PLZ HELP!

A scientist analyzes four samples of compounds. He finds the first compound contains 10 grams of sodium, 211 grams of potassium,

Zarrin [17]

Answer:

yea yea

Explanation:

6 0

2 years ago

How many moles of atoms are in 6.00 g of 13C?

Lyrx [107]

Answer: 0.462 moles

Explanation: 13C indicates an isotope of carbon and its mass number is 13. It means the mass of 1 mol of 13C is 13 gram.

The question asks to calculate the number of atoms present in 6.00 grams of 13C.

To calculate the number of moles we divide the given grams by the mass of 1 mol of the element. The set could be shown easily using dimensional analysis as:

$6.00gram(\frac{1mol}{13gram})$

= 0.462 moles

So, there will be 0.462 moles of atoms in 6.00 grams of 13C.

4 0

2 years ago

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What’a the structure of 4-ethyl-2,2-dimethylheptane?

Slav-nsk [51]

PubChem CID: 142982

Chemical Names: 4-Ethyl-2,2-dimethylhexane; Hexane, 4-ethyl-2,2-dimethyl-; 52896-99-8; 2,2-dimethyl-4-ethylhexane; AC1L3M80; Hexane,4-ethyl-2,2-dimethyl- More...

Molecular Formula: C10H22

Molecular Weight: 142.286 g/mol

InChI Key: QHLDBFLIDFTHQI-UHFFFAOYSA-N

7 0

2 years ago

What is the predicted change in the boiling point of water when 1.50 g of

dezoksy [38]

Answer:

0.00735°C

Explanation:

By seeing the question, we can see the elevation in boiling point with addition of BaCl₂ in water

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$\textsf {While} \: \sf {\Delta T_b} \: \textsf{expression is used} \\ \textsf {for elevation of boiling point}$

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The elevation in boiling point is a phenomenon in which there is increase in boiling point in solution, when the particular type of solute is added to pure solvent.

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$\sf \large \underline{The \: formula \: to \: be \: used \: in \: this \: question \: is} \\ \boxed{T_b = i \times K_b \times m}$

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Where 'i' is van't hoff factor which represents the ratio of observed osmotic pressure and the value to be expected.

and 'i' is 3 (as given in the question)

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'Kb' is molal boiling point constant. And it's value is 0.51°C/mol(given in question)

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'm' represent the molality of solution. Molatity is no. of moles of solution present in 1kg of solution.

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To find molality, we have to divide no. of moles of solute by weight of solution

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While first we need to no. of moles

$\sf \implies no. \: of \: moles = \frac{weight \: of \: solute}{molar \: mass \: of \: solute} \\ \\ \implies \sf no. \: of \: moles = \frac{1.5}{208.23} \\ \\ \sf \implies no. \: of \: moles = 0.0072$

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Now, we will find molality

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$\sf \hookrightarrow molality = \frac{no.\: of \: moles}{weight \: of \: solution} \\ \\ \sf \hookrightarrow molality = \frac{0.072}{1.5} \\ \\ \sf \hookrightarrow molality = 0.048 \: mol {kg}^{ - 1}$