An empty erlenmeyer flask that weighs 245.8 g is filled with 125 mL of carbon tetrachloride . The weight of the flask and carbon tetrachloride is found to be 444.6 g. From this information , calculate the density of carbon tetrachloride .

The density of carbon tetrachloride is 1.59 grams per milliliter.

Given that an empty erlenmeyer flask that weighs 245.8 g is filled with 125 mL of carbon tetrachloride, and the weight of the flask and carbon tetrachloride is found to be 444.6 g, to determine the density of carbon tetrachloride the following calculation must be performed:

Subtract the weight of the empty flask from the weight of the full flask, and then divide this result by 125 (the milliliters of carbon tetrachloride).

Therefore, the density of carbon tetrachloride is 1.59 grams per milliliter.

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The density is 1.59 g/mL. Therefore, the substance will NOT float in water.

Step-by-step explanation:

The following data were obtained from the question:

Mass of empty flask = 245.8 g

Mass of flask + CCl₄ = 444.6 g

Volume of CCl₄ = 125 mL

Density of CCl₄ =?

Next, we shall determine the mass of carbon tetrachloride, CCl₄. This can be obtained as follow:

Mass of empty flask = 245.8 g

Mass of flask + CCl₄ = 444.6 g

Mass of CCl₄ =.?

Mass of CCl₄ = (Mass of flask + CCl₄) – (Mass of empty flask)

Mass of CCl₄ = 444.6 – 245.8

Mass of CCl₄ = 198.8 g

Finally, we shall determine the density of carbon tetrachloride, CCl₄ as follow:

Mass of CCl₄ = 198.8 g

Volume of CCl₄ = 125 mL

Density of CCl₄ =?

Density = mass / volume

Density of CCl₄ = 198.8 / 125

Density of CCl₄ = 1.59 g/mL

The density of water is 1 g/mL. Since the density of carbon tetrachloride, CCl₄ (i.e 1.59 g/mL) is bigger than that of water, the substance will not float in water.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

the ratio of balloons to tables is 5: 2 and the constant variation is 5.

the equation is y=5x