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3 years ago

If density usually increases with decreasing temperature, why does ice float on liquid water? Explain.

Chemistry

1 answer:

zepelin [54]3 years ago

6 0

Answer:

If you continue to cool water past 4 degrees Celsius, its density starts to plummet (you can see this in the graph). At zero degrees, i.e., the temperature at which water turns into ice, the density of water is actually quite low. It turns out that ice has a lower density than water, and any object that has a lower density than the liquid form on which it’s kept (in this case, water) will be able to float!

Explanation:

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A battery has an electrical ______________ energy

aleksley [76]

Answer:

Electrical Energy

Explanation:

There are a variety of chemical and mechanical devices that are called batteries, although they operate on different physical principles. A battery for the purposes of this explanation will be a device that can store energy in a chemical form and convert that stored chemical energy into electrical energy when needed.

5 0

3 years ago

Can some answer this please?

marysya [2.9K]

Answer:

Explanation:

it's b because I just went over that frome my class and got it correct

4 0

3 years ago

Read 2 more answers

Based on the following reactants:

Doss [256]

(205) 872-9311 call me so I can help you out

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3 years ago

NEED HELP ASAP!!! TYSM!!!! What are the coefficients when the following equations are balanced?

vitfil [10]

CH4+2O2–>CO2+2H2O

4Fe+3O2–>2Fe2O3

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2 years ago

A 1.28-kg sample of water at 10.0 °C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at 215 °C into it. A

Kryger [21]

Answer:

$T_{2}=16,97^{\circ}C$

Explanation:

The specific heats of water and steel are

$Cp_{w}=4.186 \frac{KJ}{Kg^{\circ}C}$

$Cp_{s}=0.49 \frac{KJ}{Kg^{\circ}C}$

Assuming that the water and steel are into an <em>adiabatic calorimeter</em> (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium $T_{2}_{w}= T_{2}_{s}$

An energy balance can be written as

$m_{w}\times Cp_{w}\times (T_{2}- T_{1})_{w}= -m_{s}\times Cp_{s}\times (T_{2}- T_{1})_{s}$

Replacing

$1.28Kg\times 4.186\frac{KJ}{Kg^{\circ}C}\times (T_{2}-10^{\circ}C)= -0.385Kg\times 0.49 \frac{KJ}{Kg^{\circ}C} \times (T_{2}-215^{\circ}C)$

Then, the temperature $T_{2}=16,97^{\circ}C$

8 0

3 years ago