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algol [13]
4 years ago
8

How are chemical formulas of binary ionic compounds generally written? a cation on left, anion on right b anion on left, cation

on right c roman numeral first, then anion, then cation d subscripts first, then ions?
Chemistry
2 answers:
guajiro [1.7K]4 years ago
8 0

Answer:  a cation on left, anion on right

Explanation: Binary ionic compounds are compounds formed by a metal and a non metal where a metal loses electron to form positively charged cation and the non metal gains electrons to form negatively charged anion.

The chemical formula of ionic compound is written by writing the cation first and then the anion.

For Example: Binary ionic compound sodium chloride is written as :

NaCl where sodium is present as cation (Na^+) and chlorine is present as anion (Cl^-)

Margarita [4]4 years ago
3 0
<span>a) cation on left, anion on right b anion on left, for example NaCl.</span>
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Answer:

Exxplanation:

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HELP!!!
vladimir2022 [97]

Answer:

2.5 × 10⁻⁵ M H₃O⁺ and 4.0 × 10⁻¹⁰ M OH⁻.

Explanation:

<em>∵ pH = - log[H₃O⁺]</em>

∴ 4.6 = - log[H₃O⁺].

∴ log[H₃O⁺] = - 4.6.

∴ [H₃O⁺] = 2.51 x 10⁻⁵.

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

[H₃O⁺] = 2.51  x 10⁻⁵ M.

∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/(2.51  x 10⁻⁵ M) = 3.98 × 10⁻¹⁰ M ≅ 4.0 × 10⁻¹⁰ M.

<em>So, the right choice is: 2.5 × 10⁻⁵ M H₃O⁺ and 4.0 × 10⁻¹⁰ M OH⁻.</em>

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4 years ago
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' A compound has the same properties as the elements it is made from.' True or false.Why?
Vladimir [108]

Answer:

False

Explanation:

That's because elements in a compound combine and become an entirely different substance with its own unique properties.

8 0
3 years ago
In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very
SVETLANKA909090 [29]

<u>Answer:</u> The final concentration of potassium nitrate is 5.70\times 10^{-6}M

<u>Explanation:</u>

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2          .......(1)

  • <u>Calculating for first dilution:</u>

M_1\text{ and }V_1 are the molarity and volume of the concentrated KNO_3 solution

M_2\text{ and }V_2 are the molarity and volume of diluted KNO_3 solution

We are given:

M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL

Putting values in equation 1, we get:

7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M

  • <u>Calculating for second dilution:</u>

M_2\text{ and }V_2 are the molarity and volume of the concentrated KNO_3 solution

M_3\text{ and }V_3 are the molarity and volume of diluted KNO_3 solution

We are given:

M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL

Putting values in equation 1, we get:

1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M

Hence, the final concentration of potassium nitrate is 5.70\times 10^{-6}M

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