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DedPeter [7]
3 years ago
5

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 +

Chemistry
1 answer:
nikitadnepr [17]3 years ago
5 0

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⁻

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Montano1993 [528]

Answer:

\boxed {\boxed {\sf 232.9 \textdegree C}}

Explanation:

This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:

\frac {V_1}{T_1}= \frac{V_2}{T_2}

The volume of the gas starts at 250 milliliters and the temperature is 137 °C.

\frac{250 \ mL}{137 \textdegree C}= \frac{V_2}{T_2}

The volume of the gas is increased to 425 milliliters, but the temperature is unknown.

\frac{250 \ mL}{137 \textdegree C}= \frac{425 \ mL}{T_2}

We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.

250 \ mL * T_2 = 137 \textdegree C * 425 \ mL

Now the variable is being multiplied by 250 milliliters. The inverse of multiplication is division. Divide both sides of the equation by 250 mL.

\frac{250 \ mL * T_2}{250 \ mL}=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}

T_2=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}

The units of milliliters (mL) cancel.

T_2=\frac{ 137 \textdegree C * 425 }{250 }

T_2= \frac{58225}{250} \textdegree C

T_2=232.9 \textdegree C

The temperature changes to <u>232.9 degrees Celsius.</u>

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2 years ago
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mihalych1998 [28]

Answer:

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3 years ago
PLEASE HELP ME!!! NEED THE ANSWER FAST! WILL CHOOSE THE BRAINLIEST!!!
Vlad [161]
Remember that a cation will be smaller than its neutral atom, and an anion will be larger than its neutral atom. This would automatically eliminate answer choices A and D.

Also keep in mind that atomic radii decreases from left to right as you move along a periodic table. It also decreases from bottom up. 
Atomic radii increases as you move from right to left and as you go from up to down.

As bromine is higher up in the periodic table than Iodine, it would have a smaller radius. Iodine would have a larger radius.

The correct answer is B. Br
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3 years ago
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kompoz [17]
I’m pretty sure the answer is B
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Need some help, please. Explain why anions are always larger than the atoms from which they are derived, while cations are alway
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The question requires us to explain the differences in radii of neutral atoms, cations and anions.

To answer this question, we need to keep in mind that a neutral atom presents the same number of protons (positive particles) and electrons (negative particles). Another important information is that the protons are located in the nucleus of the atom, while the electrons are around the nucleus. Also, there is an electrostatic force between protons and electrons, which means that they the protons tend to attract the electrons to the nucleus.

While a neutral atom presents the same number of protons and electrons, a cation is an ion with positive charge, which means it has lost one or more electrons. In a cation, the balance between protons and electrons doesn't exist anymore: now, there is more positive than negative charge (more protons than electrons), and the overall attractive force that the protons have for the electrons is increased. As a result, the electrons stay closer to the nucleus and the radius of a cation is smaller than the neutral atom from which it was derived.

On the other side, anions present negative charge, which means they have received electrons. Similarly to cations, the balance between protons and electrons doesn't exist anymore, but in this case, there are more electrons than protons. In an anion, the overall attractive force that the protons have for the electrons is decreased. As a result, the electrons are "more free" to move and, as they are not so attracted to the nucleus, they tend to stay farther from the positive nucleus compared to the neutral atom - because of this, the radius of an anion is larger than the neutral atom from which it was derived.

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1 year ago
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