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maksim [4K]
3 years ago
10

g If potassium cyanide is dissolved in water you can say that the equilibrium concentrations of potassium and cyanide ions are:

...fill in the blank 1 ...A. High ...B. Moderate ...C. Low
Chemistry
1 answer:
Anton [14]3 years ago
3 0

Answer:

High

Explanation:

Concentration in chemistry refers to the amount of substance present in a system.

If potassium cyanide is dissolved in water, we have the following equilibrium being set up in solution;

KCN(aq) ⇄K^+(aq) + CN^-(aq)

This means that the concentration of both potassium ions and cyanide ions increases.

Hence, If potassium cyanide is dissolved in water you can say that the equilibrium concentrations of potassium and cyanide ions are high in the solution.

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Why is pluto classified as a dwarf planet?
Julli [10]

Answer:

pluto is considered a dwarf planet because it did not meet the three criteria the IAU uses to define a full sized planet

Explanation:

5 0
3 years ago
If I have one mole of sulfur, how many atoms would that be?
krok68 [10]

Answer:

Atoms of sulfur = 9.60⋅g32.06⋅g⋅mol−1×6.022×1023⋅mol−1

Explanation:

because the units all cancel out, the answer is clearly a number, ≅2×1023 as required.

7 0
4 years ago
Copper has two naturally occurring isotopes with atomic masses of 62.9296 u () and 64.9278 u (). The atomic mass of copper is 63
natta225 [31]

Answer:

The abundance of first isotope is 69.15 %

The abundance of second isotope is 30.85 %

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope:

% = x %

Mass = 62.9296 u

For second isotope:

% = 100  - x  

Mass = 64.9278 u

Given, Average Mass = 63.546 u

Thus,  

63.546=\frac{x}{100}\times {62.9296}+\frac{100-x}{100}\times {64.9278}

Solving for x, we get that:

x = 69.15 %

<u>The abundance of first isotope is 69.15 %</u>

<u>The abundance of second isotope is 100 - 69.15 % = 30.85 %</u>

4 0
4 years ago
Calculate the molar mass of nitrogen gas if 0.250 g of the gas occupies 46.65 ml at stp
Mashutka [201]
<span>pre-1982 definition STP: 120 g/mol post-1982 definition STP: 122 g/mol The answer to this question depends upon which definition of STP you're using. The definition changed in 1982 from 273.15 K at 1 atmosphere to 273.15 K at 10000 pascals. As a result the molar volume of a gas at STP changed from 22.4 L/mol to 22.7 L/mol. So let's calculate the answer using both definitions and see if your text book is 35 years obsolete. First, determine the number of moles of gas you have. Do this by dividing the volume you have by the molar volume. So pre-1982: 0.04665 / 22.4 = 0.002082589 mol post-1982: 0.04665 / 22.7 = 0.002055066 mol Now divide the mass you have by the number of moles. pre-1982: 0.250 g / 0.002082589 mol = 120.0428725 g/mol post-1982: 0.250 g / 0.002055066 mol = 121.6505895 g/mol Finally, round to 3 significant figures: pre-1982: 120 g/mol post-1982: 122 g/mol These figures are insanely large for nitrogen gas. So let's see if our input data is reasonable. Looking up the density of nitrogen gas at STP, I get a value of 1.251 grams per liter. The value of 0.250 grams in the problem would then imply a volume of about one fifth of a liter, or about 200 mL. That is over 4 times the volume given of 46.65 mL. So the verbiage in the question mentioning "nitrogen gas" is inaccurate at best. I see several possibilities. 1. The word "nitrogen" was pulled out of thin air and should be replaced with "an unknown" 2. The measurements given are incorrect and should be corrected. In any case, if #1 above is the correct reason, then you need to pick the answer based upon which definition of STP your textbook is using.</span>
6 0
3 years ago
Solution to a solution of D gives a white precipitate, F.
andreev551 [17]
Yes I belive this answer is d I. This qestion
5 0
3 years ago
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