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Lisa [10]
3 years ago
9

6.0 g of copper was heated from 20 degree c to 90 degree c . How much energy was used to heat cu?

Chemistry
1 answer:
Darina [25.2K]3 years ago
4 0
Copper heat capacity would be <span>0.385J/C*gram which means it needs 0.385 Joule of energy to increase 1 gram of copper temperature by 1 Celcius. The calculation would be:
energy= heat capacity *mass * temperature difference
energy= </span>0.385J/C*gram * 6g * (90-20)
<span>energy= 161.7J
 </span>
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A reaction mechanism has the following elementary step as the slow step: B2A What is the rate law for this elementary step? O ra
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Answer:

r = k [ B ]

Explanation:

  • B → 2A

⇒ r = k [ A ]/2 = k [ B ]

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Does temperature change during a chemical reaction?​
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Answer: yes

Explanation:

Not all the time, but temperature change can happen during a chemical reaction.

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Consider the reaction H2(g) + I2(g) &lt;---&gt; HI(g) with an equilibrium constant of 46.3 and a reaction quotient of 525. Which
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The equilibrium will shift to the left to favor the reactants.

Explanation:

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What would be the total volume of the new solution when it is changed from 0.2 M to 0.04 M?
34kurt
The question is incomplete.

You need two additional data:

1) the original volume
2) what solution you added to change the volume.

This is a molarity problem, so remember molarity definition and formula:

M = n / V in liters: number of moles per liter of solution

To give you the key to answer this kind of questions, supppose the original volumen was 1 ml and that you added only water (solvent).

The original solution was:

V= 1 ml
M = 0.2 M

Using the formula for molarity, M = n / V

n = M×V = 0.2 M × (1 / 10000)l = 0.0002 moles

For the final solution:

n = 0.0002 moles
M = 0.04

From M = n / V ⇒ V = n / M = 0.002 moles / 0.04 M = 0.05 l

Change to ml ⇒ 0.05 l × 1000 ml / l = 50 ml.  This would be the answer for the hypothetical problem that I assumed for you.

I hope this gives you all the cues you need to answer similar problems about molarity.
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3 years ago
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A useful single-displacement reaction involves thermite, which is a
Licemer1 [7]
The reaction of iron (III) oxide and aluminum is initiated by heat released from a small amount "starter mixture". This reaction is an oxidation-reduction reaction, a single replacement reaction, producing great quantities of heat (flame and sparks) and a stream of molten iron and aluminum oxide which pours out of a hole in the bottom of the pot into sand.

The balanced chemical equation for this reaction is:

2 Al(s) + Fe2O3(s) --> 2Fe(s) + Al2O3(s) + 850 kJ/mol

Curriculum Notes
This chemical reaction can be used to demonstrate an exothermic reaction, a single replacement or oxidation-reduction reaction, and the connection between ∆H calculated for this reaction using heats of formation and Hess' Law and calculating ∆H for this reaction using qrxn = mc∆T and the moles of limiting reactant. This reaction also illustrates the role of activation energy in a chemical reaction. The thermite mixture must be raised to a high temperature before it will react.

To determine how much thermal energy is released in this reaction, heats of formation values and Hess' Law can be used.

By definition, the deltaHfo of an element in its standard state is zero.

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The deltaH for this reaction is the sum of the deltaHfo's of the products - the sum of the deltaHfo's of the reactants (multiplying each by their stoichiometric coefficient in the balanced reaction equation), i.e.:

deltaHorxn = (1 mol)(deltaHfoAl2O3) + (2 mol)(deltaHfoFe) - (1 mol)(deltaHfoFe2O3) - (2 mol)(deltaHfoAl)

deltaHorxn = (1 mol)(-1,669.8 kJ/mol) + (2 mol)(0) - (1 mol)(-822.2 kJ/mol) - (2mol)(0 kJ/mol)

deltaHorxn = -847.6 kJ

The melting point of iron is 1530°C (or 2790°F).
MARK ME BRAINLIEST

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