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3 years ago

Please help I’ll mark brainleist

Chemistry

2 answers:

Setler79 [48]3 years ago

8 0

Answer:

SAme. What's le question, I will answer correctly.

Explanation:

kozerog [31]3 years ago

7 0

Answer: with what?

Explanation:

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How many moles of iron (Fe) will be produced from 6.20 moles of carbon monoxide (CO) reacting with excess iron (III) oxide (Fe2O

bonufazy [111]

Answer:

4.13 moles of Fe

Explanation:

Step 1: Write the balanced equation

3 CO + Fe₂O₃ ⇒ 2 Fe + 3 CO₂

Step 2: Establish the appropriate molar ratio

According to the balanced equation, the molar ratio of CO to Fe is 3:2.

Step 3: Calculate the moles of Fe formed from 6.20 moles of CO

We will use the previously established molar ratio.

6.20 mol CO × 2 mol Fe/3 mol CO = 4.13 mol Fe

8 0

3 years ago

Why is sulphur more reactive than chlorine

Marina86 [1]

Answer:

Because the most reactive elements are<em> HALOGENS</em>.

4 0

3 years ago

Read 2 more answers

What is the oxidation of KCL?

Assoli18 [71]

Rules:

1) The oxidation number of a monatomic ion equals the charge of the ion.

2)The oxidation number of a Group 1 element in a compound is +1.

3) The oxidation number of a Group 17 element in a binary compound is -1.

4) The sum of the oxidation numbers of all the atoms in a neutral compound is 0.

KCl is an ionic compound. Consists of K+ ions and Cl- ions.

Oxidation of K= +1 (rules 1 and 2)

Oxidation of Cl= -1 (rules 1 and 3)

Rule 4 also applies- +1 + (-1)= 0- a neutral compound

Hope this helps :)

Explanation:

3 0

4 years ago

Calculate ΔHo for the following reaction ussing the given bond dissociation energiesCH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)BOND

Mazyrski [523]

Answer:

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

Explanation:

$CH_4(g) + 2O_2(g)\rightarrow CO_2(g) + 2H_2O(g) ,$ ΔH° = ?

We are given with:

$\Delta H_{O-O}=142 kJ/mol$

$\Delta H_{O=O}=498 kJ/mol$

$\Delta H_{H-O}=459 kJ/mol$

$\Delta H_{C-H}=411 kJ/mol$

$\Delta H_{C-O}=358 kJ/mol$

$\Delta H_{C=O}=799 kJ/mol$

ΔH° =

(Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)

$\Delta H^o=(1 mol\times 4\times \Delta H_{C-H}+2 mol\times 1\times \Delta H_{O=O})-(1 mol\times 2\times \Delta H_{C=O}+2 mol\times 2\times\Delta H_{H-O})$

$\Delta H^o=(1 mol\times 4\times 411 kJ/mol+2 mol\times 1\times 498 kJ/mol)-(1 mol\times 2\times 799 kJ/mol+2 mol\times 2\times 459 kJ/mol)$

$\Delta H^o=-794kJ$

$\Delta H^o>0$ endothermic reaction

$\Delta H^o$ exothermic reaction

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

4 0

4 years ago

Given the formula 4f10, what does the 10 stand for?

egoroff_w [7]

I have not idea sorry

5 0

3 years ago