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3 years ago

Why is lewis electron-dot diagrams more suitable than electron-shell diagrams illustrating chemical bonds

Chemistry

1 answer:

castortr0y [4]3 years ago

5 0

Lewis structures show the bonding between atoms of a molecule and the lone pairs that may exist in the molecule.

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Suppose you used 0.5 M NaOH to titrate your vinegar sample instead of 0.1 M. What effect does the concentration of base added ha

laiz [17]

Answer: the reliability will be worse

Explanation:

Suppose we used 0.5 M NaOH to titrate our vinegar sample instead of 0.1 M.

Now by using 0.5M instead of 0.1M we are increasing the concentration of NaOH,

We know that the moles used = Volume x concnetration.

so for the same no of moles, if the concentration increases, the volume decreases.

Hence it will consume less NaOH.

now Since the volume decreases, the titration volume of less number will increase the % error.

Therefore the reliability will be worse.

4 0

2 years ago

How many grams are in 1.23 x 1020 atoms of arsenic?

Alina [70]

Given the data from the question, the mass of arsenic that contains 1.23×10²⁰ atoms is 0.0153 g

<h3>Avogadro's hypothesis </h3>

6.02×10²³ atoms = 1 mole of arsenic

But

1 mole of arsenic = 75 g

Thus, we can say that:

6.02×10²³ atoms = 75 g of arsenic

<h3>How to determine the mass that contains 1.23×10²⁰ atoms</h3>

6.02×10²³ atoms = 75 g of arsenic

Therefore,

1.23×10²⁰ atoms = (1.23×10²⁰ × 75) / 6.02×10²³ atoms)

1.23×10²⁰ atoms = 0.0153 g of arsenic

Thus, 1.23×10²⁰ atoms is present in 0.0153 g of arsenic

Learn more about Avogadro's number:

brainly.com/question/26141731

6 0

1 year ago

How many Joules of energy in 3565 calories? <br> Explain

nignag [31]

Answer:

14915960J

Explanation:

brendatferreira answered it on another slide this person said they used a converter

3 0

3 years ago

Which of the following could be a long-term health effect of chemical pesticide use?

Helen [10]

The correct Answer for this question is D

3 0

2 years ago

Brass is a substitutional alloy consisting of a solution of copper and zinc. A particular sample of red brass consisting of 79.0

Bas_tet [7]

<u>Answer:</u> The molality and molarity of zinc in the solution is 4.07 m and 28.11 M respectively.

<u>Explanation:</u>

Solute is the substance which is present in smaller proportion in a mixture and solvent is the substance which is present in larger proportion in a mixture.

We are given:

(m/m) % of Cu = 79 %

This means that 79 g of copper is present in 100 grams of alloy.

(m/m) % of Zn = 21 %

This means that 21 g of zinc is present in 100 grams of alloy.

As, zinc is present in smaller proportion. So, it is solute and copper is the solvent.

<u>Calculating molality of zinc:</u>

To calculate molality of the zinc, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$m_{solute}$ = Given mass of solute (Zinc ) = 21 g

$M_{solute}$ = Molar mass of solute (Zinc) = 65.3 g/mol

$W_{solvent}$ = Mass of solvent (copper) = 79 g

Putting values in above equation, we get:

$\text{Molality of Zinc}=\frac{21\times 1000}{65.3\times 79}\\\\\text{Molality of zinc}=4.07m$

<u>Calculating molarity of zinc:</u>

To calculate volume of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = $8740kg/m^3=\frac{8740kg\frac{1000g}{1kg}}{1m^3\times \frac{10^6cm^3}{1m^3}}=\frac{8740g}{1000cm^3}=8.740g/cm^3$

(Conversion factors used are: 1 kg = 1000 g & $1m^3=10^6cm^3$ )

Mass of solution = 100 g

Putting values in above equation, we get:

$8.740g/cm^3=\frac{100.0g}{\text{Volume of zinc}}\\\\\text{Volume of zinc}=11.44cm^3$

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

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

We are given:

Molar mass of zinc = 65.3 g/mol

Volume of solution = $11.44cm^3=11.44mL$ (Conversion factor: $1cm^3=1mL$ )

Mass of zinc = 21.0 g

Putting values in above equation, we get:

$\text{Molarity of zinc}=\frac{21\times 1000}{65.3\times 11.44}=28.11M$

Hence, the molality and molarity of zinc in the solution is 4.07 m and 28.11 M respectively.

4 0

3 years ago