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Mice21 [21]
3 years ago
6

What is the volume of a 5.30g piece of aluminum? The density for aluminum is 2.70mL

Chemistry
1 answer:
Makovka662 [10]3 years ago
7 0

Answer:

1.96mL

Explanation:

Density = mass/volume, and rearranged to solve for volume, volume = mass/density.

So:

volume = 5.30g/2.70g/mL = 1.96mL (assuming your unit was g/mL for density)

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A sample of 9.27 g9.27 g of solid calcium hydroxide is added to 38.5 mL38.5 mL of 0.500 M0.500 M aqueous hydrochloric acid. Writ
navik [9.2K]

<u>Answer:</u> The excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 grams.  

<u>Explanation:</u>

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

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

<u>For calcium hydroxide:</u>

Given mass of calcium hydroxide = 9.27 g

Molar mass of calcium hydroxide = 74.093 g/mol

Putting values in above equation, we get:

\text{Moles of calcium hydroxide}=\frac{9.27g}{74.093g/mol}=0.125mol

  • To calculate the moles of a solute, we use the equation:

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

We are given:

Volume of hydrochloric acid = 38.5mL = 0.0385 L   (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.500 moles/ L

Putting values in above equation, we get:

0.500mol/L=\frac{\text{Moles of hydrochloric acid}}{0.0385L}\\\\\text{Moles of hydrochloric acid}=0.01925mol

  • For the given chemical equation:

2HCl(aq.)+Ca(OH)_2(s)\rightarrow CaCl_2(s)+2H_2O(l)

Here, the solid salt is calcium chloride.

By Stoichiometry of the reaction:

2 moles of hydrochloric acid reacts with 1 mole of calcium hydroxide.

So, 0.01925 moles of hydrochloric acid will react with = \frac{1}{2}\times 0.01925=0.009625moles of calcium hydroxide.

As, given amount of calcium hydroxide is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

  • Amount of excess reagent (calcium hydroxide) left = 0.125 - 0.01925 = 0.115375 moles

By Stoichiometry of the reaction:

2 moles of hydrochloric acid produces 1 mole of calcium chloride.

So, 0.01925 moles of hydrochloric acid will produce = \frac{1}{2}\times 0.01925=0.009625moles of calcium chloride.

Now, calculating the mass of calcium chloride from equation 1, we get:

Molar mass of calcium chloride = 110.98 g/mol

Moles of calcium chloride = 0.009625 moles

Putting values in equation 1, we get:

0.009625mol=\frac{\text{Mass of calcium chloride}}{110.98g/mol}\\\\\text{Mass of calcium chloride}=1.068g

Hence, the excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 grams.

5 0
3 years ago
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A 85.4 lb child has a Streptococcus infection. Amoxicillin is prescribed at a dosage of 45 mg per kg of body weight per day give
bixtya [17]

The amount of Amoxicillin dose given to the 85.4 lb child daily is determined as 1,743.3 mg.

<h3>What is the amount of Amoxicillin dose given to the child?</h3>

The amount of  Amoxicillin  dose given to the child is calculated as follows;

amount of Amoxicillin dose = weight of the child x dosage prescribed

<h3>What is the weight of the child in pounds (lb) </h3>

The weight of the child in pounds (lb) is calculated as follows;

1 lb = 0.453592 kg

85.4 lb = ?

= 85.4 x 0.453592 kg

= 38.74 kg

amount of Amoxicillin dose = 38.74 kg x 45 mg/kg

amount of Amoxicillin dose = 1,743.3 mg

Thus, the amount of Amoxicillin dose given to the 85.4 lb child daily is determined as 1,743.3 mg.

Learn more about amount of dose here: brainly.com/question/11185154

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The complete question is below:

A 85.4 lb child has a Streptococcus infection. Amoxicillin is prescribed at a dosage of 45 mg per kg of body weight per day given b.i.d. Calculate the daily dose of the child.

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B. It is shortwave.....
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Which of the following is an example of an Arrhenius base? Select one: A. HNO3(aq) B. H3PO4(aq) C. H2CO3(aq) D. H2O(l) E. none o
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Answer:

The correct choice is e. Two of the above.

Explanation:

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