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Brums [2.3K]
3 years ago
12

Is rubbing alcohol polar, non-polar, or ionic?

Chemistry
1 answer:
Karolina [17]3 years ago
7 0

Answer: Rubbing alcohol molecules have a polar and nonpolar part, which means they are able to form hydrogen bonds with water and therefore able to mix with it.

Explanation:

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The diagram shows a carrier wave before and after modulation. The diagram best represents which type of modulation? FM phase AM
bixtya [17]

Answer: On Edge it's A. FM            

5 0
3 years ago
Read 2 more answers
Some table salt substitutes have potassium chloride in them. If you consume potassium chloride, are you truly eating potassium a
4vir4ik [10]

Answer:

You are consuming both

Explaination

2Na(s)+Cl two(g)

That gives us 2NaCl(s)

8 0
3 years ago
A solution of toilet bowl cleaner has PH of 13 the solution is?
AlladinOne [14]
The solution is an alkali.

Usually with the pH value range of 14, substances with pH 7 can be called neutral. Meanwhile substances lower than pH 7 are acids, the lower the pH is, the more acidic it is. Such as cola, it has a pH 2, which is very acidic.
In opposite, the substances with pH over 7 are called alkalis. Again, the larger the pH value is, the more alkaline it is. So pH 13 is a strong alkaline therefore it it corrosive and can clean the toilet well.
8 0
3 years ago
In order to prepare 50.0 mL of 0.100 M NaOH you will add _____ mL of 1.00 M NaOH to _____ mL of water
FinnZ [79.3K]

The question requires us to complete the sentence regarding the preparation of a more dilute NaOH solution (0.100 M, 50.0 mL) from a more concentrated NaOH solution (1.00 M).

Analyzing the blank spaces that we need to fill in the sentence, we can see that we must provide the volume of the more concentrated solution and the volume of water necessary to prepare the solution.

We can use the following equation to calculate the volume of more concentrated solution required:

\begin{gathered} C_1\times V_1=C_2\times V_2 \\ V_1=\frac{C_2\times V_2}{C_1} \end{gathered}

where C1 is the concentration of the initial solution (C1 = 1.00 M), V1 is the volume required of the inital solution (that we'll calculate), C2 is the concentration of the final solution (C2 = 0.100 M) and V2 is the volume of the final solution (V2 = 50.0 mL).

Applying the values given by the question to the equation above, we'll have:

\begin{gathered} V_1=\frac{C_2\times V_2}{C_1} \\ V_1=\frac{0.100M_{}\times50.0mL_{}}{1.00M_{}}=5.00mL \end{gathered}

Thus, we would need 5.00 mL of the more concentrated solution.

Since the volume of the final solution is 50.0 mL and it corresponds to the volume of initial solution + volume of water, we can calculate the volume of water necessary as:

\begin{gathered} \text{final volume = volume of initial solution + volume of water} \\ 50.0mL=5.00mL\text{ + volume of water} \\ \text{volume of water = 45.0 mL} \end{gathered}

Thus, we would need 45.0 mL of water to prepare the solution.

Therefore, we can complete the sentence given as:

<em>"In order to prepare 50.0 mL of 0.100 M NaOH you will add </em>5.00 mL<em> of 1.00 M NaOH to </em>45.0 mL<em> of water"</em>

5 0
1 year ago
Propenoic acid, C3H4O2, is a reactive organic liquid that is used in the manufacturing of plastics, coatings, and adhesives. An
Shkiper50 [21]

Answer:

C = 39%

H = 5.43%

Explanation:

Firstly, we calculate the number of moles of each.

For carbon, we use carbon iv oxide

Mass here = 0.636g

Number of moles of carbon iv oxide = mass of carbon iv oxide ÷ molar mass of carbon iv oxide. Molar mass = 44g/mol

Number of moles = 0.636 ÷ 44 = 0.014mole

Since 1 mole of carbon iv oxide contains 1 mole carbon, 0.014 mole of carbon iv oxide is also produced.

Mass of carbon = 0.014 × 12 = 0.173g, where 12 is the a.m.u of carbon.

For hydrogen, we use water .

Mass of water = 0.220g

No of moles of water = 0.22 ÷ 18 = 0.012 mole

1 mole of water has two moles of hydrogen, thus the amount of hydrogen produced = 0.012 × 2 = 0.024mole

Mass of hydrogen produced = 0.024 × 1 = 0.024g

The percentage composition are as follows:

Carbon = 0.173/0.442 × 100 = 39%

Hydrogen = 0.024/0.442 × 100 = 5.43%

5 0
3 years ago
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