Am i the only one getting a red text saying my question is mean?

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horsena [70]

Answer: ok ill just take my points in dip

Explanation:

3 0

3 years ago

Naphthalene, commonly found in moth balls, is composed of 93.7% carbon and 6.3% hydrogen. The molar mass of naphthalene is 128 g

Vikki [24]

Answer:

Molecular formula naphthalene → C₁₀H₈

Empirical formula naphthalene → C₅H₄

Explanation:

Centesimal composition means that in 100 g of compound we have x g of the element. Therefore in 100 g of naphthalene we have:

93.7 g of C

6.3 g of H

Let's make a rule of three:

In 100 g of naphthalene we have 93.7 g of C and 6.3 g of H

In 128 g of naphthalene we would have:

128 . 93.7 / 100 = 120 g of C

128. 6.3 / 100 = 8 g of H

We convert the mass to moles, by molar mass:

120 g . 1mol / 12 g = 10 moles C

8 g . 1mol/ 1g = 8 moles H

Molecular formula naphthalene → C₁₀H₈

Empirical formula naphthalene → C₅H₄

(The sub-index of each element is divided by the largest possible number)

6 0

3 years ago

Read 2 more answers

If you made a three-dimensional model of an atom and its nucleus, how would you represent the atom? 7th grade

Gemiola [76]

Answer:

it shows the breakdown of the atom

Explanation:

it will show it molecularly

7 0

3 years ago

If a pharmacist dissolves 1.2 grams of a medicinal agent in 60 ml of a cough syrup having a specific gravity of 1.20, what is th

guapka [62]

Mass of medicinal agent taken = 1.2 g

the volume is 60 mL

Specific gravity = 1.20

So the mass of solution = specific gravity X volume = 1.20 * 60 = 72g

Now if we have increased the volume by 0.2 so the new volume = 60.2

New mass = 72 + 1.2 = 73.2

Specific gravity = mass / volume = 73.2 / 60.2 = 1.22 g/mL

7 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