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Mariulka [41]
2 years ago
8

Use the PhET simulation to identify what happens to the concentrations of a dilute Drink mix mixture or a pure Drink mix solutio

n for the following scenarios. A pure Drink mix solution can be obtained by draining all the mixture, then adding Drink mix to the container as a Solution (radio button in the upper right area of the simulation). Note that the Concentration meter can be placed toward the bottom of the container or in the stream of the Drink mix. Drag the appropriate items to their respective bins.

Chemistry
1 answer:
elena-s [515]2 years ago
7 0

Hello user, you did not add a picture of an attachment to enable me help you solve this problem. Please check the attachment I added i believe it is the question that needs to be solved.

Explanation:

1. In the first column

<u>INCREASE CONCENTRATION</u>: these are the items that needs to be placed in this bin

a. Add drink mix solid to a diluted mixture of drink mix in pure water

b. Add drink mix solution to a diluted mixture of drink mix in pure water

c. Add drink mix solid to pure drink mix solution

d Evaporate water from a diluted mixture of drink mix in pure water

e Evaporate water from pure drink mix solution

2. The items below are what should be placed in this bin,<u>DECREASE CONCENTRATION:</u>

a. Add water to a diluted mixture of drink mix in pure water

b. Add water to pure drink mix solution

3. These items here should be placed here. <u>DOES NOT AFFECT CONCENTRATION:</u>

a. Add pure drink mix solution to pure drink mix solution

b. Drain the pure drink mix solution

c. Drain the diluted mixture

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Determine the number of bonding electrons and the number of nonbonding electrons in the structure of xef2. enter the number of b
lakkis [162]

Answer : The number of bonding electrons and the number of non-bonding electrons are (4, 18).

Explanation :

The number of bonding electrons and non-bonding electrons in the structure of XeF_2 is determined by the Lewis-dot structure.

Lewis-dot structure : It tell us about the number of valence electrons of an atom within a molecule and it is also shows the bonding between the atoms of a molecule and the lone-pair of electrons.

In the given structure, 'Xe' is the central atom and 'F' is the terminal atom.

Xenon has 8 valence electrons and fluorine has 7 valence electrons.

Total number of valence electrons in XeF_2 = 8 + 2(7) = 22 electrons

From the Lewis-dot structure, we conclude that

The number of electrons used in bonding = 4

The number of electrons used in non-bonding (lone-pairs) = 22 - 4 = 18

Therefore, the number of bonding electrons and the number of non-bonding electrons are (4, 18).

The Lewis-dot structure of XeF_2 is shown below.

4 0
2 years ago
Read 2 more answers
Metals are considered this if they can be made into sheets
Korvikt [17]

They are considered malleable. They can be made into sheets

Happy to help! Please mark me as the brainliest!

6 0
2 years ago
When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [C
UNO [17]

Answer:

The new molar concentration of CO at equilibrium will be  :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M

Equilibrium constant of the reaction :

K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}

K = 4

CO_2(g) + H_2(g) \rightleftharpoons H_2O(g) + CO(g)

Concentration at eq'm:

0.24 M          0.24 M                 0.48 M            0.48 M

After addition of 0.34 moles per liter of CO_2 and H_2 are added.

(0.24+0.34) M    (0.24+0.34) M  (0.48+x)M         (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}

4=\frac{(0.48+x)^2}{(0.24+0.34)^2}

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

3 0
2 years ago
A sample of 0.53 g of carbon dioxide was obtained by heating 1.31 g of calcium carbonate. what is the percent yield for this rea
Masja [62]

CaCO3(s) ⟶ CaO(s)+CO2(s) 

<span>
moles CaCO3: 1.31 g/100 g/mole CaCO3= 0.0131 </span>

<span>
From stoichiometry, 1 mole of CO2 is formed per 1 mole CaCO3, therefore 0.0131 moles CO2 should also be formed. 
0.0131 moles CO2 x 44 g/mole CO2 = 0.576 g CO2 </span>

Therefore:<span>
<span>% Yield: 0.53/.576 x100= 92 percent yield</span></span>

4 0
2 years ago
Read 2 more answers
A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (
djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

Kb = \frac{[ HSO3-][0H-]}{SO3-2}

Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation Kb=\frac{[HSO3-][OH-]}{[SO3-2]}

We noe have:

1.485*10^-^7=\frac{x*x}{(0.056-x)}

x = [OH-] = 9.11*10^-^5

pOH = -log(OH) = -log(9.11*10^-^5)

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0
2 years ago
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