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posledela
3 years ago
10

Please fill in the blanks.

Chemistry
1 answer:
svetoff [14.1K]3 years ago
4 0

photosynthesis: the process when plants make their own food using sunlight

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1. Calculate how many moles of glycine are in a 130.0-g sample of glycine.2. Calculate the percent nitrogen by mass in glycine.
Alexxx [7]

Answer:

n=1.732mol

\% N=18.7\%

Explanation:

Hello!

In this case, since the molecular formula of glycine is C₂H₅NO₂, we realize that the molar mass is 75.07 g/mol; thus, the moles in 130.0 g of glycine are:

n=130.0g*\frac{1mol}{75.07 g}\\\\ n=1.732mol

Furthermore, we can notice 75.07 grams of glycine contains 14.01 grams of nitrogen; thus, the percent nitrogen turns out:

\% N=\frac{14.01}{75.07}*100\% \\\\\% N=18.7\%

Best regards!

4 0
3 years ago
Which one is an accurate description of an erosion?
Sergeeva-Olga [200]

the process of eroding or being eroded by wind, water, or other natural agents

6 0
3 years ago
Ob] The electrodes of a fuel cellare in contact with water and all
iogann1982 [59]
Iron rusts when in contact with water.
8 0
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

4 0
3 years ago
What is the pH of a solution made by mixing 30.00 mL 0.10 M HCl with 40.00 mL of 0.10 M KOH? Assume that the volumes of the solu
nirvana33 [79]

Answer:

pH = 12.15

Explanation:

To determine the pH of the HCl and KOH mixture, we need to know that the reaction is  a neutralization type.

HCl  +  KOH  →  H₂O  +  KCl

We need to determine the moles of each compound

M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl

M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH

The base is in excess, so the HCl will completely react and we would produce the same mmoles of KCl

HCl  +  KOH  →  H₂O  +  KCl

3 m       4 m                       -

             1 m                      3 m

As the KCl is a neutral salt, it does not have any effect on the pH, so the pH will be affected, by the strong base.

1 mmol of KOH has 1 mmol of OH⁻, so the [OH⁻] will be 1 mmol / Tot volume

[OH⁻] 1 mmol / 70 mL = 0.014285 M

- log [OH⁻] = 1.85 → pH = 14 - pOH → 14 - 1.85 = 12.15

3 0
3 years ago
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