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jeyben [28]
2 years ago
5

A compound that contains only carbon, hydrogen, and oxygen is 48.64% C and 8.16% H by mass. What is the empirical formula of thi

s substance? 84. The most common form
Chemistry
1 answer:
TiliK225 [7]2 years ago
6 0

Answer:

The empirical formula of this substance is:

C_3H_6O_2

Explanation:

To find the empirical formula of this substance we need the molecular weight of the elements Carbon, Hydrogen and Oxygen, we can find this information in the periodic table:

- C: 12.01 g/mol

- H: 1.00 g/mol

- O: 15.99 g/mol

With the information in this exercise we can suppose in 100 g of the substance we have:

C: 48.64 g

H: 8.16 g

O: 43.2 g (100 g - 48.64g - 8.16g= 43.2 g)

Now, we need to divide these grams by the molecular weight:

C=\frac{48.64g}{12.01 g/mol} =4.05 mol\\H=\frac{8.16g}{1.00g/mol}= 8.16 mol\\O=\frac{43.2g}{15.99 g/mol} = 2.70 mol

We need to divide these results by the minor result, in this case O=2.70 mol

C=\frac{4.05mol}{2.70mol}= 1.5\\H= \frac{8.16mol}{2.70 mol} = 3.02 \\O=\frac{2.70mol}{2.70mol} = 1

We need to find integer numbers to find the empirical formula, for this reason we multiply by 2:

C= 1.5*2=3\\H= 3.02*2= 6.04 \\O= 1*2=2

This numbers are very close to integer numbers, so we can find the empirical formula as  subscripts in the chemical formula:

C_3H_6O_2

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lana66690 [7]

Answer:

Wavelength of this beam of light: \rm 4.39\times 10^{-7}\; m.

Explanation:

The speed of light in vacuum is approximately \rm 2.998\times 10^{8}\;m \cdot s^{-1}.

Light behaves like a wave. The wavelength of a wave is equal to the distance that it travels (in the given medium) in each period of oscillation.

On the other hand, the frequency of a wave is the number of periods in unit time. 1\rm \; Hz means one oscillation per second. The frequency of this particular wave is \rm 6.83\times 10^{14}\; Hz. In other words, there are 6.83\times 10^{14} oscillations in each second.

The period of oscillation will be equal to

\displaystyle t = \frac{1}{f} = \frac{1}{\rm 6.83\times 10^{14}\; s^{-1}}.

In that period of time, a beam of light in vacuum would have traveled  

\displaystyle \rm 2.998\times 10^{8}\; m\cdot s^{-1} \times \frac{1}{\rm 6.83\times 10^{14}\; s^{-1}} = 4.39\times 10^{-7}\; m.

In other words, if this beam of light of frequency \rm 6.83\times 10^{14}\; Hz is in vacuum, its wavelength will be equal to \rm 4.39\times 10^{-7}\; m.

8 0
2 years ago
Find the ions in the periodic table that have an electron configuration of nd8 (n = 3, 4, 5...).
Alla [95]
 <span>Cu⁺ is the only one of the ions in the list that will show 8 electrons in a d sublevel....its configuration will be Ar| 4s² 3d⁸
hope this helps</span>
4 0
2 years ago
Read 2 more answers
Danny sails a boat downstream. The wind pushes the boat along at 21 km/hr. The current runs downstream at 15 km/hr. What is the
deff fn [24]

Answer: D) 36.0 km/hr, downstream

Explanation:

For downstream motion of the boat, the actual velocity of the boat is the sum of velocity of the water current and the velocity of the boat due to pushed by wind.

Velocity of water current, v = 15 km/h

Velocity of the boat going downstream, u = 21 km/h

Actual velocity of the boat = v'

v' = v + u

⇒v' = 15 km/h + 21 km/h

⇒u = 21 km/h +15 km/h = 36.0 km/h downstream

Thus, the correct answer is option D.

7 0
2 years ago
Read 2 more answers
Quinine, an antimalarial drug, is 8.63% nitrogen. There are two nitrogen atoms per molecule. What is the molecular weight of qui
Furkat [3]

Answer:

324.18 g/mol

Explanation:

Let the molecular mass of the antimalarial drug, Quinine is x g/mol

According to question,

Nitrogen present in the drug is 8.63% of x

So, mass of nitrogen = \frac {8.63}{100}\times x

Also, according to the question,

2 atoms are present in 1 molecule of the drug.

Mass of nitrogen = 14.01 amu = 14.01 g/mol (grams for 1 mole)

So, mass of nitrogen = 14.01×2 = 28.02

These 2 must be equal so,

\frac {8.63}{100}\times x=28.02

solving for x, we get:

<u>x = 324.18 g/mol</u>

6 0
2 years ago
Consider the dissolution of MnS in water (Ksp = 3.0 × 10–14). MnS(s) + H2O(l) Mn2+(aq) + HS–(aq) + OH–(aq) How is the solubility
Mademuasel [1]

Answer:

The solubility of MnS will decrease on addition of KOH solution.

Explanation:

As per the equation given:

MnS(s)+H_{2}O(l) -->Mn^{+2}(aq)+HS^{-}(aq)+OH^{-}(aq)

On dissolution of MnS in water it gives a basic solution as it gives hydroxide ions.

Now when the we are adding aqueous KOH solution, it will dissociate as:

KOH(aq)--->K^{+}(aq)+OH^{-}(aq)

Thus it will further furnish more hydroxide ion,

This will increase the concentration of hydroxide ions (present of product side), the system will try to decrease its concentration by shifting towards reactant side.

Thus the solubility of MnS will decrease on addition of KOH solution.

7 0
2 years ago
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