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sashaice [31]
3 years ago
5

Any change that alters a substance without changing it into another substance.

Chemistry
1 answer:
Ber [7]3 years ago
7 0
A physical change.

Physical change = can be reverted + doesn't change substance
Chemical = changes substance
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HELP FAST 75 PTS Calculate the amount of heat needed to melt 35.0 g of ice at 0 ºC. must show work
victus00 [196]

Answer:

It takes 12,000 Joules of energy to melt 35 grams of ice at 0 °C

Explanation:

Good Luck!

7 0
3 years ago
Calculate the standard potential for the following galvanic cell: Fe(s) | Fe2+(aq) || Ag+(aq) | Ag(s) which has the overall bala
Nataly [62]

Answer:

E° = 1.24 V

Explanation:

Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)

According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:

Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻

Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)

The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an

E° = 0.80 V - (-0.44 V) = 1.24 V

6 0
3 years ago
Ductile means...
balu736 [363]

Answer:

A

Explanation:

3 0
2 years ago
What is the mass % of iron in Fe(CH3CO2)2<br><br> (Fe=iron=55.85)
Serjik [45]
The mass percentage is 15.1465%.
5 0
3 years ago
Suppose an ice cube weighing 36.0 g at a temperature of 10°C is placed in 360 g water at a temperature of 20°C. Calculate the te
Scilla [17]

Answer:

10.44 °C

Explanation:

When the thermal equilibrium is reached, both of the substances have the same final temperature (T). The liquid water will lose heat, and the ice cube will absorb this heat. The temperature of the ice will increase until it reaches 0°C, at this temperature, it will change of phase for liquid, absorbing heat, but without a change in the temperature. Then the temperature will increase until the equilibrium.

By the energy conservation, the total amount of heat must be equal to 0:

Qice + Qmelting + Qliquid1 + Qliquid2 = 0

Liquid 1 is the ice after melting, and liquid 2 the liquid that was already at the flask. When there's a change of temperature:

Q = n*c*ΔT, where n is the number of moles, c is the heat capacity and ΔT is the temperature change (final - initial). The temperature variation in °C is equal in K, so the temperature may be used in °C.

The melting heat is:

Q = n*Hfus, Hfus = 6007 J/mol

The molar mass of the water is 18 g/mol, so the number of moles of the water and the ice are:

nwater = nliquid1 = 360/18 = 20 moles

nice = 36/18 = 2 moles

Qice + Qmelting + Qliquid1 + Qliquid2 = 0

2*38*(0 - (-10)) + 2*6007 + 2*75*(T - 0) + 20*75*(T - 20) = 0

760 + 12014 + 150T + 1500T - 30000 = 0

1650T = 17226

T = 10.44 °C

4 0
3 years ago
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