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2 years ago

Post- Assessment:

Chemistry

1 answer:

joja [24]2 years ago

6 0

Answer:

the answer is C

I just took the test

Explanation:

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What is heating and cooling

melisa1 [442]

make or become hot or warm is heating.

cooling is to become colder or cool down from a higher temp.

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3 years ago

A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after

choli [55]

Answer:

9.25

Explanation:

Let first find the moles of $NH_4Cl$ and $NaOH$

number of moles of $NH_4Cl$ = 0.40 mol/L × 200 × 10⁻³L

= 0.08 mole

number of moles of $NaOH$ = 0.80 mol/L × 50 × 10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

$NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}$

The ICE Table is shown below as follows:

$NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}$

Initial (M) 0.08 0.04 0

Change (M) - 0.04 -0.04 + 0.04

Equilibrium (M) 0.04 0 0.04

$K_a*K_b = 10^{-14} \ at \ 25^0C$

$K_a = \frac{10^{-14}}{K_b}$

$K_a = \frac{10^{-14}}{1.76*10^{-5}}$

$K_a= 5.68*10^{10}$

$pK_a = - log \ (K_a)$

$pK_a = - log \ (5.68*10^{-10})$

$pK_a = 9.25$

$pH = pKa + \ log (\frac{HB}{HA} )$ for buffer solutions

$pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} )$ since they are in the same solution

$pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )$

$pH = 9.25$

8 0

2 years ago

Which change occurs when an atom in excited state returns to ground state? 1 energy is emitted 2energy is absorbed 3 the number

kozerog [31]

The answer is number 1 - energy is emitted

Energy is released when an atom in an excited state returns to the ground state.

7 0

2 years ago

Read 2 more answers

A chemist measures the amount of chlorine gas produced during an experiment. She finds that of chlorine gas is produced. Calcula

Trava [24]

Answer:

0.779 moles.

Explanation:

Mass of chlorine/number of moles = molar mass of chlorine.

Molar mass of chlorine = (35.5*2)

= 71 g/mol.

Mass = 55.3 g of chlorine.

Number of moles = 55.3/71

= 0.779 moles.

8 0

2 years ago

In an experiment, a student needs 250.0 mL of a 0.100 M copper (II) chloride solution. A stock solution of 2.00 M copper (II) ch

LekaFEV [45]

Answer : The volume of stock solution needed are, 12.5 mL

Explanation :

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the initial molarity and volume of copper (II) chloride.

$M_2\text{ and }V_2$ are the final molarity and volume of stock solution of copper (II) chloride.

We are given:

$M_1=0.100M\\V_1=250.0mL\\M_2=2.00M\\V_2=?$

Putting values in above equation, we get:

$0.100M\times 250.0mL=2.00M\times V_2\\\\V_2=12.5mL$

Hence, the volume of stock solution needed are, 12.5 mL

8 0

2 years ago