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dsp73
3 years ago
14

Can someone help me name this hydrocarbon?

Chemistry
1 answer:
ser-zykov [4K]3 years ago
3 0
2-ethyl-4,4 -dimethyl hex-1-ene.
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New Moon, First Quarter, Full Moon, Last Quarter, Waxing Crescent, Waxing Gibbous, Waning Gibbous, and Waning Crescent.

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Which would be attracted most to a magnet?
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Either iron or metal
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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
The concentration of a biomolecule inside a rod‑shaped prokaryotic cell is 0.0035 M . Calculate the number of molecules inside t
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Answer: This rod-shape prokaryotic cell has 3.740,734725‬ molecules

Explanation:

<u>Step 1 :</u> given data

Molarity of the prokaryotic cell = 0.0035 M

Length of the cell = 4.2 μm = 4.2 * 10^-6 m

diameter of the cell = 1.3 μm = 1.3 * 10^-6 m

<u>Step 2: </u>calculate volume

To calculate volume of a rod, weneef to know the radius.

V = r ² × l

The radius = half of the diameter : r = d/2 ⇒ (1.3 * 10^-6 m)/2 = 0.65 * 10^-6 m

V= (0.65 * 10^-6 m)² * 4.2 * 10^-6 m = 1.7745 * 10 ^-18 L

<u>Step 3:</u> Calculating number of moles

Number of moles = Concentration * Volume

moles = 0.0035 M * 1.7745 * 10 ^-18 L = 6.21075 * 10^-21 moles‬

<u>Step 4:</u> calculating number of molecules

1 mole contains 6.023 * 10 ^-23 molecules

6.21075 * 10^-21 moles contain : 6.21075 * 10^-21 * 6.023 * 10 ^-23 molecules = 3.740,734725‬ molecules

This rod-shape prokaryotic cell has 3.740,734725‬ molecules

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