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3 years ago

How many formula units are in 0.25 mole of NaCl?

Chemistry

2 answers:

BlackZzzverrR [31]3 years ago

6 0

Formula Units = mole x Avogadro's Number

Formula Units of NaCl = mole NaCl x Avogadro's Number

Formula Units of NaCl = 0,25 mole x 6,022 x 10²³ molecule

Formula Units of NaCl = 1,5055 x 10²³ molecule

goblinko [34]3 years ago

3 0

Answer:

The meaning and usefulness of the mole Page 2 2 • One mole of NaCl contains 6.022 x 1023 NaCl formula units. Use the mole quantity to count formulas by weighing them. The mass of an atom in amu is numerically the same as the mass of one mole of atoms of the element in grams.

Explanation:

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Margaret [11]

Answer:

sodium oxide Nao
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3 years ago

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All of the substances on the periodic table are classified as elements except

zzz [600]

Answer: all elements in the periodic table is classified as elements

Explanation:

The structure of the table shows periodic trends. The seven rows of the table, called periods, generally have metals on the left and nonmetals on the right. The columns, called groups, contain elements with similar chemical behaviours. Six groups have accepted names as well as assigned numbers: for example, group 17 elements are the halogens; and group 18 are the noble gases

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4 years ago

You can practice converting between the mass of a solution and mass of solute when the mass percent concentration of a solution

LUCKY_DIMON [66]

Answer: The mass of solution having 768 mg of KCN is 426.66 grams.

Explanation:

We are given:

0.180 mass % of KCN solution.

0.180 %(m/m) KCN solution means that 0.180 grams of KCN is present in 100 gram of solution.

To calculate the mass of solution having 768 mg of KCN or 0.786 g of KCN (Conversion factor: 1 g = 1000 mg)

Using unitary method:

If 0.180 grams of KCN is present in 100 g of solution.

So, 0.768 grams of KCN will be present in = $\frac{100g}{0.180g}\times 0.768=426.66g$ of solution.

Hence, the mass of solution having 768 mg of KCN is 426.66 grams.

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4 years ago

Which is a compound that contains carbon?

Anastaziya [24]

carbon dioxide, that is what I found

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3 years ago

Two students are given the starting material benzoic acid and are asked to prepare benzaldehyde. The first student starts by ref

Likurg_2 [28]

Answer:

Compound A: Benzoyl chloride

Compound B: Benzaldehyde - (tBuO)₃Al complex

Compound C: Benzaldehyde

Compound D: Benzyl alcohol

Explanation:

The lithium tri-tert-butoxyaluminum hydride that the first student used is a milder reagent than LAH and will stop reacting at the aldehyde.

The LAH that the second student used is much more reactive and will continue to reduce the benzoic acid as far as possible, going all the way to the alcohol.

See the attachment for the reaction steps.

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3 years ago