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Artemon [7]
3 years ago
10

What is the pH when [H+] = 0.000032 M?

Chemistry
1 answer:
notka56 [123]3 years ago
5 0

Explanation:

pH =  -  log[H {}^{ + } ] \\  =  -  log(0.000032)  \\  = 4.5

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Balloons may contain tiny microscopic holes, which allow the gas particles within to escape via effusion. Suppose a 4.30 L ballo
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Answer:

The balloon filled with Helium will effuse more quickly.

Explanation:

Step 1: Data given

Balloon 1: Has a volume 4.3L , a pressure of 1.68 atm and is filled with 1.21g of Helium

Balloon 2: Has also a volume of 4.3 L and a pressure of 1.68 atm but is filled with 39.63 grams of Xenon

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The balloon filled with Helium will effuse more quickly.

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3 years ago
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Alcohol is poured into a beaker that contains 2% saltwater. Which are the solutes in the new solution?
Vadim26 [7]
A 2% saltwater is a dilute solution in which the solvent is water and the solute is salt. The solvent is the component of the solution which comprise a large percentage while the solute comprise the smaller part. When alcohol is added, the solute still is salt and the solvent is the mixture of water and alcohol.
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When do you identify your variables?
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Option numberrrr C I THINK
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Which of these properties would be possessed by both acids and bases? note: we do not mean mixing both in the same solution. we
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Answer is: <span>3. they conduct electricity.
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3) Acids and bases give ions in water solutions, so <span>they conduct electricity.
</span>

An Arrhenius base is a substance that dissociates in water to form hydroxide ions (OH⁻<span>). 
In this example lithium hydroxide is an Arrhenius base:</span>

LiOH(aq) → Li⁺(aq) + OH⁻(aq).

An Arrhenius acid is a substance that dissociates in water to form hydrogen ions or protons (H⁺<span>). 
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5) bases have pH greater than 7, acids have pH less than 7.

5 0
3 years ago
2 moles of a gas have a pressure of 2 atm at a temperature of 300 K. What is the volume of the gas?
qwelly [4]
  • No of moles=2mol=n
  • Pressure=2atm=P
  • Temperature=300K=T
  • Volume=V

\\ \tt\hookrightarrow PV=nRT

\\ \tt\hookrightarrow V=\dfrac{nRT}{P}

\\ \tt\hookrightarrow V=\dfrac{2(8.314)(300)}{2}

\\ \tt\hookrightarrow V=8.314(300)

\\ \tt\hookrightarrow V=2494.2mL

\\ \tt\hookrightarrow V=24.94dm^3

4 0
2 years ago
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