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3 years ago

Emma strikes a match and sets some logs on fire. As the fire burns, which two forms of energy are released?

Chemistry

1 answer:

Ivanshal [37]3 years ago

7 0

Answer:

Explanation:

There are changes in the chemical composition of the word.

There's light energy given off.

There's heat energy given up.

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anastassius [24]

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5 0

3 years ago

If you pour water into a hot metal pan, the water will absorb heat energy from the pan. Will the water come to the same temperat

Marina CMI [18]

Answer:

Yes

Explanation:

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4 0

3 years ago

How can one determine that a redox reaction will be nonspontaneous?

fenix001 [56]

Answer:

A redox reaction is spontaneous if the standard electrode potential for the redox reaction, Eo(redox reaction), is positive. ...

If Eo(redox reaction) is positive, the reaction will proceed in the forward direction (spontaneous)

Explanation:

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8 0

3 years ago

Let us assume that fe(oh)2(s) is completely insoluble, which signifies that the precipitation reaction with naoh(aq) (presented

Yuki888 [10]

17.8 mL NaOH

Step 1. Write the chemical equation

Fe^(2+) + 2NaOH → Fe(OH)2 + 2Na^(+)

Step 2. Calculate the moles of Fe^(2+)

Moles of Fe^(2+) = 500 mL Fe^(2+) × [0.0230 mmol Fe^(2+)]/[1 mL Fe^(2+)]

= 11.50 mmol Fe^(2+)

Step 3. Calculate the moles of NaOH

Moles of NaOH = 11.50 mmol Fe^(2+) × [2 mmol NaOH]/[1 mmol Fe^(2+)]

= 23.00 mmol NaOH

Step 4. Calculate the volume of NaOH

Volume of NaOH = 23.00 mmol NaOH × (1 mL NaOH/1.29 mmol NaOH)

= 17.8 mL NaOH

3 0

4 years ago

Consider the following reaction:

zzz [600]

Answer:

Question 1: Both the rate of reaction of H₂O₂ and the rate of formation of H₂O are:

$6.6\cdot 10^{-3}mol/(liter.s)$

Question 2: 0.60 moles of O₂ are formed in the first 50 s of reaction.

Explanation:

The coefficients of the balanced chemical equation tells the relation of formation of the reactants into the products, which is also the relation between their rates of reaction.

The balanced chemical equation (given) is:

$2H_2O_2(aq)\rightarrow 2H_2O(l)+O_2(g)$

Then, in the same time that 2 moles of H₂O₂ react, 2 moles of H₂O and 1 mol of O₂ are formed.

1. Question 1.

That means that the rate of formation of O₂ is half the rate of reaction of H₂O₂ and half the rate of formation of H₂O.

Hence, if the instantaneous rate of formation of O₂ is 3.3×10⁻³moles/(liters×seconds), then the rate of reaction of H₂O₂ and the rate of formation of H₂O is twice:

$2\times 3.3\cdot 10^{-3}mol/(liter.s)=6.6\cdot 10^{-3}mol/(liter.s)$

2. Question 2.

First you must find how much the concentration has changed.

You can read the initial concentration of H₂O₂ on the graph. It is the y-intercept. It is 1.0M.

You can read the approixmate concentration of H₂O₂ at the time 50 s. It is about 0.2M.

Then the change in concentration in the first 50 s of reaction is about 1.0M - 0.2M = 0.8M.

Now you can convert the concentration 0.8M into number of moles, using the volume (1.5 liters), assuming the liquid volume does not change, and the molarity definition (molarity is the number of moles of solute in one liter of solution)

$\Delta [H_2O_2]=\Delta moles/Volume(liters)$

$0.8M=\Delta moles/1.5liter\\\\\Delta moles=0.8M\times 1.5liter=1.2mol$

Hence, 1.2 mol of H₂O₂ have reacted in the first 50s of reaction.

From the coefficients of the chemical equation, the number of moles of O₂ formed is half the number of moles of H₂O₂ that react.

Thus, 1.2mol/2 = 0.60 moles of O₂ are formed in the first 50 s of reaction.

7 0

3 years ago