Percocet contains oxycodone at varying strengths.Every strength also contains 325 mg of

How does fabric softener affect the flammability of clothes?

Brrunno [24]

It makes your clothes smell great

6 0

3 years ago

Read 2 more answers

The boiling point of a substance is affected by

aev [14]

How strong the bonds are between the atoms

7 0

3 years ago

Rank these species by their ability to act as an oxidizing agent. cu+ mg2+ fe2+

natita [175]

Explanation:

The species or elements which gain electrons and reduces itself are known as oxidizing agent or oxidant.

Ability of an element to act as an oxidizing agent depends on its electrode potential.

The electrode potential of $Cu^{+}$ is 0.52 V.

The electrode potential of $Fe^2+$ is -0.41 V.

The electrode potential of $Mg^{2+}$ is -2.38 V.

Greater is the value of electrode potential, stronger will be the oxidizing agent.

Therefore, rank of these species by their ability to act as an oxidizing agent are as follows.

$Cu^{+}$ > $Fe^2+$ > $Mg^{2+}$

7 0

3 years ago

1. What type of wave is shown below?

svlad2 [7]

Answer : The correct option is, transverse wave.

Explanation :

Longitudinal waves : It is defined as the waves in which the particles of the medium move in the direction of the wave.

Transverse wave : It is defined as the waves in which the particles of the medium travel perpendicularly to the direction of the wave.

Surface wave : It is defined as a combination of transverse and longitudinal waves.

From the given image we conclude that, this illustration depict the transverse wave because the particles of the medium move perpendicularly to the direction of the wave.

Hence, the type of wave is, transverse wave

3 0

3 years ago

Determine whether or not the mixing of each of the two solutions indicated below will result in a buffer.

WARRIOR [948]

Part A

75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF

This combination will form a buffer.

Explanation

Here, weak acid HF and its conjugate base F- is available in the solution

Part B

150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl

This combination cannot form a buffer.

Explanation

Here, moles of HF = 0.15 x 0.1 = 0.015 moles

Moles of HCl = 0.135 x 0.175 = 0.023

Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution

Part C

165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH

This combination will form a buffer.

Explanation

Moles of HF = 0.165 x 0.1 = 0.0165 moles

Moles of KOH = 0.135 x 0.05 = 0.00675 moles

Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer

Part D

125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl

This combination will form a buffer

Explanation

Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer

Part E

105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl

This combination will form a buffer

Explanation

Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles

Moles of HCl = 0.095 x 0.1 = 0.0095 moles

Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer

5 0

3 years ago