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2 years ago

I need help with this pls help

Chemistry

1 answer:

dusya [7]2 years ago

3 0

Answer:

L X W

Explanation:

Area is the amount of space in a 2-dimensional figure. You do this by Length x width, which is the same as Base x width. Take a rectangle for instance. The length is 5 feet and the width is 7 feet. To find the area you do 5 by 7/ 5 x 7. 5 x 7 = 35. The area is 35 square feet/ 35 sq².

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The diameter of the He He atom is approximately 0.10 nm nm . Calculate the density of the He atom in g/cm 3 g/cm3 (assuming that

Sladkaya [172]

Answer:

Density of the He atom = 12.69 g/cm³

Explanation:

From the information given:

Since 1 mole of an atom = 6.022x 10²³ atoms)

1 atom of He = $1 \ atom \times (\dfrac{1 \ mole}{ 6.022 \times 10^{23} \ atoms}) \times ( \dfrac{4.003 \ grams}{ 1 \ mole})$

$=6.647 \times 10^{-24} \ grams$

The volume can be determined as folows:

since the diameter of the He atom is approximately 0.10 nm

the radius of the He = $\dfrac{0.10}{2}$ = 0.05 nm

Converting it into cm, we have:

$0.05 nm \times \dfrac{10^{-9} \ meters}{ 1 nm} \times \dfrac{ 1 cm }{10^{-2} \ meters}$

$=5 \times 10^{-9} \ cm$

Assuming that it is a sphere, the volume of a sphere is

= $\dfrac{4}{3}\pi r^3$

= $\dfrac{4}{3}\pi \times (5\times 10^{-9})^3$

= $5.236 \times 10^{-25} \ cm^3$

Finally, the density can be calcuated by using the formula :

$Density = \dfrac{mass}{volume}$

$D = \dfrac{6.647 \times 10^{-24} \ grams }{ 5.236 \times 10^{-25} \ cm^3}$

D = 12.69 g/cm³

Density of the He atom = 12.69 g/cm³

4 0

3 years ago

Read 2 more answers

Muscles, bones, skin, and

ziro4ka [17]

Answer:

The answer is Protein.

8 0

3 years ago

The combustion of 135 mg of a hydrocarbon produces 440 mg of CO2 and 135 mg H2O. The molar mass of the hydrocarbon is 270 g/mol.

NeX [460]

Answer:

Molecular formula = C20H30

Explanation:

NB 440mg = 0.44g, 135mg= 0.135g

From the question, moles of CO2= 0.44/44= 0.01mol

Since 1 mol of CO2 contains 1mol of C, it implies mol of C = 0.01

Also from the question, moles of H2O = 0.135/18= 0.0075mole

Since 1 mol of H2O contains 2mol of H, it implies mol of H = 0.0075×2= 0.015 mol of H

To get the empirical formula, divide by smallest number of mole

Mol of C = 0.01/0.01=1

Mol of H = 0.015/0.01= 1.5

Multiply both by 2 to obtain a whole number

Mol of C =1×2 = 2

Mol of H= 1.5×2 = 3

Empirical formula= C2H3

[C2H3] not = 270

[ (2×12) + 3]n = 270

27n = 270

n=10

Molecular formula= [C2H3]10= C20H30

5 0

3 years ago

How many mL of 3.0M HCl are needed to make 300.0 mL of a 0.10M HCl?

nalin [4]

Answer:

V₁ = 10 mL

Explanation:

Given data:

Initial volume of HCl = ?

Initial molarity = 3.0 M

Final molarity = 0.10 M

Final volume = 300.0 mL

Solution:

Formula:

M₁V₁ = M₂V₂

M₁ = Initial molarity

V₁ = Initial volume of HCl

M₂ =Final molarity

V₂ = Final volume

Now we will put the values.

3.0 M ×V₁ = 0.10 M×300.0 mL

3.0 M ×V₁ = 30 M.mL

V₁ = 30 M.mL /3.0 M

V₁ = 10 mL

3 0

3 years ago

Place the following transitions of the hydrogen atom in order from longest to shortest wavelength of the photon emitted. Rank fr

gizmo_the_mogwai [7]

Answer:

7 to 4 (higher $\lambda$ )

5 to 3

3 to 2

4 to 2 (lower $\lambda$ )

Explanation:

We can use the Rydberg formula which relates the wavelength of the photon emissions to the principle quantum numbers involved in the transition:

$\frac{1}{\lambda}=R((\frac{1}{n_1^2})-(\frac{1}{n_2^2}))$

with $n_1$ final n, and $n_2$ initial n

evaluating for each transition:

7 to 4 $\frac{1}{\lambda} = R((\frac{1}{4^2})-(\frac{1}{7^2}))= R(1/16-1/49)=R(0.042)$

5 to 3 $\frac{1}{\lambda}= R(1/9-1/25)=R(0.071)$

4 to 2 $\frac{1}{\lambda}= R(1/4-1/16)=R(0.1875)$

3 to 2 $\frac{1}{\lambda}= R(1/4-1/9)=R(0.139)$

Note that the above formula is written for $\frac{1}{\lambda}$ , so lower $\frac{1}{\lambda}$ value obtained involves higher $\lambda$ .

So we should order from lower to higher $\frac{1}{\lambda}$

7 to 4 (higher $\lambda$ )

5 to 3

3 to 2

4 to 2 (lower $\lambda$ )

Note: Take into account that longer wavelength involves lower energy ( $E=\frac{hc}{\lambda}$ ).

3 0

3 years ago