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Elis [28]
3 years ago
13

Woh want to do a long distance relationship with me

Chemistry
1 answer:
nikitadnepr [17]3 years ago
5 0
Love how the subject is chemistry HAHA. I live in FL :/
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Moles are in 4.3 * 10^22 molecules of H3PO4
Helga [31]

Answer:

The answer is 0.073

Explanation:

Final step: 0.73 x 10^-1

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What is the average density of the Earth's inner core?
eduard
The answer is A hopefully this help you
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90 grams of glucose, C6H12O6, is dissolved in 2 liters of solution. Determine the molarity (M).
Masja [62]

Answer:

0.25 M

Explanation:

The first step is to determine how many moles of glucose there are, which means first finding out the molar mass of glucose. Carbon has a molar mass of about 12, hydrogen 1, and oxygen 16, so the total molar mass of glucose is about 6(12)+12(1)+6(16)=180 grams. This means that 90 grams is half a mole. Half a mole of substance divided by two liters of solution is (1/2)/2=1/4, or a molarity of 0.25. Hope this helps!

7 0
4 years ago
Pls help i am in 8th grade k12
zheka24 [161]
It’s definitely persuade!
4 0
3 years ago
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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