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Lemur [1.5K]
2 years ago
15

Density (D) is defined as the ratio of mass (m) to volume (V) and can be determined from the expression D = . Find the density o

f a sample of pure aluminum if a rectangular prism of volume 32.32 cm3 occupies a mass of 87.329 g. (Remember to put a space between your numerical value and the unit.)
Chemistry
1 answer:
Usimov [2.4K]2 years ago
7 0

That is a question about density.

To solve that question, it is important to know how we can obtain the Density of an object. Below, you can see the formula:

\boxed{D=\frac{m}{v} }

D = Density\\m = Mass\\V = Volume

Now, it is easy to answer, right? We just replace the values in the formula. The mass is 87.329g and the volume is 32.32cm³. So,

D=\frac{m}{v} \\\\D = \frac{87.329g}{32.32cm^3} \\\\D \approx 2.70g/cm^3

Therefore, the density of the the aluminium is approximately 2.7 g/cm³.

I hope I've helped. ^^

Enjoy your studies \o/

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What does it mean to tare a balance and why do you think it is important to complete this before you begin measuring mass?
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Answer:

allows for you to account for only the weight of the substance being measured and not the vessel it’s being measured in.

Explanation:

What does it mean to tare a balance and why do you think it is important to complete this before you begin measuring mass? Explanation: The term tare is used when weighing chemicals on a balance, using the tare button allows for you to account for only the weight of the substance being measured and not the vessel it’s being measured in.

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Answer: 23 g

the mol of Fe is 11,5/56 = 23/112 (mol)

the mol of CO is \frac{2,63.10^{24} }{6.10^{23} }= \frac{263}{60} (mol)

we have:

Fe2O3 + 3CO => 2Fe + 3CO2

23/112    69/112     23/56

=> the mol of Fe is 23/56 mol

=> the mass of Fe is 23/56.56 = 23 (g)

Explanation:

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The point at where the water is changing phase
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The ideal gas heat capacity of nitrogen varies with temperature. It is given by:
hammer [34]

Answer:

A)  1059 J/mol

B)  17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

C_p=C_v+R

We can determine C_v from above if we make C_v the subject of the formula as:

C_v=C_p-R

C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314

C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4

A).

The formula for calculating change in internal energy is given as:

dU=C_vdT

If we integrate above data into the equation; it implies that:

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 1059J/mol

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 17,920 J/mol

3 0
4 years ago
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