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Ivan
3 years ago
9

PLEASEEEE HELLPPP I CANT FAILL PLSS

Chemistry
1 answer:
sdas [7]3 years ago
4 0
It’s the second won the air layer one
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Carmen lives on a farm in an area far away from the city and surround by forests. New families are starting to move into the are
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Answer:

Carmen and her family would not have to interest crops where the land has very little rapid recovery after one always, or even in those productions where their neighbors produce the same, the ideal would be the cultivation of trees to make wood, or crops where the harvest is fast and the land can regenerate quickly.

Explanation:

Within the fields, agricultural competition is very important, since it tends to happen within commerce or in rural areas, generating diversity of productions among farmers.

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What is the reason for the trend in lonization Energy? Select all that apply.
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Explanation:

A, D,E and B i think is the correct answers  

7 0
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What is the difference between elements, mixtures, and compounds?
MAXImum [283]

Answer:

If it is pure, the substance is either an element or a compound. If a substance can be separated into its elements, it is a compound. If a substance is not chemically pure, it is either a heterogeneous mixture or a homogeneous mixture. If its composition is uniform throughout, it is a homogeneous mixture.

6 0
4 years ago
Be sure to answer all parts. For the titration of 10.0 mL of 0.250 M acetic acid with 0.200 M sodium hydroxide, determine the pH
DaniilM [7]

Explanation:

Molarity=\frac{moles}{Volume(L)}

Molarity of the acetic acid = 0.250 M

Volume of the acetic acid solution = 10.0 mL = 0.010 L( 1 mL =0.001L)

Moles of acetic acid ;

n=0.250 M\times 0.010 L=0.0025 mol

Molarity of the NaOH = 0.200 M

a) Volume of the NaOH solution = 10.0 mL = 0.010 L( 1 mL =0.001L)

Moles of NaOH : 0.200M\times 0.010 L=0.002 mol

CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O

1 mole NaOH neutralizes 1 mole of acetic acid , then 0.002 moles of NaOH will neutralize 0.002 mol of acetic acid.

Moles of acetic acid left un-neutralized = 0.0025 mol - 0.002 = 0.0005 mol

1 mole of acetic acid gives 1 mole of hydrogen ion, then 0.0005 mole of acetic acid will give 0.0005 mole of hydrogen ions.

Moles of hydrogen ion= 0.0005 mol

Volume of the solution = 0.010 L+ 0.010 L = 0.020 L

[H^+]=\frac{0.0005 mol}{0.020 L}=0.025 M

The pH of the 10.0 mL of base added to acetic acid solution :

pH=-\log[H^+]=-\log[0.025 M]=1.60

b) Volume of the NaOH solution = 12.0 mL = 0.012 L( 1 mL =0.001L)

Moles of NaOH : 0.200M\times 0.012 L=0.0024 mol

CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O

1 mole NaOH neutralizes 1 mole of acetic acid , then 0.0024 moles of NaOH will neutralize 0.0024 mol of acetic acid.

Moles of acetic acid left un-neutralized = 0.0025 mol - 0.0024 = 0.0001 mol

1 mole of acetic acid gives 1 mole of hydrogen ion, then 0.0001 mole of acetic acid will give 0.0001 mole of hydrogen ions.

Moles of hydrogen ion= 0.0001 mol

Volume of the solution = 0.010 L+ 0.012 L = 0.022 L

[H^+]=\frac{0.0001 mol}{0.022 L}=0.0045 M

The pH of the 12.0 mL of base added to acetic acid solution :

pH=-\log[H^+]=-\log[0.0045 M]=2.34

c) Volume of the NaOH solution = 15.0 mL = 0.015 L( 1 mL =0.001L)

Moles of NaOH : 0.200M\times 0.015 L=0.003 mol

CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O

1 mole NaOH neutralizes 1 mole of acetic acid , then 0.003 moles of NaOH will neutralize 0.003 mol of acetic acid.

All the moles of acetic acid will get neutralized by NaOH and un-neutralized sodium hydroxide will left over.

Moles of NaOH left un-neutralized = 0.003 mol - 0.0025 = 0.0005 mol

1 mole of NaOH gives 1 mole of hydroxide ion, then 0.0005 mole of NaOH acid will give 0.0005 mole of hydroxide ions.

Moles of hydroxide ion= 0.0005 mol

Volume of the solution = 0.010 L+ 0.015 L = 0.025 L

[OH^-]=\frac{0.0005 mol}{0.025 L}=0.02 M

The pOH of the 15.0 mL of base added to acetic acid solution :

pOH=-\log[OH^-]=-\log[0.02 M]=1.70

The pH of the 15.0 mL of base added to acetic acid solution :

pH=14-pOH=14-1.70=12.3

7 0
3 years ago
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lyudmila [28]

elements in the periodic table are arranged by the atomic number of the element

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