All categories

3 years ago

A wave with a higher frequency has a longer wavelength TRUE FALSE

Chemistry

2 answers:

BartSMP [9]3 years ago

7 0

Answer:

true

Explanation:

A wave with a higher frequency has a longer wavelength

givi [52]3 years ago

3 0

Answer:

<h2>False</h2>

A wave with a higher frequency have a shorter wavelength, the higher the sound it, the closer the arcs of the wavelength are, the faster the sound is, the shorter they are.

According to scied.ucar.edu ~ "The frequency of a wave is inversely proportional to its wavelength. That means that waves with a high frequency have a short wavelength, while waves with a low frequency have a longer wavelength. Light waves have very, very short wavelengths."

Hope this helps

Explanation:

May I have brainliest please? :)

You might be interested in

Which of the following properly describes the movement of atoms as they represent the states of matter? Atoms making liquids hav

7nadin3 [17]

Answer:

Atoms making liquids have less attraction than solids, but more than gases

Explanation:

The attraction between atoms in different molecules in a solid is very strong due to strong intermolecular forces present in a solid. However, such intermolecular forces are weaker in liquids than in solids.

This implies that the solid has higher intermolecular forces of attraction compared to gases and liquids. Based on the negligible degree of intermolecular forces between them, a gas has the weakest intermolecular forces hence the atom has very minimal interaction between them.

7 0

3 years ago

What concentration of monosodium phosphate would be required to make a solution of pH 7.4 with 0.2 M disodium phosphate ( pKa

denis-greek [22]

Answer:

The concentration of monosodium phosphate is 0.1262M

Explanation:

The buffer of H₂PO₄⁻ / HPO₄²⁻ (Monobasic phosphate and dibasic phosphate has a pKa of 7.2

To determine the pH you must use Henderson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is molarity of the conjugate base of the weak acid, [HA].

For H₂PO₄⁻ / HPO₄⁻ buffer:

pH = 7.2 + log [HPO₄⁻² ] / [H₂PO₄⁻]

As molarity of the dibasic phosphate is 0.2M and you want a pH of 7.4:

7.4 = 7.2 + log [0.2] / [H₂PO₄⁻]

0.2 = log [0.2] / [H₂PO₄⁻]

1.58489 = [0.2] / [H₂PO₄⁻]

[H₂PO₄⁻] = 0.1262M

<h3>The concentration of monosodium phosphate is 0.1262M</h3>

8 0

3 years ago

Give the IUPAC name for the following structure: 3-chloro-6-methylcyclohexanol 2-methyl-5-chlorocyclohexanol 1-chloro-4-methylcy

Shtirlitz [24]

Answer:

5-chloro-2-methylcyclohexanol

Explanation:

There is no structure for the compound, but we can analyze the proposed options using the IUPAC rules to name organic compounds.

IUPAC rules state that to name an organic compound, first we have to identify the priorities for the functional groups present in the compound. In this case, the priority functional group is the alcohol group, so we will start the counting of the carbons in this group. Then, the counting of carbon atoms is followed by the next substituents so they have the lowest possible numbers, in this case, we can assign the number 2 to the methyl group and 5 to the chloride group, and name the compound in alphabetical order, using commas to separate the words from the numbers and with no space between the words.

Since the other options involve: high countings for the susbtituents groups (3-chloro-6-methylcyclohexanol), wrong assignation of priority functional group (1-chloro-4-methylcyclohexanol), wrong sequence of counting in the compound (2-methyl-3-chlorocyclohexanol) and no alphabetical order to name the compound (2-methyl-5-chlorocyclohexanol), the correct option is:

5-chloro-2-methylcyclohexanol

Have a nice day!

4 0

4 years ago

Ethanedioic acid, a compound that is present in many vegetables, has a molar mass of 90.04 g/mol and a composition of 26.7% Carb

aleksandrvk [35]

Answer:

Empirical CHO2

Molecular C2H2O4

Explanation:

To determine the formulas, firstly, we need to divide the percentage compositions by the atomic masses.

Kindly note that the atomic mass of carbon, oxygen and hydrogen are 12, 16 and 1 respectively. We proceed with the division as follows:

C = 26.7/12 = 2.225

H = 2.2/1 = 2.2

O = 71.1/16 = 4.44375

We then proceed to divide by the smallest value which is 2.2 in this case

C = 2.25/2.2 = 1

H = 2.2/2.2 = 1

O = 4.44375/2.2 = 2

Thus, the empirical formula is CHO2

We now proceed to get the molecular formula as follows

[12+ 1 + 16(2) ]n = 90.04

45n = 90.04

n = 90.04/45 = 2

The molecular formula is :

C2H2O4

4 0

3 years ago

Pls help me with this

Zina [86]

Answer:

$[I_2]=[Br]=0.31M$

Explanation:

Hello there!

In this case, according to the given information, it is possible for us to set up the following chemical equation at equilibrium:

$I_2+Br_2\rightleftharpoons 2IBr$

Now, we can set up the equilibrium expression in terms of x (reaction extent) whereas the initial concentration of both iodine and bromine is 0.5mol/0.250L=2.0M:

$K=\frac{[IBr]^2}{[I_2][Br_2]} \\\\1.2x10^2=\frac{(2x)^2}{(2.0-x)^2}$

Thus, we solve for x as show below:

$\sqrt{1.2x10^2} =\sqrt{\frac{(2x)^2}{(2.0-x)^2}} \\\\10.95=\frac{2x}{2.0-x}\\\\21.91-10.95x=2x\\\\21.91=12.95x\\\\x=\frac{21.91}{12.95} \\\\x=1.69M$

Therefore, the concentrations of both bromine and iodine are:

$[I_2]=[Br]=2.0M-1.69M=0.31M$

Regards!

8 0

3 years ago