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3 years ago

Identify the term that applies to each definition.

Chemistry

1 answer:

Illusion [34]3 years ago

6 0

Answer:

A. Reference blank

B. Cuvettes

C. Transmittance

D. Absorbance

E. Wavelength

Explanation:

A reference blank is a sample prepared using the solvent and any other chemicals in the sample solutions, but not the absorbing substance.

A square-shaped container, typically made of quartz, designed to hold samples in a spectrophotometer is known as Cuvettes.

A measurement of the amount of light that passes through a sample or percentage of light transmitted by the sample, with the respective intensities of the incident and transmitted beams is called Transmittance.

The measurement of the amount of light taken in by a sample is known as Absorbance

The wavelength is also the distance travelled by the wave during a period of oscillation. In spectrophotometry, the unit is inversely proportional to energy and commonly measured in nanometers

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An aqueous 0.300 M glucose solution is prepared with a total volume of 0.150 L. The molecular weight of

kherson [118]

Answer:

We need 8.11 grams of glucose for this solution

Explanation:

Step 1: Data given

Molarity of the glucose solution = 0.300 M

Total volume = 0.150 L

The molecular weight of glucose = 180.16 g/mol

Step 2: Calculate moles of glucose in the solution

Moles glucose = molarity solution * volume

Moles glucose = 0.300 M * 0.150 L

Moles glucose = 0.045 moles glucose

Step 3: Calculate mass of glucose

MAss glucose = moles glucose* molecular weight of glucose

MAss glucose = 0.045 moles * 180.16 g/mol

MAss glucose = 8.11 grams

We need 8.11 grams of glucose for this solution

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3 years ago

Which of the following is true about most predators? (Select all that apply.)

-Dominant- [34]

Answer:

select all of them except they are the biggest

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2 years ago

Which statement about the atomic nucleus is correct

Sladkaya [172]

The nucleus is made of protons and neutrons. It has a positive charge.

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3 years ago

Four different objects are placed on a number line at 0. The chart describes the motion of each object. Using the information in

mote1985 [20]

The chart with the information to answer this question may be found in other Brainly's link, so I attach it now.

Answer:

Explanation:

To facilitate the accompaniment of the reasoning, I copy the data here:

Object Motion

X 3 units left, 3 units right

Y 6 units right, 18 units right

W 8 units left, 24 units right

Z 16 units right, 8 units left

The distance is the scalar magnitude that indicates the total length of the path traveled.

The displacement is the vector that indicates how far the end point and the initial point are.

In order to calculate the distance, the displacement and the ratio r = distance/displacement, to check which object has a distance that is three times as great as its displacement, i.e. r = 3, we use this new chart:

Object Motion Distance Displacement ratio

X 3 units left, 3 units right 3 + 3 = 6 3 - 3 = 0 ∞

Y 6 units right, 18 units right 6 + 18 = 24 6 + 18 = 24 1

W 8 units left, 24 units right 8 + 24 = 32 -8 + 24 = 16 2

Z 16 units right, 8 units left 16 + 8 = 24 16 - 8 = 8 3

The ratios were calculated dividing the distance by the displacement: distance / displacement.

For the object Z, that was 24 / 8 = 3. Thus, the object that has a distance that is three times as great as its displacement is the object Z.

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3 years ago

Read 2 more answers

Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)

Dovator [93]

Answer:

1. 0.97 V

2. $Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)$

Explanation:

In this case, we can start with the half-reactions:

$Ag^+~_(_a_q_)->~Ag_(_s_)$

$Al_(_s_)~->~Al^+^3~_(_a_q_)$

With this in mind we can add the electrons:

$Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)$ Reduction

$Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~$ Oxidation

The reduction potential values for each half-reaction are:

$Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_)$ - 0.69 V

$Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_)$ -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to flip the reduction potential value:

$Al_(_s_)~->~Al^+^3~+~2e^-~$ +1.66 V

Finally, to calculate the overall potential we have to add the two values:

1.66 V - 0.69 V = 0.97 V

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

$Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)$

I hope it helps!

3 0

3 years ago