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2 years ago

What mass of O2 is needed to completely react with 500 g of Fe

Chemistry

1 answer:

Zina [86]2 years ago

8 0

Answer:

solution given:

Explanation:

4Fe +3O2------------>2Fe2O3

4mole. 3mole. 2 mole

224g. 96g. 320g

we have

224 g of fe needed to react completely with 96 g of O2

500g of fe needed to react completely with

96 ×500/224=214.3g of O2

214.3g of O2 is a required answer.

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aleksandr82 [10.1K]

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2 years ago

According to the text, which amino acid(s) contains a side chain..

solniwko [45]

Answer:

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Explanation:

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2 years ago

Read 2 more answers

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

⠀

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

⠀

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.

⠀

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

⠀

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)

⠀

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

⠀

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

⠀

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

⠀

While first we need to no. of moles

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

⠀

Now, we will find molality

⠀

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

⠀

$\textsf{ \large{ \underline{Now substituting the required values}}}$

⠀

$\sf \longmapsto \Delta T_b = 3 \times 0.51 \times 0.0048 \\ \\ \\ \boxed{ \tt{ \longmapsto \Delta T_b =0.00735{ \degree}C}}$

⠀

Henceforth, the change in boiling point is 0.00735°C.

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1 year ago

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Law Incorporation [45]

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2 years ago

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Answer:

it dependes on the material

Explanation:

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