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tatuchka [14]
3 years ago
5

Question number 1 they are asking for highest possible theoretical temperature?

Chemistry
1 answer:
den301095 [7]3 years ago
7 0

I think the highest theoretical temperature would be 36.5°C.

Hope I helped, sorry if not.

You might be interested in
What do scientists call organisms, such as plants, which make their own food?
larisa [96]

Most plants are called autotrophs they make their own food

5 0
3 years ago
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
How is liming a lake similar to a doctor prescribing medicine for a patient?
Jlenok [28]
An anti-acid medication
3 0
2 years ago
A 3.82 g piece of limestone contains 2.62 g of CaCO3
Kobotan [32]

Considering the definition of percentage by mass, the mass percentage of CaCO₃ is 68.59%.

<h3>What is mass percentage</h3>

The percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.

In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.

The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:

mass percentage=\frac{mass of solute}{mass of solution}x100

<h3>Mass percentage of CaCO₃</h3>

In this case, you know:

  • mass of CaCO₃: 2.62 grams
  • mass of limestone: 3.82 grams

Replacing in the definition of mass percentage:

mass percentage=\frac{2.62 grams}{3.82 grams}x100

<u><em>mass percentage= 68.59 %</em></u>

Finally, the mass percentage of CaCO₃ is 68.59%.

Learn more about percentage by mass:

brainly.com/question/24201923

brainly.com/question/9779410

brainly.com/question/17030163

#SPJ1

7 0
1 year ago
How would you prepare 100.0 ml of.400 m CaCl2 from a stock solution of 2.00 M CaCl2?
docker41 [41]
C₀=2 mol/l
c₁=0.400 mol/l
v₁=100.0 ml = 0.1 l

c₁v₁=c₀v₀

v₀=c₁v₁/c₀
v(H₂O)=v₁-v₀

v₀=0.1*0.400/2=0.02 l = 20 ml
v(H₂O)=100 - 20 = 80 ml

It is necessary to mix 20 ml of the feed solution and 80 ml of water.

4 0
3 years ago
Read 2 more answers
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