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3 years ago

What is the Scientific notation for 0.10050

Chemistry

2 answers:

klasskru [66]3 years ago

4 0

Answer:

1.005 x $10^{1}$

Explanation:

attashe74 [19]3 years ago

3 0

1.005 x 10^1
Move the decimal once to the right

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Explanation:

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3 years ago

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Which 5.0-milliliter sample of NH3 will take the shape of and completely fill a closed 100.0-milliliter container?(1) NH3(s) (3)

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NH3(g) will take the shape of and completely fill a closed 100.0 milliliter container.

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3 years ago

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3 years ago

In an aqueous solution, 42% of a substance dissociates to release hydronium ions. Which of the following statements is true for

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Answer: The statement it is a weak acid is true for the substance.

Explanation:

An acid that dissociates completely when dissolved in water to give hydrogen $(H^{+})$ or hydronium $(H_{3}O^{+})$ ions is called a strong acid.

For example, HCl is a strong acid.

$HCl + H_{2}O \rightarrow H_{3}O^{+} + Cl^{-}$

An acid that dissociates partially or weakly when dissolved in water to given hydrogen or hydronium ions is called a weak acid.

For example, $CH_{3}COOH$ is a weak acid.

$CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H_{3}O^{+}$

A strong base is a base which when dissolved in water then it dissociates completely to give hydroxide ions.

For example, NaOH is a strong acid.

A weak base is a base which when dissolved in water then it dissociates partially or weakly to give hydroxide ions.

For example, $NH_{3}$ is a weak base.

Hence, in an aqueous solution where 42% of a substance dissociates to release hydronium ions shows that the dissociation is less than 50%. This means that substance is dissociating weakly so, it is a weak acid.

Thus, we can conclude that the statement it is a weak acid is true for the substance.

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3 years ago

3) During the day at 27°C a cylinder with a sliding top contains 20.0 liters

tatiyna

Answer:

$T_2=12\°C$

Explanation:

Hello there!

In this case, according to the Charles' law equation which help us to understand the directly proportional relationship between volume and temperature:

$\frac{T_2}{V_2}=\frac{T_1}{V_1}$

Thus, by solving for the final temperature, T2, and making sure we use the temperatures in Kelvin, we can calculate the final temperature as shown below:

$T_2=\frac{T_1V_2}{V_1} \\\\T_2=\frac{(27+273)K*19L}{20.0L}\\\\T_2=285-273\\\\T_2=12\°C$

Best regards!

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3 years ago