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JulsSmile [24]
2 years ago
10

4. The vet instructed Manuel to give his

Chemistry
1 answer:
djverab [1.8K]2 years ago
3 0

Answer:

=60 milligrams

Explanation:

12 x 5

=60 milligrams

Have a nice day!!!!!!! :-)

<u>KA</u>

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One glass of water is 85 degrees F and another is 40 degrees F. When an Alka Seltzer tablet is dropped into each glass, at the s
Andreyy89

Answer:

the cold will get hot and the hot will get cold

Explanation:

7 0
2 years ago
Calculate the mass of aluminum in 500 g of Al(C2H3O2)3
Luden [163]
Answer: 66.2 g

Explanation:

1) The ratio of Al in the molecule is  1 mol to 1 mol .

2) The mass of 1 mol of molecules of Al (CH2H3O2)3 is the molar mass of the compound.

3) You calculate the molar mass of the compound using the atomic masses of each atom, in this way:

Al: 27 g/mol
C: 2 * 3 * 12 g/mol = 72 g/mol
H: 3 * 3 * 1 g/mol = 9 g/mol
O: 2 * 3 * 16 g/mol = 96 g/mol

Molar mass = 27 g/mol + 72 g/mol + 9 g/mol + 96 g/mol = 204 g/mol

4) Set a proportion:

    27 g/mol                x
-------------------- =  ----------
   204 g/mol            500 g

5) Solve for x:

x = 500 g * 27 g/mol / 204 g/mol = 66.2 g
5 0
3 years ago
By titration, it is found that 28.5 mL of 0.183 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the
maksim [4K]

Answer:

0.209M

Explanation:

M1V1=M2V2

(28.5 mL)(0.183M)=(25.0mL)(M)

M2= 0.209M

*Text me at 561-400-5105 for private tutoring if interested: I can do homework, labs, and other assignments :)

4 0
2 years ago
A student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equa
Elina [12.6K]

<u>Answer:</u> The identity of the unknown acid is butanoic acid or ascorbic acid.

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}

Molarity of NaOH solution = 0.570 M

Volume of solution = 39.55 mL

Putting values in above equation, we get:

0.570M=\frac{\text{Moles of NaOH}\times 1000}{39.55}\\\\\text{Moles of NaOH}=\frac{0.570\times 39.55}{1000}=0.0225mol

The chemical equation for the reaction of NaOH and monoprotic acid follows:

NaOH+HX\rightarrow NaX+H_2O

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of HX

So, moles of monoprotic acid = 0.0225 moles

The chemical equation for the reaction of NaOH and diprotic acid follows:

2NaOH+H_2X\rightarrow 2NaX+2H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of diprotic acid

So, moles of diprotic acid = \frac{0.0225}{2}=0.01125moles

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

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

  • <u>For butanoic acid:</u>

Mass of butanoic acid = 2.002 g

Molar mass of butanoic acid = 88 g/mol

Putting values in above equation, we get:

\text{Moles of butanoic acid}=\frac{2.002g}{88g/mol}=0.02275mol

  • <u>For L-tartaric acid:</u>

Mass of L-tartaric acid = 2.002 g

Molar mass of L-tartaric acid = 150 g/mol

Putting values in above equation, we get:

\text{Moles of L-tartaric acid}=\frac{2.002g}{150g/mol}=0.0133mol

  • <u>For ascorbic acid:</u>

Mass of ascorbic acid = 2.002 g

Molar mass of ascorbic acid = 176 g/mol

Putting values in above equation, we get:

\text{Moles of ascorbic acid}=\frac{2.002g}{176g/mol}=0.01137mol

As, the number of moles of butanoic acid and ascorbic acid is equal to the number of moles of acid getting neutralized.

Hence, the identity of the unknown acid is butanoic acid or ascorbic acid.

5 0
3 years ago
The volume of air in a person’s lungs is 615 mL at a pressure of 760. mmHg. Inhalation occurs as the pressure in the lungs drops
chubhunter [2.5K]

<u>Answer:</u> The final volume of lungs is 621.5 mL

<u>Explanation:</u>

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.

The equation given by this law is:

P_1V_1=P_2V_2

where,

P_1\text{ and }V_1 are initial pressure and volume.

P_2\text{ and }V_2 are final pressure and volume.

We are given:

P_1=760mmHg\\V_1=615mL\\P_2=752mmHg\\V_2=?mL

Putting values in above equation, we get:

760mmHg\times 615mL=752mmHg\times V_2\\\\V_2=\frac{760\times 615}{752}=621.5mL

Hence, the final volume of lungs is 621.5 mL

8 0
3 years ago
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