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2 years ago

What is the [OH-] in a solution with a pol of 7.86?

Chemistry

1 answer:

yulyashka [42]2 years ago

3 0

Answer:

pH of 7.86, then the [OH-] is equal to 10^(14-7.86) = 10^6.14 M

Explanation:

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What is the name of the following compound? A line-angle formula shows a ring with six vertices and a circle inscribed. A Cl ato

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Answer: Option C. p-dichlorobenzene and 1,4-dichlorobenzene.

Explanation:

A line-angle formula with six vertices and a circle inscribed corresponds to the compound known as benzene.

Further, according to the IUPAC standards for naming benzene derivatives, you must first number the position of the substituent. In this case, the substituents (chloros) are located at the positions 1 and 4; also, for the benzene derivatives when they have 2 substituents and the positions are 1 and 4, this configuration is known as <em>para </em>or <em>p </em>configuration.

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2 years ago

I have 2 samples of solid chalk (aka calcium carbonate). Sample A has a total mass of 4.12 g and Sample B has a total mass of 19

IRINA_888 [86]

Answer:

A) Sample B has more calcium carbonate molecules

Explanation:

M = Molar mass of calcium carbonate = 100.0869 g/mol

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

For the 4.12 g sample

Moles of a substance is given by

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{4.12}{100.0869}\\\Rightarrow n=0.0411\ \text{mol}$

Number of molecules is given by

$nN_A=0.0411\times 6.022\times 10^{23}=2.48\times 10^{22}\ \text{molecules}$

For the 19.37 g sample

$n=\dfrac{19.37}{100.0869}\\\Rightarrow n=0.193\ \text{mol}$

Number of molecules is given by

$nN_A=0.193\times 6.022\times 10^{23}=1.16\times 10^{23}\ \text{molecules}$

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So, sample B has more calcium carbonate molecules.

The ratio of the elements of carbon, oxygen, calcium atoms, ions, has to be same in both the samples otherwise the samples cannot be considered as calcium carbonate. Same is applicable for impurities. If there are impurites then the sample cannot be considered as calcium carbonate.

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3 years ago

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