What is the mass of six of these marbles? What is the volume? What is the density?

Chemistry

1 answer:

chubhunter [2.5K]3 years ago

5 0

Answer:

All right. So let's calculate the density of a glass marble. Remember that the formula for density is mass over volume. So if I know that the masses 18.5 g. And I know that the um volume is 6.45 cubic centimeters. I can go ahead and answer this to three significant figures. So it's going to be 2.87 grams per cubic centimeter. Okay, that's our density. Now, density is an intensive process. Okay. We're an intensive property. I really should say. It doesn't depend on how much you have. Mhm. If I have one marble, its density is going to be 2.87 g per cubic centimeter. If I have two marbles, the density will be the same because I'll double the mass and I'll also double the volume. So when I divide them I'll get the same number. Okay, that's what makes it an intensive property. No matter how many marbles I have, they'll have the same density. Mass though is not an intensive property. So if I have six marbles and I want to know what the massive six marbles is. Well, I know the mass of each marble is 18.5 g. So the mass of six marbles Is going to be 100 11 g. Because mass is an extensive property. It depends on how much you have. If I change the number of marbles, I'm going to change the mass. That's an extensive property. All right. So we've calculated the density. We've calculated the mass and then what happens to the density of one marble compared to six marbles as we mentioned before. Since densities and intensive property, the densities will be the same, no matter how may.

Explanation:

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3. A 31.2-g piece of silver (s = 0.237 J/(g · °C)), initially at 277.2°C, is added to 185.8 g of a liquid, initially at 24.4°C,

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Answer:

$Cp_{liquid}=2.54\frac{J}{g\°C}$

Explanation:

Hello,

In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

$Q_{Ag}=-Q_{liquid}$

That in terms of the heat capacities, masses and temperature changes turns out:

$m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})$

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

$Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}$