (diving underwater without the use of supplemental oxygen)
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Answer:

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

Explanation:

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 3.6 L  

V₂ = ?

P₁ = 1.0 atm

P₂ = 13.3 atm

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{1.0\ atm}\times {3.6\ L}={13.3\ atm}\times {V_2}

{V_2}=\frac{{1.0}\times {3.6}}{13.3}\ L

{V_2}=0.27\ L

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3 years ago
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