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Aliun [14]
4 years ago
4

How could you safely determine if a base is stronger than an acid?​

Chemistry
1 answer:
JulsSmile [24]4 years ago
6 0

Answer:

If you compare an acid's pKa and a base's pKb, the thing with the lower value is “stronger” in terms of acid or base. If you mix equal amounts of a base with a lower pKb with a substance that has a higher pKa, the resulting solution would be acidic.

Explanation:

If you’re trying to observe a battle of wills between an acid and a base, whichever one forms a solution (of equal molarity and volume, in deionized water) with a pH further away from 7 is stronger. I say equal molarity and volume because this means that an equal number of ions of each substance would be floating around in an equal amount of water.

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How much of a sample remains after three half-lives have occurred? 1/16 of the original sample 1/9 of the original sample 1/8 of
Drupady [299]
For this question, assume that you have 1 compound. This compound is divided in half once, so you are left with 0.5. That 0.5 that remains is divided in half again, this is the second half-life, and you are left with 0.25. The final half life involves dividing 0.25 in half, which means you are left with 0.125. For the answer to make sense, you need to know your conversions between decimals and fractions. To make it simple, if you have 0.125 and you times it by 8, you are left with your initial value of 1. Therefore, after three half-lives, you are left with 1/8th of the compound.
5 0
3 years ago
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for the substituted cyclohexane compound shown, identify the atoms that are trans to the bromo substituent.
hodyreva [135]

Substituents are trans to each other if the appear on opposite sides in the structure. C, E, G, H and K are trans to the bromo substituent.

The substituted cyclohexane compound is shown in the image attached. Two substituents in a compound are trans to each other if they appear on opposite sides of the structure. Cis substituents appear on the same side of the structure.

Considering the structure as shown in the image attached, the substituents that are trans to the bromo substituent are the substituents shown as letters C, E, G, H and K. They lie opposite to the bromo substituent.

Learn more: brainly.com/question/8155254

6 0
3 years ago
If a 22.4 L volume of a sample of gas has a density of 0.900 grams/L at 1.00 atm and 0.00°C. Given
disa [49]

Answer:

Formula Weight of gas sample = 20.1 g/mole => Neon (Ne)

Explanation:

Use Ideal Gas Law formula to determine formula weight and compare to formula weights of answer choices.

PV = nRT = (mass/fwt)RT => fwt = (mass/Volume)RT = Density x R x T

Density = 0.900 grams/L

R = 0.08206 L·atm/mole·K

T = 0.00°C = 273Kfwt = (0.900g/L)(0.08206L·atm/mole·K )(273K)

= 20.1 g/mol => Neon (Ne)

4 0
3 years ago
A tank at is filled with of dinitrogen monoxide gas and of boron trifluoride gas. You can assume both gases behave as ideal gase
blsea [12.9K]

Answer:

(1). Mole fraction = 0.152 for sulfur tetrafluoride gas.

Mole fraction = 0.848 For dinitrogen monoxide gas.

(2). Partial Pressure for dinitrogen monoxide gas = 187 kPa

Partial Pressure for sulfur tetrafluoride gas = 33.4 kpa.

(3). Total Partial Pressure = 220.4 kpa.

Explanation:

So, we are given the following data or parameters or information in the question above;

• Volume of the tank = 5.00L per tank;

• Temperature of the tank = 7.03°C;

• The mass of the content in the tank =

17.7g of dinitrogen monoxide gas and

7.77g of sulfur tetrafluoride gas.

So, we will be making use of the formulae below to calculate the MOLE FRACTION:

Moles, n= mass/molar mass and mole fraction = n(1)/ n(1) + n(2) per each constituents.

Moles, n1 = 17.7g of dinitrogen monoxide gas/ 44 grams per mole. =0.4023 moles.

Moles, n2 = 7.77g of sulfur tetrafluoride gas/ 108.1 grams per mole. = 0.07188 moles.

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Mole fraction = 0.07188/0.47415 = 0.152 of sulfur tetrafluoride gas.

PART TWO: CALCULATE THE PARTIAL PRESSURE AND TOTAL PRESSURE BY USING THE FORMULA BELOW;

pressure × volume = number of moles × gas constant, R × temperature.

Pressure = n × R × T/ V.

For dinitrogen monoxide gas. ;

Partial Pressure = 0.4023 × 8.314 × 280.03 / 5 × 10^-3 = 187 kPa.

For sulfur tetrafluoride gas

Partial Pressure = 0.07188 × 8.314 ( × 280.03 / 5 × 10^-3. = 33.4 kpa.

(3). Total pressure = (187 + 33.4)kpa = 220.4 kpa

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3 years ago
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