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Sliva [168]
2 years ago
12

Why would a table be more suitable than a graph to show data

Chemistry
1 answer:
kow [346]2 years ago
6 0
It might be more suitable because some data cannot use a graph to show data because there is too much data.
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What is the balanced form of the following equation? Br 2 + S 2 O 3 2– + H 2 O → Br 1– + SO 4 2– + H +
nikitadnepr [17]

Answer:

4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

Explanation:

Br₂ +  S₂O₃²⁻  + H₂O  → Br⁻ + SO₄²⁻ + H⁺

This is a redox reaction:

Br₂ changes the oxidation state from 0 to -1, so it was reduced

In the S₂O₃⁻² anion S changes the oxidation state from +2 to +6 in sulfate anion. (S₂O₃⁻², it is called thiosulfate)

We have protons in the main equation, so we assume we are in acidic medium:

Br₂ + 2e⁻ → 2Br⁻         Reduction

We balanced the bromide with 2, so the bromine has gained 2 electrons.

<u>5H₂O</u> + S₂O₃²⁻ → 2SO₄²⁻ + <u>10H⁺</u> + <em>8e</em>-  Oxidation

First of all, we add 2 to the sulfate anion in the product side, in order to balance the S.

As we have 8 O in right side, and 3 O in left side, we must add 5 O. We add 5 water in the place where the O are lower (reactant side).

Now, we have 10 H, in the reactant side, so we balance the product side with protons (10 H⁺).

Sulfur changed the oxidation state from +2 to +6, so it released 4 electrons, but, if you see thiosulfate anion you have 2 sulfurs so finally it has released 8 electrons.

Electrons are unbalanced so we multiply reduction x4, and oxidation x1.

(Br₂ + 2e⁻ → 2Br⁻) . 4 = 4Br₂ + 8e⁻ → 8Br⁻

(5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + <em>8e</em>-) . 1 = STAYS THE SAME.

We sum both half reactions, to cancel the elecetrons:

4Br₂ + 8e⁻ + 5H₂O + S₂O₃²⁻  → 2SO₄²⁻ + 10H⁺ + <em>8e</em>- + 8Br⁻

Finally the balanced reaction is: 4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

5 0
3 years ago
Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying
PIT_PIT [208]

The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (K_a for HF is 6.8\times 10^{-4}.)

[HF] = 0.280 M

Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = K_a=6.8\times 10^{-4}

HF\rightleftharpoons H^++F^-

Initially

c          0            0

At equilibrium :

(c-x)      x             x

The expression of disassociation constant is given as:

K_a=\frac{[H^+][F^-]}{[HF]}

K_a=\frac{x\times x}{(c-x)}

6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}

Solving for x, we get:

x = 0.01346 M

So, the concentration of hydrogen ion at equilibrium is :

[H^+]=x=0.01346 M

The pH of the solution is ;

pH=-\log[H^+]=-\log[0.01346 M]=1.87

The pH of an 0.280 M HF solution is 1.87.

6 0
2 years ago
___ Au₂S₃ + ___ H₂ → ___ Au + ___ H₂S
Natalija [7]

Answer:

2Au₂S₃ +  6H₂ → 4Au + 6H₂S

Explanation:

Balancing:

2Au₂S₃ +  6H₂ → 4Au + 6H₂S

6 0
3 years ago
How many control(s) are in an experiment
weqwewe [10]
You can have as many controls as necessary, But they must remain equal at all times in order to get the most accurate results
6 0
3 years ago
Which of these statements are not true about a chemical reaction rate?
Licemer1 [7]
A) Temperature does not affect reaction rate
6 0
3 years ago
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