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Elis [28]
11 months ago
9

WILL GIVE BRAINIEST ANSWER A chemical equation has different numbers of atoms on the left and right sides.

Chemistry
1 answer:
Alinara [238K]11 months ago
4 0

Answer:

Unbalanced

Explanation:

You need to have the same number of atoms on both sides for it to be balanced

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The ionization constant (Ka) of HF is 6.7 X 10 ^-4. Which of the following is true in a 0.1 M solution of this acid?
spin [16.1K]
The ionization equation is:

HF ⇄ H(+) + F(-)

The ionization constant is Ka = [H(+)] * [H(-)] / [HF]

=> [H(+)] * [F(-)] = Ka * [HF]

Given that Ka < 1

[H(+)] * [F(-)] < [HF]

Which is [HF] >  [H(+)] * [F(-)] the option a. fo the list of choices.
8 0
2 years ago
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HELPPPP ANSWERSS!!!<br><br> A. <br> 13<br><br> B. <br> 8<br><br> C. <br> 15<br><br> D. <br> 5
KATRIN_1 [288]

Answer:

C

Explnation:

D

4 0
2 years ago
An iron chloride compound contains 55.85 grams of iron and 106.5 grams of chlorine. What is the most likely empirical formula fo
mart [117]

Answer:

FeCl_{3}

Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

mass of Fe = 55.85 g

Molar mass of Fe = 55.85 g/mol

<u>Moles of Fe = 55.85 / 55.85 = 1</u>

mass of Cl = 106.5 g

Molar mass of Cl = 35.5 g/mol

Moles of Cl = 106.5 / 35.5 = 3

Taking the simplest ratio for Fe and Cl as:

1 : 3

The empirical formula is = FeCl_{3}

3 0
3 years ago
The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when
Natasha_Volkova [10]

Answer: 0.52V

Explanation:

Ecell = Ecell(standard) - [(0.0592 logQ)/n]

Q = product of the quotient

n = no of electrons transferred = 2

Ecell = 0.63 - [(0.0592*Log(1 / 2.0 * 10-4) / 2]

Ecell = 0.63 - 0.0194

Ecell = 0.5205V

5 0
3 years ago
For the reaction N2(g) + 2H2(g) → N2H4(l), if the percent yield for this reaction is 100.0%, what is the actual mass of hydrazin
barxatty [35]

Answer:

53.6 g of N₂H₄

Explanation:

The begining is in the reaction:

N₂(g) + 2H₂(g) → N₂H₄(l)

We determine the moles of each reactant:

59.20 g / 28.01 g/mol = 2.11 moles of nitrogen

6.750 g / 2.016 g/mol = 3.35 moles of H₂

1 mol of N₂ react to 2 moles of H₂

Our 2.11 moles of N₂ may react to (2.11 . 2) /1 = 4.22 moles of H₂, but we only have 3.35 moles. The hydrogen is the limiting reactant.

2 moles of H₂ produce at 100 % yield, 1 mol of hydrazine

Then, 3.35 moles, may produce (3.35 . 1)/2 = 1.67 moles of N₂H₄

Let's convert the moles to mass:

1.67 mol . 32.05 g/mol = 53.6 g

4 0
3 years ago
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