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Aleks04 [339]
3 years ago
10

Which of the following statements about fats is FALSE?

Chemistry
2 answers:
Assoli18 [71]3 years ago
8 0

Answer:

They dissolve easily in the body’s fluids

vodka [1.7K]3 years ago
7 0
The answer is: The dissolve easily in the body’s fluids.

Fats don’t dissolve easily.
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What factor can limit the buffering capacity of a molecule?
Bad White [126]

Answer:

The concentration and pKa of an acid and its conjugate base can limit the buffering capacity of a molecule.

Explanation:

A buffer is an acid and its conjugate base and the quality of it depends on its buffer capacity. This buffer capacity is the resistance to change the pH of the solution when strong acids or bases are added. The buffer capacity is related to the buffer concentration, the concentration of the acid end its conjugate base. Also, every acid has a pKa and the buffer capacity is at its maximum at the pKa value and can buffer the solution between ± 1 the pKa value.

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A chemistry student was asked to calculate the number of moles of iron required to react with 1.20 mol of oxygen to produce iron
nikitadnepr [17]
I’m on the same one smh
6 0
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What are benthos?
WINSTONCH [101]
Benthos are "organisms inhabiting the seafloor" according to one source, so I would say the answer is C.
4 0
3 years ago
Read 2 more answers
Which type of heat exchanger shell and tube or plate heat exchanger is suitable for evaporation?
klasskru [66]

Answer:

shell and tube type heat exchanger

Explanation:

for evaporation the shell and tube type heat exchanger is best suited.

  • in the plate heat exchanger there is gaskets in between every part so this part become weak part in heat echanger and there is possibilities of leakage through this part, there is no such problem in shell and tube type.
  • the plate type cant be used when there is high temperature and high pressure drop but shell and tube type can be used
  • in evaporation there the liquids change into vapors due to which there is sudden change in pressure and in which plate type is not used because there is chances of leakage
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
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