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Gnesinka [82]
3 years ago
9

How are Kelvin scale and celcius scale similar ​

Chemistry
1 answer:
Alekssandra [29.7K]3 years ago
7 0

Answer: The Kelvin scale is related to the Celsius scale. The difference between the freezing and boiling points of water is 100 degrees in each, so that the kelvin has the same magnitude as the degree Celsius.

Explanation:

Celsius is, or relates to, the Celsius temperature scale (previously known as the centigrade scale). The degree Celsius (symbol: °C) can refer to a specific temperature on the Celsius scale as well as serve as a unit increment to indicate a temperature interval(a difference between two temperatures or an uncertainty). “Celsius” is named after the Swedish astronomer Anders Celsius (1701-1744), who developed a similar temperature scale two years before his death.

K = °C + 273.15

°C = K − 273.15

Until 1954, 0 °C on the Celsius scale was defined as the melting point of ice and 100 °C was defined as the boiling point of water under a pressure of one standard atmosphere; this close equivalence is taught in schools today. However, the unit “degree Celsius” and the Celsius scale are currently, by international agreement, defined by two different points: absolute zero, and the triple point of specially prepared water. This definition also precisely relates the Celsius scale to the Kelvin scale, which is the SI base unit of temperature (symbol: K). Absolute zero—the temperature at which nothing could be colder and no heat energy remains in a substance—is defined as being precisely 0 K and −273.15 °C. The triple point of water is defined as being precisely 273.16 K and 0.01 °C.

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Assoli18 [71]
I believe distillation is used to separate solvents of different kind as long as it is heated to its boiling temp
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2 years ago
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0.415 g of an unknown triprotic acid are used to make a 100.00 mL solution. Then 25.00 mL of this solution is transferred to an
Arturiano [62]

Answer:

Explanation:

Initial burette reading = 1.81 mL

final burette reading = 39.7 mL

volume of NaOH used = 39.7 - 1.81 = 37.89 mL .  

37.89 mL of .1029 M NaOH is used to neutralise triprotic acid

No of moles contained by 37.89 mL of .1029 M NaOH

= .03789 x .1029 moles

= 3.89 x 10⁻³ moles

Since acid is triprotic ,  its equivalent weight = molecular weight / 3

No of moles of triprotic acid = 3.89 x 10⁻³ / 3

= 1.30   x 10⁻³ moles .

8 0
3 years ago
The two balanced equations (1) and (2) are for reactions in which gaseous carbon dioxide is produced from the combustion of (1)
V125BC [204]

Answer:

2.0 mol of oxygen are consumed.

Step-by-step explanation:

You know that you will need a balanced equation with masses, moles, and molar masses, so gather all the information in one place.

M_r:   28.0                 44.0

          CO  + ½O₂ ⟶ CO₂ + 67.6 kcal

m/g:   112

<em>Step 1</em>. Convert grams of CO to moles of CO

1 mol CO = 28.0 g CO

Moles of CO = 112 × 1/28.0

<em>Step 2.</em> Convert moles of CO to moles of CO₂.

The molar ratio is 1 mol CO₂ to 1 mol CO

Moles of CO₂ = 4.000 × 1/1

Moles of CO₂ = 4.00 mol CO₂

Option A is <em>wrong</em>.

<em>Step 3.</em> Calculate the amount of heat generated.

q = ΔH

The conversion factor is 67.6 kcal/1 mol CO₂

q = 4.00 × 67.6

q = 270 kJ

Option B is <em>wrong</em>, because it gives the heat generated by 1 mol of CO.

<em>Step 4. </em>Calculate the moles of O₂ consumed

Moles of O₂ = 2.00 mol O₂

Option C is correct.

<em>Step 5.</em> Calculate the moles of CO₂ formed

Already done in Step 2.

Moles of CO₂ = 4.00 mol CO₂

Option D is <em>wrong. </em>

<em> </em>

<em>Step 6</em>. Calculate the moles of O₂ produced

Already done in Step 4.

Moles of O₂ = 2.00 mol O₂

Option E is <em>wrong. </em>

3 0
4 years ago
: Citric acid, H3C6H5O7, is a triprotic acid. Consider a buffer system comprising H2C6H5O7 - and HC6H5O7 2- ions. What is the ne
olchik [2.2K]

Answer:

The Net reaction is    

  1.    H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction
  2.    H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH
  3.    HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH  

Explanation:

From the Question we are told that the buffers are

                H_2C_6H_5O_7^ - and  HC_6H_5O_7 ^{ 2-}

When NaOH is added the Net ionic reaction would be

  1.    H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction
  2.    H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH
  3.    HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH  

             

8 0
3 years ago
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Based on what you’ve determined regarding one mole of sand and one mole of water, which statements must be correct? Select all t
Igoryamba

Answer:

-The mole is appropriate only for counting things that are very small.

-One mole is a very large number of something.

-A molecule of water is much smaller than a grain of sand.

Explanation:

8 0
3 years ago
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