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4 years ago

Consider the following balanced equation:

1 answer:

4 0

Zn⁰ + 2H⁺ ------> Zn²⁺ + H2⁰

H⁺ ion has oxidation number +1.
Zn²⁺ ion has oxidation number +2.
Atom of Zn has electric charge 0, and each hydrogen atom in the molecule H2 have oxidation number 0. So, Zn and each hydrogen atom in H2 have oxidation numbers equals "0".

Answer is
D. Zn and each hydrogen atom in H2

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3 years ago

Which of the following is not true about elements? 1. Elements are made of atoms of the same type. 2. Each element has a unique

Marat540 [252]

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2 years ago

200.0 mL of 3.85 M HCl is added to 100.0 mL of 4.6 M barium hydroxide. The reaction goes to completion. What is the concentratio

Ede4ka [16]

Answer:

2.387 mol/L

Explanation:

The reaction that takes place is:

2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

First we calculate how many moles of each reagent were added:

HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl

Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂

460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.

Now we calculate how many moles of Ba(OH)₂ reacted, by converting the total number of HCl moles to Ba(OH)₂ moles:

203.85 mmol HCl * $\frac{1mmolBa(OH)_{2}}{2mmolHCl}$ = 101.925 mmol Ba(OH)₂

This means the remaining Ba(OH)₂ is:

460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂

There are two OH⁻ moles per Ba(OH)₂ mol:

OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻

Finally we divide the number of OH⁻ moles by the total volume (100 mL + 200 mL):

716.15 mmol OH⁻ / 300.0 mL = 2.387 M

So the answer is 2.387 mol/L

7 0

3 years ago

Please help me. Quickly!!!

Tomtit [17]

Answer:

500

Explanation:

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3 years ago

So I saw this question: If 28.0 grams of Pb(NO3)2 react with 18.0 grams of NaI, what mass of PbI2 can be produced? Pb(NO3)2 + Na

Nataly_w [17]

mass of PbI₂ = 27.6606 g

<h3>Further explanation</h3>

Given

Pb(NO₃)₂ + NaI → PbI₂ + NaNO₃

28.0 grams of Pb(NO₃)₂ react with 18.0 grams of NaI

Required

mass of PbI₂

Solution

Balanced equation

Pb(NO₃)₂ + 2NaI → PbI₂ + 2NaNO₃

The principle of a balanced reaction is the number of atoms in the reactants = the number of atoms in the product

mol Pb(NO₃)₂ :

= 28 : 331,2 g/mol

= 0.0845

mol NaI :

= 18 : 149,89 g/mol

= 0.12

Limiting reactant : mol : coefficient

Pb(NO₃)₂ : 0.0845 : 1 = 0.0845

NaI : 0.12 : 2 = 0.06

NaI limiting reactant (smaller ratio)

mol PbI₂ based on NaI

= 1/2 x 0.12 = 0.06

Mass PbI₂ :

= 0.06 x 461,01 g/mol

= 27.6606 g

4 0

3 years ago