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3 years ago

How has Chemistry contributed to what we Eat?

1 answer:

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Food as Chemicals
Food scientists think of cooking in terms of the combination of food chemicals and the reactions they undergo to form a new food product. ... Proteins, carbohydrates, and fats are particular types of molecules and amino acids that combine in predictable ways to make up a food.

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What is the arrangement of water molecules in the gaseous , liquid and soild states ?

const2013 [10]

Gas, would be spread out everywhere, liquid would be loose, and solid would be compact tightly.

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3 years ago

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A solution is made by adding 200 g table salt to 1

pochemuha

We say that the solution is unsaturated.

Explanation:

If the salt solubility is 36 g in 0.1 L of water then we can dissolve 360 g of salt in 1 L of water.

Because the solution contains 200 g of salt in 1 L of water, the solution is unsaturated because more salt can be added until we reach the saturation point.

We call the solution dilute when we compare the concentration of a solution with the concentration of another solution, but here we do not compare different solutions.

Learn more about:

solubility

brainly.com/question/9126711

#learnwithBrainly

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4 years ago

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What is the volume, in liters, of 5.500 mol of C3H3 gas at STP

ZanzabumX [31]

Answer:

123.2 Liters.

Explanation:

At STP (T = 273K & P = 1atm)<em>, one mol of any gas will occupy 22.4 liters</em>.

With the above information in mind, we can <u>calculate how many liters would 5.500 mol of gas occupy</u>:

5.500 mol * 22.4 L / mol = 123.2 L

So 5.500 moles of C₃H₃ would have a volume of 123.2 liters at STP.

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3 years ago

What is the difference between a helium atom and an alpha particle

artcher [175]

Answer:

???????

????

Explanation:

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3 years ago

A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0

3 years ago