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KIM [24]
3 years ago
13

SORRY BUGGING ALL YALL AGAIN NEED HELP DUMB SORRY IF YOUR READING THIS U KNOW WHO U R D

Chemistry
2 answers:
natita [175]3 years ago
8 0

Answer:

hope this helps

please like and Mark as brainliest

Explanation:

  1. reduce eliminate
  2. interdependent
  3. mutually beneficial, competitive and predatory
garik1379 [7]3 years ago
7 0

Answer:

Dont think your dumb because your not you simply just dont understand the question

Explanation:

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mina [271]

Answer:

85

Explanation:

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3 years ago
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

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If you add a solution of NaOH to a solution of H₂CO₃, two reactions occur, one after the other. Write the chemical equations for
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We have a solution of NaOH and H₂CO₃

First, NaOH will dissociate into Na⁺ and OH⁻ ions

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The H⁺ released from the substitution will bond with the OH⁻ ion to form a water molecule

If there were to be another NaOH molecule, a similar substitution will take place, substituting the second hydrogen from H₂CO₃ as well to form Na₂CO₃

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Solve the following equation and check for extraneous solutions. Show all work. Thank you.

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The molecule I2 has a lower melting point than Cl2.<br><br> TRUE<br><br> FALSE
Gala2k [10]
True. <span><span>Melting points decrease down a group and increase across a period.</span> </span>
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