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3 years ago

plz help asap (plz don't give helpless answers) and also ill give brainist to those who answer the questing right

Chemistry

1 answer:

Lera25 [3.4K]3 years ago

5 0

1.) Particle B has a greater kinetic energy because when a particle is at higher temperature it contains more energy.

2.) Diagram B best shows the overflow of heat between the particles. It does because of the enclosed space that it is in and how it will circulate in it.

3.) Heat is transferred between object A and B by having object B (warmer) make a reaction with object A (colder).

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Which bond type is found in ammonium chloride. Select one: a. metallic b. covalent c. ionic

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Answer:

ionic

Explanation:

In NH4Cl molecule, ionic bond is formed between NH4+ and Cl– ions, 3 covalent bonds are formed between N and three H atoms and one coordinate bond is formed between N and 1 H atom.

HOPE IT HELPS :)

PLEASE MARK IT THE BRAINLIEST!

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Rainforests may be located in what latitude range?

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How many molecules of CaCL2 are equivalent to 75.9g CaCl2

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In an acid-base neutralization reaction 38.74 ml of 0.500 m potassium hydroxide (ki) reacts with 50.00 ml of sulfuric acid solut

g100num [7]

Answer: 0.19M

Explanation: $2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O$

Using Molarity equation:

$n_1M_1V_1=n_2M_2V_2$

$M_1$ = molarity of acid

$V_1$ = volume of acid

$M_2$ = molarity of base

$V_2$ = volume of base

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$M_1=0.19M$

Thus the concentration of the $H_2SO_4$ solution is 0.19M.

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3 years ago

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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