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4 years ago

Select the correct answer.

Chemistry

1 answer:

konstantin123 [22]4 years ago

6 0

Answer:

A. not wearing goggles when breaking a rock with a hammer

Explanation:

You could get extremely hurt if you are not wearing goggles while doing lab work.

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Plz help i need the answer ASAP If 6.20 cm3 of a 6.75 M (M = mol/dm3) solution are diluted to 85.6 cm3 with water, what is the c

Brilliant_brown [7]

Answer:

Option A. 6.75 (6.20/85.6)

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 6.2cm³

Initial concentration (C1) = 6.75M

Final volume (V2) = 85.6cm³

Final concentration (C2) =.?

The final concentration can be obtained by using the dilution formula as show below:

C1V1 = C2V2

6.75 x 6.2 = C2 x 85.6

Divide both side by 85.6

C2 = (6.75 x 6.2) /85.6

C2 = 6.75 (6.2 / 85.6).

6 0

3 years ago

Write the orbital notation and electron configuration for the following elements. (Worksheet #2)

Mademuasel [1]

Answer:

the answer is B

Explanation:

4 0

3 years ago

Read 2 more answers

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant o

sattari [20]

Answer:

4.26 %

Explanation:

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

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10 ⁻⁴.

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50 × 10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

$[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M$

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

$\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%$

4 0

4 years ago

Aspirin (C9H8O4) is synthesized by reacting salicylic acid (C7H6O3) with acetic anhydride (C4H6O3). The balanced equation isC7H6

prisoha [69]

Answer:

For a: The mass of acetic anhydride needed is 73.91 grams.

For b: The theoretical yield of aspirin is 130.43 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For a:

Given mass of salicylic acid = $1.00\times 10^2g=100g$

Molar mass of salicylic acid = 138.121 g/mol

Putting values in equation 1, we get:

$\text{Moles of salicylic acid}=\frac{100g}{138.121g/mol}=0.724mol$

For the given chemical reaction:

$C_7H_6O_3+C_4H_6O_3\rightarrow C_9H_8O_4+CH_3COOH$

By stoichiometry of the reaction:

1 mole of salicylic acid reacts with 1 mole of acetic anhydride.

So, 0.724 moles of salicylic acid will react with = $\frac{1}{1}\times 0.724=0.724mol$ of acetic anhydride.

Now, to calculate the mass of acetic anhydride, we use equation 1:

Moles of acetic anhydride = 0.724 moles

Molar mass of acetic anhydride = 102.09 g/mol

Putting values in equation 1, we get:

$0.724mol=\frac{\text{Mass of acetic anhydride}}{102.09g/mol}\\\\\text{Mass of acetic anhydride}=73.91g$

Hence, the mass of acetic anhydride needed is 73.91 grams.

For b:

By stoichiometry of the reaction:

1 mole of salicylic acid is producing 1 mole of aspirin.

So, 0.724 moles of salicylic acid will produce = $\frac{1}{1}\times 0.724=0.724mol$ of aspirin.

Now, to calculate the mass of aspirin, we use equation 1:

Moles of aspirin = 0.724 moles

Molar mass of aspirin = 180.158 g/mol

Putting values in equation 1, we get:

$0.724mol=\frac{\text{Mass of aspirin}}{180.158g/mol}\\\\\text{Mass of aspirin}=130.43g$

Hence, the theoretical yield of aspirin is 130.43 grams.

4 0

3 years ago

Write a balanced equation and ka expression for the bronsted-lowry acid hcooh in water

statuscvo [17]

The given acid is HCOOH, it is dissociated in water. This balanced equation represents the dissociation of HCOOH in water:

HCOOH + H2O ---> H+ + COOH-

The Ka of HCOOH is:

Ka = concentration of products / concentration of reactants

Ka = [COOH-] [H+] / [HCOOH]

3 0

3 years ago

Read 2 more answers