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3 years ago

Iron has a density of 7.86 g/cm3 (1 cm3=1 mL). Calculate the volume (in dL) of a piece of iron having a mass of 3.26 kg . Note t

hat the density is provided in different units of volume and mass than the desired units of volume (dL) and the given units of mass (kg). You will need to express the density in kg/dL (1 cm3 = 1 mL) before calculating the volume for the piece of iron.

Chemistry

1 answer:

Mkey [24]3 years ago

7 0

$d=7,86\frac{g}{cm^{3}}=7,86\frac{0,001kg}{0,01dL}=0,786\frac{kg}{dL}\\
m=3,26kg\\\\\\
d=\frac{m}{V} \ \ \ \ \Rightarrow \ \ \ \ V=\frac{m}{d}=\frac{3,26kg}{0,786\frac{kg}{dL}}\approx4,15dL$

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the red line observed in the line spectrum for hydrogen has 3.03x10^-19J, what is the wavelength, in nm, of this?

djverab [1.8K]

Answer:

λ = 6.5604 x 1016 nm

Explanation:

Given Data:

The energy of the red line in Hydrogen Spectra = 3.03 x 10-19

Formula to calculate Wave length

E= hv

Where E is Energy

h is Planks Constant = 6.626 x 10–34 J s

v is frequency

In turn

v= c/ λ

where c is speed of light = 3.00 x 108 m s–1

λ is wavelength = to find

Solution:

Formula to be Used:

E= hv………………………… (1)

Putting the value v in equation 1

E= h c/ λ…………………… (2)

Put the value in equation 2

3.03 x 10-19 J = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) / λ ……………………….(3)

By rearranging equation 3

λ = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) /3.03 x 10-19 J

λ = 6.5604 x 107 m

The answer is in “m”

So we have to convert it into nm

So for this to convert “m” to “nm” multiply the answer with 109

λ = 6.5604 x 107 x 109

λ = 6.5604 x 1016 nm

6 0

3 years ago

A substance that is ______ will NOT dissolve in a solvent. A) freezing B) insoluble C) evaporating D) soluble

Alexxandr [17]

Answer:

insoluble

Explanation:

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3 years ago

Read 2 more answers

A copper cylinder, 12.0 cm in radius, is 44.0 cm long. If the density of commoner is 8.90 g/cm3, calculate the mass in grams of

inna [77]

The mass of the copper cylinder is 177065.856g

Given:

Radius of the copper cylinder R=12cm

Height of the copper cylinder H=44cm

Density of the cylinder=8.90 $\frac{g}{c m^{3}}$

To find:

Mass of the copper cylinder

Step by Step by explanation:

Solution:

According to the formula, Mass can be calculated as

$\rho=\frac{m}{v}$ and from this

$m=\rho \times v$

Where, m=mass of the cylinder

$\rho$ =density of the cylinder

v=volume of the cylinder

And also cylinder is provided with radius and height value.

So volume of the cylinder is calculated as

$v=\pi r^{2} h$

Where $\pi$ =3.14

r=radius of the cylinder=12cm

h=height of the cylinder=44cm

Thus, $v=3.14 \times 12^{2} \times 44$

v=3.14 \times 144 \times 44

$v=19895.04 \mathrm{cm}^{3}$

And we know that, $m=\rho \times v$

Substitute the known values in the above equation we get

$m=8.90 \times 19895.04$

m=177065.856g or 177.065kg

Result:

Thus the mass of the copper cylinder is 177065.856g

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3 years ago

Four springs are stretched to the same distance from the equilibrium position. The spring constants are listed in the table. A 2

svp [43]

Answer:

The correct option is;

X, W, Y, Z

Explanation:

The parameters given are;

Spring (S), Spring Constant (N/m)

W, 24

X, 35

Y, 22

Z, 15

The equation for elastic potential energy, $E_e$ , is $E_e = 0.5 \times k \times x^2$

The above equation can also be written as $E_e =\dfrac{1}{2} \times k \times x^2$

Where:

k = The spring constant in (N/m)

x = The spring extension

Therefore, since the elastic potential energy, $E_e$ , of the spring is directly proportional to the spring constant, k, we have the springs with higher spring constant will have higher elastic potential energy, $E_e$ , therefore the correct order is as follows;

X > W > Y > Z

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3 years ago

Suppose there are two known compounds containing the generic elements X and Y. You have a 1.00-g sample of each compound. One sa

Lapatulllka [165]

I think the correct answers are X2Y and X3Y, X2Y5 and X3Y5, and X4Y2 and X3Y, for the following reason:

If you look at the combining masses of X and Y in each of the two compounds,

The first compound contains 0.25g of X combined with 0.75g of Y
so the ratio (by mass) of X to Y = 1 : 3

The second compound contains 0.33 g of X combined with 0.67 g of Y
so the ratio (by mass) of X to Y = 1 : 2

Now, you suppose to prepare each of these two compounds, starting with the same fixed mass of element Y ( I will choose 12g of Y for an easy calculation!)

The first compound will then contain 4g of X and 12g of Y
The second compound will then contain 6g of X and 12g of Y

The ratio which combined the masses of X and the fixed mass (12g) of Y
= 4 : 6
or 2 : 3 

So, the ratio of MOLES of X which combined with the fixed amount of Y in the two compounds is also = 2 : 3

The two compounds given with the plausible formula must therefore contain the same ratio.

8 0

3 years ago