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3 years ago

5

Determine a piece of copper with a volume of 100 cm power 3 has a mass of 890 g. Explain how you could use this information to f

ind out how much mass would be in 13 cm power 3 of copper

1 answer:

balandron [24]3 years ago

8 0

You will have to multiply 13 by 100 and then divide 3 and get 60

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3 years ago

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How much heat is needed to raise the temperature of a 5.0 gram block of aluminum from 22.0°c to 37.0°c?

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You'll need the specific heat capacity of aluminium to solve this question.
H=(0.005)(37-22)(specific heat capacity of aluminium)

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3 years ago

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Generally how does the first ionization energy vary as the atomic number increases going across a period

Vlad1618 [11]

Across a period I.E increases progressively from left to right

Explanation:

The trend of the first ionization energy is such that across a period I.E increases from left to right due to the decreasing atomic radii caused by the increasing nuclear charge. This not compensated for by successive electronic shells.

Ionization energy is a measure of the readiness of an atom to lose an electron.
The lower the value, the easier it is for an atom to lose an electron.
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4 years ago

A 37.2 g sample of copper at 99.8 °C is carefully placed into an insulated container containing 188 g of water at 18.5 °C. Calcu

klasskru [66]

Answer:

T₂ = 19.95°C

Explanation:

From the law of conservation of energy:

$Heat\ Lost\ by\ Copper = Heat\ Gained\ by\ Water\\m_cC_c\Delta T_c = m_wC_w\Delta T_w$

where,

mc = mass of copper = 37.2 g

Cc = specific heat of copper = 0.385 J/g.°C

mw = mass of water = 188 g

Cw = specific heat of water = 4.184 J/g.°C

ΔTc = Change in temperature of copper = 99.8°C - T₂

ΔTw = Change in temperature of water = T₂ - 18.5°C

T₂ = Final Temperature at Equilibrium = ?

Therefore,

$(37.2\ g)(0.385\ J/g.^oC)(99.8\ ^oC-T_2)=(188\ g)(4.184\ J/g.^oC)(T_2-18.5\ ^oC)\\99.8\ ^oC-T_2 = \frac{(188\ g)(4.184\ J/g.^oC)}{(37.2\ g)(0.385\ J/g.^oC)}(T_2-18.5\ ^oC)\\\\99.8\ ^oC-T_2 = (54.92) (T_2-18.5\ ^oC)\\54.92T_2+T_2 = 99.8\ ^oC + 1016.02\ ^oC\\\\T_2 = \frac{1115.82\ ^oC}{55.92}$

<u>T₂ = 19.95°C</u>

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3 years ago

What is the standard potential, e∘celle∘cell, for this galvanic cell? use the given standard reduction potentials in your calcul

Olin [163]

The standard potential for the given galvanic cell is 0.477 V

<h3>What is electrode potential?</h3>

The electrode potential is the electromotive force of a galvanic cell built using a standard reference electrode and another electrode whose potential is to be found.

There are two types of electrode potential

Oxidation potential - The potential associated with oxidation reaction is known as oxidation potential

Reduction potential - The potential associated with reduction reaction is known as reduction potential

At the anode, oxidation occurs

$Sn(s)\rightarrow Sn^{2+}(aq)+2e^-$

At the cathode, reduction occurs

$Cu^{2+}(aq)+2e^-\rightarrow Cu(s)$

$E^o_{cell} =E^o_{cathode} -E^o_{anode}$

= 0.337 - (-0.140)

= 0.477 V

Thus, The standard potential for the given galvanic cell is 0.477 V

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Disclaimer: The question was given incomplete on the portal. Here is the complete question

Question: What is the standard potential, E∘cell, for this galvanic cell? Use the given standard reduction potentials in your calculation as appropriate.

$Sn^{2+}(aq)+2e^-\rightarrow Sn(s)$ , E°red=−0.140 V

$Cu^{2+}(aq)+2e^-\rightarrow Cu(s)$ , E°red=+0.337 V

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2 years ago

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