Which of the following measurements is equal to 150 dL?

1 answer:

6 0

This is a case of metric system of measurements. The scale of the metric system only differs by a factor of 10. The scale (from greatest to least) is kiloliter, hectoliter, dekaliter, liter, deciliter, centiliter and milliliter. If the unit is 150 dL, and you want to find the equivalent dkL measurement, just move the decimal point 2 decimals places to the left (just follow the scale). The same procedure is done for the other metric units.

150 dL = 1.5 dkL
150 dL = 15 L
150 dL = 1,500 cL
150 dL = 15,000 mL

From the choices, the answer is letter C.

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If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

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Answer:

$T_{eq}=28.9\°C$

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$