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Degger [83]
2 years ago
5

Students investigating chemical reactions conducted the experiment you see here. An antacid tablet was dropped into a plastic ba

g containing 10 ml distilled water at 22oC. The bag was immediately zipped closed and the chemical reaction proceeded until completion. Based on your knowledge of chemical reactions, choose all of the statements below would apply?
A) the tablet completely disappeared.
B) The total mass at the completion of the experiment is 16.72g.
C) The mass of the products is slightly less than the mass of the reactants.
D) A decrease in the mass of the products in the result of the production of a gas.
E) Evidence of a chemical reaction includes the production of something new: a gas
Chemistry
1 answer:
zhannawk [14.2K]2 years ago
7 0

Answer: B and E

Explanation:

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Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t
Rasek [7]

Answer:

4.6*10^{-14} M

Explanation:

Concentration of Cu^{2+} = [Cu(NO_3)_2] = 0.020 M

Constructing an ICE table;we have:

                                 Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}

Initial (M)             0.020          0.40                        0

Change (M)         -  x                - 4 x                       x

Equilibrium (M)  0.020 -x        0.40 - 4 x              x

Given that: K_f =1.7*10^{13}

K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}

1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}

Since x is so small; 0.40 -4x = 0.40

Then:

1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}

1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}

1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x

8.704*10^9-4.352*10^{11}x =x

8.704*10^9 = 4.352*10^{11}x

x = \frac{8.704*10^9}{4.352*10^{11}}

x = 0.0199999999999540

Cu^{2+}= 0.020 - 0.019999999999954

Cu^{2+} = 4.6*10^{-14} M

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3 years ago
Choose the substance with the highest vapor pressure at a given temperature.
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<span>SiS2 ..... solid </span>


<span>Ethanol solid --> ethanol melts --> ethanol liquid </span>
<span>-135C ---------------> -114C --------------> -50C </span>
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3 years ago
Bromine (63 g) and fluorine (60 g) are mixed to give bromine trifluoride. a) Write a balanced chemical reaction. b) What is the
fiasKO [112]

Answer:

a) Br2 + 3F2 → 2BrF3

<u>b) Br2 is the limiting reactant</u>.

c) There will be formed 86.3 grams of BrF3

d) There will remain <u>0.326 moles of F2 = 12.97 grams</u>

<u />

Explanation:

Step 1: The balanced equation

Br2 + 3F2 → 2BrF3

Step 2: Given data

Mass of Bromine = 63grams

Mass of fluorine = 60 grams

percent yield = 80%

Molar mass of bromine = 79.9 g/mol =

Molar mass of fluorine = 19 g/mol

Molar mass of bromine trifluoride = 136.9 g/mol

Step 3: Calculating moles

Moles Br2 = 63 grams / (2*79.9)

Moles Br2 = 0.394 moles

Moles F2 = 60 grams / (2*19.9)

Moles F2 = 1.508

For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

<u>Br2 is the limiting reactant</u>. It will completely be consumed (0394 moles).

There will react 3*0.394 = 1.182 moles of F2

There will remain 1.508 - 1.182 = <u>0.326 moles of F2 = 12.97 grams</u>

Step 5: Calculate moles of BrF3

For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

So there is 2*0.394 moles = 0.788 moles of BrF3 moles produced

Step 6: Calculate mass of BrF3

mass = Moles * Molar mass

mass of BrF3 = 0.788 moles * 136.9 g/mol = 107.88 grams = Theoretical yield

Step 7: Calculate actual yield

% yield = 0.80 = actual yield / theoretical yield

actual yield = 0.80 * 107.88 grams = 86.3 grams

actual yield = <u>86.3 grams</u>

<u />

<u />

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3 years ago
What is the molarity of ammonia if it is 1.0 ionized?
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Answer:

A (0.0018 M

Explanation:

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3 years ago
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