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2 years ago

Which equation represents a reduction half-reaction?

Chemistry

1 answer:

horsena [70]2 years ago

7 0

Answer:

Fe3+ + 3e– --------> Fe

Explanation:

Reduction refers to a gain of electrons. A specie is said to be reduced when it gains electrons or when it experiences a decrease in oxidation number. The both sentences above can be clearly seen when inspecting a reduction half equation.

Consider the redox reaction half equation;

Fe3+ + 3e– ------>Fe

We can see that Fe^3+ accepted three electrons, the oxidation number of iron thereby decreased from +3 to zero from left to right. This implies that the Fe^3+ was actually reduced according to the equation shown.

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suppose that during the icy hot lab that 65 kj of energy were transferred to 450 g of water at 20 C. What would have been the fi

balandron [24]

Answer:

The final temperature of water is 54.5 °C.

Explanation:

Given data:

Energy transferred = 65 Kj

Mass of water = 450 g

Initial temperature = T1 = 20 °C

Final temperature= T2 = ?

Solution:

First of all we will convert the heat in Kj to joule.

1 Kj = 1000 j

65× 1000 = 65000 j

specific heat of water is 4.186 J /g. °C

Formula:

q = m × c × ΔT

ΔT = T2 - T1

Now we will put the values in Formula.

65000 j = 450 g × 4.186 J /g. °C × (T2 - 20°C )

65000 j = 1883.7 j /°C × (T2 - 20°C )

65000 j/ 1883.7 j /°C = T2 - 20°C

34.51 °C = T2 - 20°C

34.51 °C + 20 °C = T2

T2 = 54.5 °C

5 0

2 years ago

Marianne designs an experiment involving electrically charged objects. She wants to know which objects will be attracted to a ne

svet-max [94.6K]

Answer:

When one object is rubbed against another, static electricity can be created. This is because the rubbing creates a negative charge that is carried by electrons. The electrons can build up to produce static electricity. For example, when you shuffle your feet across a carpet, you are creating many surface contacts between your feet and the carpet, allowing electrons to transfer to you, thereby building up a static charge on your skin. When you touch another person or an object, you can suddenly discharge the static as an electrical shock.

Similarly, when you rub a balloon on your head it causes opposite static charges to build up both on your hair and the balloon. Consequently, when you pull the balloon slowly away from your head, you can see these two opposite static charges attracting one another and making your hair stand up.

Materials

• Balloon

• An object made out of wool (such as a sweater, scarf, blanket or ball of yarn)

• Stopwatch

• A wall

• A partner (optional)

Preparation

• Blow up the balloon and tie off the end.

• Have your partner prepare to use the stopwatch.

Procedure

• Hold the balloon in a way that your hand covers as little of its surface area as possible, such as by using only your thumb and pointer finger or by gripping the balloon by its neck where it is tied off.

• Rub the balloon on the woolly object once, in one direction.

• Hold the balloon up on the wall with the side that was rubbed against the wool facing the wall, then release it. Does the balloon stay stuck on the wall? If the balloon stays stuck, have your partner immediately start the stopwatch to time how long the balloon remains bound to the wall. If the balloon does not stick, move to the next step.

• Touch the balloon to a metal object. Why do you think this is important to do?

• Repeat the above process but each time increase the number of times you rub the balloon on the woolly object. Rub the balloon in the same direction each time. (Do not rub the balloon back and forth.)

Observations and results

In general, did the balloon stick to the wall for a longer amount of time as you increased the number of times you rubbed the balloon on the woolly object?

Wool is a conductive material, which means it readily gives away its electrons. Consequently, when you rub a balloon on wool, this causes the electrons to move from the wool to the balloon's surface. The rubbed part of the balloon now has a negative charge. Objects made of rubber, such as the balloon, are electrical insulators, meaning that they resist electric charges flowing through them. This is why only part of the balloon may have a negative charge (where the wool rubbed it) and the rest may remain neutral.

When the balloon has been rubbed enough times to gain a sufficient negative charge, it will be attracted to the wall. Although the wall should normally have a neutral charge, the charges within it can rearrange so that a positively charged area attracts the negatively charged balloon. Because the wall is also an electrical insulator, the charge is not immediately discharged. However, because metal is an electrical conductor, when you rub the balloon against metal the extra electrons in the balloon quickly leave the balloon and move into the metal so the balloon is no longer attracted and does not adhere.

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6 0

2 years ago

Read 2 more answers

A mixture of CH4 and H2O is passed over a nickel catalyst at 1000 K. The emerging gas is collected in a 5.00L flask and is found

Harman [31]

Answer:

Kc = 3.74*10⁻³

Kp = 25.21

Explanation:

Step 1: Data given

Temperature = 1000 K

Volume = 5.00 L

Mass of CO = 8.62 grams

Mass of H2 = 2.60 grams

Mass of CH4 = 43.0 grams

Mass of H2O = 48.4 grams

Kc = [CO]*[H₂]³ / ([CH₄]∙*H₂O])

Kp = p(CO)*p(H₂)³ / (p(CH₄)*p(H₂O) )

Step 2: The balanced equation

CH₄ + H₂O ⇄ CO + 3 H₂

Step 3: Calculate number of moles

The number of moles of each compund in the equilibrium mixture are:

Moles = mass / molar mass

n(CH₄) = 43.0g / 16g/mol = 2.688mol

n(H₂O) = 48.4g / 18g/mol = 2.689mol

n(CO) = 8.62g/28g/mol = 0.308mol

n(H₂) = 2.60g / 2g/mol = 1.3mol

Step 4: Calculate concentrations at equilibrium

So the equilibrium concentrations are:

Concentration = moles / volume

[CH₄] = 2.688mol/5L = 0.5376 M

[H₂O] = 2.689mol/5L = 0.5378M

[CO] = 0.308mol/5L = 0.0616M

[H₂) = 1.3mol/5L = 0.26M

Step 5: Calculate Kc

Kc = 0.0616 ∙ (0.26)³ / (0.5376∙0.5378) = 3.74*10⁻³

Step 5: Calculate partial pressure

Partial pressures in equilibrium can be found from ideal gas law:

p(X) = n(X)∙R∙T/V = [X]∙R∙T

=> p(CH₄) = [CH₄]∙R∙T = 0.5376mol/L * 0.082 06Latm/molK ∙ 1000K = 44.11 atm

p(H₂O) = [H₂O]∙R∙T = 0.5738mol/L * 0.082 06Latm/molK * 1000K = 44.13 atm

p(CO) = [CO]∙R∙T = 0.0616mol/L * 0.082 06Latm/molK * 1000K = 5.05atm

p(H₂) = [CO]∙R∙T = 0.26mol/L * 0.082 06Latm/molK * 1000K = 21.34atm

Step 5: Calculate Kp

Kp = p(CO)*p(H₂)³ / (p(CH₄)*p(H₂O) )

Kp = 5.05*21.34³ / (44.11*44.13 ) = 25.21

8 0

2 years ago

Strontium nitrate, Sr(NO3)2, is used in fireworks to produce brilliant red colors, Suppose we need to prepare 366.6 ml. of 0.115

natulia [17]

Answer:

$\boxed{\text{8.91 g}}$

Explanation:

1. Calculate the moles of Sr(NO₃)₂

$n = \text{366 mL} \times \dfrac{\text{0.115 mmol}}{\text{1 mL}}= \text{42.09 mmol}$

2. Calculate the mass of SrNO₃)₂

$m = \text{42.09 mmol} \times \dfrac{\text{211.63 mg}}{\text{1 mol}}= \text{8910 mg} = \text{8.91 g}\\\text{The mass of strontium nitrate required is }\boxed{\textbf{8.91 g}}$

3 0

2 years ago

Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces

Alchen [17]

The question is incomplete, here is the complete question:

Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces only nitrogen and water vapor, and in the liquid form it is easily transported. An industrial chemist studying this reaction fills a 5.0 L flask with 2.2 atm of ammonia gas and 2.4 atm of oxygen gas at 44.0°C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of nitrogen gas to be 0.99 atm.

Calculate the pressure equilibrium constant for the combustion of ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.

Answer: The pressure equilibrium constant for the reaction is 32908.46

Explanation:

We are given

Initial partial pressure of ammonia = 2.2 atm

Initial partial pressure of oxygen gas = 2.4 atm

Equilibrium partial pressure of nitrogen gas = 0.99 atm

The chemical equation for the reaction of ammonia and oxygen gas follows:

$4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(g)$