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luda_lava [24]
3 years ago
10

How many moles of electrons is required to deposit 5.6g of iron from a solution of iron (2) tetraoxosulphate(6)

Chemistry
1 answer:
yuradex [85]3 years ago
7 0

Answer:

0.20 mol

Explanation:

Let's consider the reduction of iron from an aqueous solution of iron (II).

Fe²⁺ + 2 e⁻ ⇒ Fe

The molar mass of Fe is 55.85 g/mol. The moles corresponding to 5.6 g of Fe are:

5.6 g × 1 mol/55.85 g = 0.10 mol

2 moles of electrons are required to deposit 1 mole of Fe. The moles of electrons required to deposit 0.10 moles of Fe are

0.10 mol Fe × 2 mol e⁻/1 mol Fe = 0.20 mol e⁻

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An element crystallizes in a face-centered cubic lattice. If the length of an edge of the unit cell is 0.408 nm, and the density
V125BC [204]

Answer:

Au

Explanation:

For the density of a face-centered cubic:

Density = \dfrac{4 \times M_w}{N_A \times a^3}

where

M_w = molar mass of the compound

N_A= avogadro's constant

a^3 = the volume of a unit cell

Given that:

Density (\rho) = 19.30 g/cm³

a = 0.408 nm

a = 0.408 \times 10^{-9} \times 10^{2} \ cm

a = 4.08 \times 10^ {-8} \ cm

∴

19.3 = \dfrac{4 \times M_w}{(6.023 \tmes 10^{23})\times (4.08 \times 10^{-8})^3}

M_w= \dfrac{19.3\times (6.023 \times 10^{23})\times (4.08 \times 10^{-8})^3}{4}

M_w=197.37 \ g/mol

Thus, the molar mass of 197.37 g/mol element is Gold (Au).

4 0
3 years ago
A chemical reaction was carried out by mixing 25 g of pure CaCO3 and 0.75 mole of pure HCl to give CaCl2, H2O and CO2. a. Which
Naily [24]

hola, esta pregunta es bastante difícil pero está bien, no lo sé, lo siento :) :)

4 0
3 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

⠀

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

⠀

⠀

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

⠀

⠀

\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.

⠀

⠀

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

⠀

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

⠀

⠀

<u>Now</u><u>,</u><u> </u><u>we</u><u> </u><u>will</u><u> </u><u>find</u><u> </u><u>molality</u>

⠀

\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}}

⠀

⠀

⠀

<u>Henceforth</u><u>,</u><u> </u><u>the</u><u> </u><u>change</u><u> </u><u>in</u><u> </u><u>boiling</u><u> </u><u>point</u><u> </u><u>is</u><u> </u><u>0</u><u>.</u><u>0</u><u>0</u><u>7</u><u>3</u><u>5</u><u>°</u><u>C</u><u>.</u>

7 0
1 year ago
The ka of hf is 6.8 x 10-4. what is the ph of a 0.35 m solution of hf?
damaskus [11]
When the reaction equation is:

HF ↔ H+   +   F-

and when the Ka expression
= concentration of products/concentration of reactions

so, Ka = [H+][F-]/[HF]

when we assume:

[H+] = [F-] = X

and [HF] = 0.35 - X

So, by substitution:

6.8 x 10^-4 = X^2 / (0.35 - X) by solving for X

∴ X = 0.015 M

∴[H+] = X = 0.015

when PH = -㏒[H+]

∴PH = -㏒0.015

        = 1.8
6 0
3 years ago
What do acid formulas have in common?
maks197457 [2]

the answer is H is the cation

Explanation :

The general formula of an acid is represented as,  in which 'H' is hydrogen cation and 'X' is a non-metal or a poly-atomic anion.

For example : etc.

All the acids produces hydrogen ion,  in an aqueous solution while the base produces hydroxide ion,  in an aqueous solution.

5 0
3 years ago
Read 2 more answers
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