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3 years ago

Question 13 of 32

Chemistry

1 answer:

jarptica [38.1K]3 years ago

3 0

Answer:

Explanation:

this should be obvious if you read it haha

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Use tabulated standard electrode potentials to calculate the standard cell potential for the following reaction occurring in an

SVETLANKA909090 [29]

Answer : The standard cell potential of the reaction is, -1.46 V

Explanation :

The given balanced cell reaction is,

$3Pb^{2+}(aq)+2Cr(s)\rightarrow 2Cr^{3+}(aq)+3Pb(s)$

Here, chromium (Cr) undergoes oxidation by loss of electrons and act as an anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

The standard values of cell potentials are:

Standard reduction potential of lead $E^0_{[Pb^{2+}/Pb]}=-0.13V$

Standard reduction potential of chromium $E^0_{[Cr^{3+}/Cr]}=1.33V$

Now we have to calculate the standard cell potential for the following reaction.

$E^0=E^0_{cathode}-E^0_{anode}$

$E^0=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cr^{3+}/Cr]}$

$E^0=(-0.13V)-1.33V=-1.46V$

Therefore, the standard cell potential of the reaction is, -1.46 V

6 0

4 years ago

How many grams of Fe will be produced if 705.0g of H20 are produced?

fredd [130]

<h3>Answer:</h3>

2265 g Fe₃O₄

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Fe₃O₄ + 4H₂ → 3Fe + 4H₂O

[Given] 705.0 g H₂O

<u>Step 2: Identify Conversions</u>

[RxN] 4 mol H₂O → 1 mol Fe₃O₄

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Fe - 55.85 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Molar Mass of Fe₃O₄ - 3(55.85) + 4(16.00) = 231.55 g/mol

<u>Step 3: Convert</u>

Set up stoich: $\displaystyle 705.0 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{1 \ mol \ Fe_3O_4}{4 \ mol \ H_2O})(\frac{231.55 \ g \ Fe_3O_4}{1 \ mol \ Fe_3O_4})$
Multiply/Divide/Cancel units: $\displaystyle 2264.74 \ g \ Fe_3O_4$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

2264.74 g Fe₃O₄ ≈ 2265 g Fe₃O₄

5 0

3 years ago

What mass of CH3COOH is present in a 250 mL cup of 1.25 mol/L solution of vinegar?

aleksley [76]

If 1000 ml (1 L) of CH₃COOH contain 1.25 mol
let 250 ml of CH₃COOH contain x

⇒ x = $\frac{250 ml * 1.25mol}{1000 ml}$

= 0.3125 mol

∴ moles of CH₃COOH in 250ml is 0.3125 mol

Now, Mass = mole × molar mass

= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol

= 18.75 g

∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>

4 0

3 years ago

Is this statement true or false?

Sauron [17]

1. true
2. false
3. false
4. true
5.false
6. It's true but can be false because a change in state can be a physical change as well.

5 0

4 years ago

Read 2 more answers

Write 5.5 x 104 in standard form

taurus [48]

Answer: 55,000

<u>Explanation:</u>