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olga nikolaevna [1]
3 years ago
7

What is meant by "citing a source"?remembering where you found the materialnaming the source of the informationkeeping the note

card on filenone of the above?
Chemistry
1 answer:
Semmy [17]3 years ago
5 0
Naming the source of information
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nadezda [96]
Primary hope this helps
8 0
3 years ago
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Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after
Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

(c) pH = 1.93

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

4 0
3 years ago
How many fluorine atoms bond with calcium to form calcium fluoride? one two three four five
Elena-2011 [213]
Calcium fluoride.
Ca is metal, F is non-metal, so they form ionic bond.
Ca as metal can form only positive ion. Ca in the second group, so the charge of Ca ion is 2+.   Ca²⁺
F is in the 17th group, so it has 7 electrons on the last level. It is non-metal, non-metal, so it has negative charge -(8-7)=-1. "8" because on the last level cannot be more than 8 electrons. F-ion is F¹⁻.

Ca²⁺  F¹⁻
Number of positive charges should be equal to number of negative charges,
Formula of calcium fluoride
CaF2.
2 atoms Fluorine bond with Calcium.
7 0
3 years ago
How will the concentration of H+ and OH− ions change when a substance with a pH 3.2 is added to water? Both H+ and OH− will incr
Lunna [17]
Also water H2O is made of H+ and OH- ions. so when an acidic substance is added to water the concentration of H+ ions increase.

8 0
2 years ago
Read 2 more answers
How much heat is added if 0.0318g of water is increased in temperature by 0.364 degrees C?
SSSSS [86.1K]

Answer:

0.04838J

Explanation:

Heat is a form of energy that is transferred from one body to another as the result of a difference in temperature between the bodies , here heat is added to the water as a result of temperature change of 0.364 degreesC

Given:change in temperature=0.364

Mass of water=0.0318g

But we need specific heat capacity of water which is

4.2 J/g°C

Then we can calculate How much heat is added by using below formula

Energy = Mass * specific heat capacity *(change in temperature)

energy =0.0318g* 4.18g*0.364

=0.04838J

8 0
3 years ago
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