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3 years ago

How much energy will it take to raise the temperature of 75.0 g of water from 20.0°C to 55.0°C?

Chemistry

1 answer:

Hatshy [7]3 years ago

6 0

Answer:

We can use heat = mcΔT to determine the amount of heat, but first we need to determine ΔT. Because the final temperature of the water is 55°C and the initial temperature is 20.0°C, ΔT is as follows:

ΔT = Tfinal − Tinitial = 55.0°C − 20.0°C = 35.0°C

given the specific heat of water as 1 cal/g·°C. Substitute the known values into heat = mcΔT and solve for amount of heat:

= heat=(75.0 g)(1 cal/ g· °C )(35.0°C) =

= 75x1x35=2625 cal

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3 years ago

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3 years ago

Calculate the mass of water produced when 1.92 g of butane reacts with excess oxygen.

never [62]

C4h10+6.5o2=4co2+5h2o
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3 years ago

Consider an oxygen-concentration cell consisting of two zinc electrodes. One is immersed in a water solution with a low oxygen c

blsea [12.9K]

Answer:

(a) The anode electrode which comprises the zinc electrode being placed in a water solution with low oxygen concentration.

(b) Cathodic reaction is: $O_{2} + 2H_{2}O + 4e^{-}$ ⇒ $4OH^{-}$

Anodic reaction is: $Zn$ ⇒ $Zn^{2+} + 2e^{-}$

Explanation:

In the given problem, we have an oxygen-concentration cell consisting of two zinc electrodes. One is immersed in a water solution with a low oxygen concentration and the other in a water solution with a high oxygen concentration. The zinc electrodes are connected by an external copper wire.

(a) Which electrode will corrode?

The electrode that will corrode is the anode electrode which comprises the zinc electrode being placed in a water solution with low oxygen concentration.

(b) Write half-cell reactions for the anodic reaction and the cathodic reaction.

Cathodic reaction is: $O_{2} + 2H_{2}O + 4e^{-}$ ⇒ $4OH^{-}$

Anodic reaction is: $Zn$ ⇒ $Zn^{2+} + 2e^{-}$

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3 years ago

If your lawn is 21.0 ft wide and 20.0 ft long, and each square foot of lawn accumulates 1350 new snow flakes every minute. How m

GaryK [48]

<span>First we can calculate the area of the rectangular lawn using the formula:
Area = Width x Length = 21 ft x 20 ft = 420 square feet

And the total number of snow flakes per minute on the entire lawn is:

(1350 snowflakes per minute per square foot) x (420 square feet) = 567,000 snowflakes per minute

In one hour (or 60 minutes) we get a total of:

(567,000 snowflakes per minute) x (60 minutes / 1 hour) = 34,020,000 snowflakes

The total mass of which would be:

34,020,000 snowflakes x 1.60 mg = 54,432,000 mg = 54.432 kg (as 1 kg = 1,000,000 mg).

So 54.432 kg of snow accumulates every hour on the lawn.</span>

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3 years ago