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Kazeer [188]
3 years ago
13

Name the unit being abbreviated in each measurement (spelling counts). Only write the name of the unit, do not include the numbe

r. 0.8 L litre 3.6 cm centimeter 4.0 kg kilogram 3.5 s 373 K
Chemistry
1 answer:
givi [52]3 years ago
5 0

Answer:

Litre (L) , Centimetre (cm) , Kilogram (Kg), Seconds (s) and Kelvin (K)

Explanation:

The units are used for the following measurement;

Litre = Volume

Centimetre = Length

Kilogram = Mass

Seconds = Time

Kelvin = Temperature

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Which of the following elements has the largest atomic radius?
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nitrogen

Explanation:

as the group number increases, the atomic radius decreases

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A science student observes that a dissolvable antacid tablet acts faster in relieving stomach pain when taken with warm water th
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How do volcanos such as ojos del salado form?
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Answer:

Deep inside Earth, between the molten iron core and the thin crust at the surface, there is a solid body of rock called the mantle. When rock from the mantle melts, moves to the surface through the crust, and releases pent-up gases, volcanoes erupt.

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3 years ago
Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t
Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

Q_1=m\times \Delta H_{fusion}

where,

Q_1 = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

\Delta H_{fusion} = enthalpy change for fusion = 3.35\times 10^5J/Kg

Now put all the given values in Q_1, we get:

Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J

<u>For process 2 :</u>

Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})

where,

Q_2 = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

c_{p,l} = specific heat of liquid water = 4186J/Kg^oC

T_1 = initial temperature = 0^oC

T_2 = final temperature = 100^oC

Now put all the given values in Q_2, we get:

Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC

Q_2=4.186\times 10^5J

From this we conclude that, Q_1 that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

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