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Sergeu [11.5K]
3 years ago
9

What’s the answer to the last one?

Chemistry
1 answer:
Nata [24]3 years ago
6 0

There isn't any one answer to this question, as it's asking you to create a mnemonic device to help you remember the names of these types of energies. In case you are confused on what a mnemonic device is, it is a phrase/acronym used to help you remember words with the same first letters of each word in the acronym. Examples:

"Never Eat Soggy Waffles" for "North, East, South, West"

"King Henry Died by drinking chocolate milk" for the metric prefixed "Kilo, Hecto, Deka, base unit, deci, centi, milli".

So, to answer your question, you should come up with a phrase that will help you remember those words. The phrase should have six words with one beginning with each of these letters "N, C, R, E, M, T" in any order.

To give you an example of one I just made up, you could say "Never (Nuclear) Create (Chemical) Metals (Mechanical) That (Thermal) React (Radiant) Excessively (Electrical)".

You might be interested in
consider the titration of hclo4 with koh. what is the ph after 17.0 ml of 0.15 m koh has been added to 15 ml of 0.20 m hclo4?
bezimeni [28]

The ph after 17.0 ml of 0.15 m Koh has been added to 15 ml of 0.20 m hclo4 is  <u>3.347</u>.

Titration is a commonplace laboratory technique of quantitative chemical analysis to determine the attention of an identified analyte. A reagent, termed the titrant or titrator, is ready as a trendy answer of recognized awareness and extent.

<u>Calculation:-</u>

Normality of acid                                               Normality of base

= nMV                                                                        nMV

= 1 × 0. 15 × 0.017                                              1 ×  0. 20 ×0.015 L

= 2.55 × 10⁻³                                                             = 3 × 10⁻³

The overall base will be high

net concentration = 3× 10⁻³ - 2.55 × 10⁻³

                             = 0.45 × 10⁻³

                             = 4.5× 10⁻⁴

pH = -log[4.5 × 10⁻⁴]

    = 4 - log4.4

     = <u>3.347</u>

A titration is defined as 'the manner of determining the amount of a substance A by using adding measured increments of substance B, the titrant, with which it reacts till precise chemical equivalence is completed the equivalence factor.

Learn more about titration here:-brainly.com/question/186765

#SPJ4

7 0
2 years ago
It has been hypothesized that a chemical known as BW prevents colds. To test this hypothesis, 20,000 volunteers were divided int
Amiraneli [1.4K]

Grams of BW

i think thats irtu9rgirg

8 0
3 years ago
How many kL does a 9.51 ´ 109 cL sample contain?
elena-14-01-66 [18.8K]

Answer:

9.51 × 10⁴ kL

Explanation:

Step 1: Given data

Volume of the sample (V): 9.51 × 10⁹ cL

Step 2: Convert "V" to liters

We will use the conversion factor 1 L = 100 cL.

9.51 × 10⁹ cL × (1 L / 100 cL) = 9.51 × 10⁷ L

Step 3: Convert "V" to kL

We will use the conversion factor 1 kL = 1000 L.

9.51 × 10⁷ L × (1 kL / 1000 L) = 9.51 × 10⁴ kL

9.51 × 10⁹ cL is equal to 9.51 × 10⁴ kL.

4 0
3 years ago
Robidium and cesium have similar chemical properties because in the ground it states the atoms of both elements each have
anzhelika [568]
It would be one electron in the outermost shell. Hope this helps!
5 0
3 years ago
(a) The student dissolves the entire impure sample of CuSO4(s) in enough distilled water to make 100.mL of solution. Then the st
Talja [164]

Answer:

Explanation:

Given parameters :

Volume of solution  = 100mL

Absorbance of solution  = 0.30

Unknown:

Concentration of CuSO₄ in the solution = ?

Solution:

There is relationship between the absorbance and concentration of a solution. They are directly proportional to one another.

A graph of absorbance against concentration gives a value of 0.15M at an absorbance of 0.30.

The concentration is 0.15M

Also, we can use:  Beer-Lambert's law;

                    A  = ε mC l

where εm is the molar extinction coefficient

             C is the concentration

             l is the path length

Since the εm is not given and assuming path length is 1;

         Then we solve for the concentration.

3 0
4 years ago
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