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Vinvika [58]
3 years ago
8

8.03 Solutions Lab Report Does anyone have a PDF or Document of FLVS 8.03 Solutions Lab

Chemistry
2 answers:
GarryVolchara [31]3 years ago
7 0

8.03 solutions report is described below.

Explanation:

8.03 Solutions Lab Report

In this laboratory activity, you will investigate how temperature, agitation, particle size, and dilution affect the taste of a drink. Fill in each section of this lab report and submit it and your pre-lab answers to your instructor for grading.  

Pre-lab Questions:

In this lab, you will make fruit drinks with powdered drink mix. Complete the pre-lab questions to get the values you need for your drink solutions.  

Calculate the molar mass of powered fruit drink mix, made from sucrose (C12H22O11).

Using stoichiometry, determine the mass of powdered drink mix needed to make a 1.0 M solution of 100 mL.

-BARSIC- [3]3 years ago
5 0

Answer:

I need it too.

Explanation:

Did you ever do it?

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Lady bird [3.3K]

Answer:

Molecules vibrate in a fixed position.

Explanation:

first one is just wrong.

third is liquid

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4 0
3 years ago
A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi
Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of Na = 5.0 g

Mass of Cl_2 = 10.0 g

Molar mass of Na = 23 g/mol

Molar mass of Cl_2 = 71 g/mol

First we have to calculate the moles of Na and Cl_2.

\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}

\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol

and,

\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}

\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

2Na+Cl_2\rightarrow 2NaCl

From the balanced reaction we conclude that

As, 2 mole of Na react with 1 mole of Cl_2

So, 0.217 moles of Na react with \frac{0.217}{2}=0.108 moles of Cl_2

From this we conclude that, Cl_2 is an excess reagent because the given moles are greater than the required moles and Na is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaCl

From the reaction, we conclude that

As, 2 mole of Na react to give 2 mole of NaCl

So, 0.217 mole of HCl react to give 0.217 mole of NaCl

Now we have to calculate the mass of NaCl

\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl

Molar mass of NaCl = 58.5 g/mole

\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g

Now we have to calculate the percent yield of this reaction.

Percent yield = \frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = \frac{5.24g}{12.7g}\times 100

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

4 0
3 years ago
What is the concentration of a sodium hydroxide solution if 14.5 mL of it are exactly neutralized by 30.0 mL of a 0.500 M hydroc
rusak2 [61]

Answer:

1.03 M

Explanation:

Step 1: Write the balanced equation

NaOH + HCl ⇒ NaCl + H₂O

Step 2: Calculate the reacting moles of HCl

30.0 mL (0.0300 L) of 0.500 M HCl react.

0.0300 L × 0.500 mol/L = 0.0150 mol

Step 3: Calculate the moles of NaOH that react with 0.0150 moles of HCl

The molar ratio of NaOH to HCl is 1:1. The moles of NaOH that react are 1/1 × 0.0150 mol = 0.0150 mol.

Step 4: Calculate the molar concentration of NaOH

0.0150 moles of NaOH are in 14.5 mL (0.0145 L).

M = 0.0150 mol/0.0145 L = 1.03 M

4 0
3 years ago
How many moles of oxygen are required to produce 37.15 g CO2?
denpristay [2]
Carbon = 12.010. Oxygen = 15.999 x 2 15.999 x 2 = 31.998 + 12.010 = 44.008 \frac{37.15 grams * 1 mole CO2}{44.008 grams}
8 0
3 years ago
Read 2 more answers
Which of the following is the SI unit used to measure mass? a. kilogram b. liter c. meter d. Kelvin Please select the best answe
Ket [755]

Answer: d

Explanation: because it’s right

5 0
3 years ago
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