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nalin [4]
3 years ago
14

Is mixing salt and pepper physical change or chemical change ?

Chemistry
1 answer:
Rom4ik [11]3 years ago
6 0
Yes mixing salt with pepper change
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28. The IUPAC name for OH is
Jet001 [13]

Answer:

A

Explanation:

5 0
3 years ago
Identify the type of intermolecular force for water, vegetable oil, and corn syrup.
aleksandrvk [35]

Answer:

Intermolecular force for oil the dipole-dipole.

Intermolecular force for water hydrogen bonds.

Intermolecular force for vegetable oil nonpolar compound.

Intermolecular force for corn syrup fructose.  

5 0
3 years ago
Air in a closed cylinder is heated from 25 degree to 36 degree. If the initial pressure is 3.80 atm what is the final pressure ?
aleksklad [387]
Aww man, I love these types of problems!!!
So, it's a simple formula:
P1*V1/T1=P2*V2/T2

In this case, we remove V(volume) as there is none listed, leaving this:
P1/T1=P2=T2

Think you got it from here? All you gotta do is plug in the numbers and solve
If you have any further problems, just let me know!!
3 0
3 years ago
Read 2 more answers
The compound sodium thiosulfate pentahydrate, Na2S2O3 • 5H2O
Mnenie [13.5K]

The theoretical yield is 204.4 g while the percent yield is 2.57%.

<h3>What is theoretical yield?</h3>

Theoretical yield is the amount of product obtained based on the stoichiometry of the reaction.

S8(s) + 8 Na2SO3(aq) + 40 H2O(l) --->8 Na2S2O3·5 H2O(s)

Number of moles of sulfur =  3.25 g /8(32) = 0.013 moles

Number of moles of sodium sulfite =  13.1 g/126 g/mol = 0.103 moles

Since 1 moles of sulfur reacts with 8 moles of sodium sulfite

0.013 moles reacts with 0.013 moles ×  8 moles /1 mole = 0.104 moles

There is not enough sodium sulfite hence it is the limiting reactant.

1 mole of sodium sulfite yields 8 moles of product

0.103 moles of sodium sulfite yields  0.103 moles × 8 moles /1 mole = 0.824 moles

Mass of product = 0.824 moles × 248 g/mol = 204.4 g

percent yield =  5.26 g /204.4 g × 100/1

= 2.57%

Learn more about percent yield: brainly.com/question/2506978

4 0
2 years ago
A 0.2722 g sample of a pure carbonate, X n CO 3 ( s ) , was dissolved in 50.0 mL of 0.1200 M HCl ( aq ) . The excess HCl ( aq )
Alex73 [517]

Answer:

0.00369 moles of HCl react with carbonate.

Explanation:

Number of moles of HCl present initially = \frac{0.1200}{1000}\times 50.0 moles = 0.00600 moles

Neutralization reaction (back titration): NaOH+HCl\rightarrow NaCl+H_{2}O

According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.

So, excess number of moles of HCl present = number of NaOH added for back titration = \frac{0.0980}{1000}\times 23.60 moles = 0.00231 moles

So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles

Hence, 0.00369 moles of HCl react with carbonate.

3 0
3 years ago
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