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ira [324]
3 years ago
12

How are sucrose (table sugar) and the starch in flour chemically related? what makes them different from each other?

Chemistry
1 answer:
marysya [2.9K]3 years ago
3 0
Both the sucrose ( table sugar)  and the starch is composed of glucose molecules C6H12O6. They are made up of monosaccharides. Both are covalently bonded with glucose molecules. The only difference is that sucrose<span> is made up of the disaccharide sucrose but starch</span><span> is a polysaccharide.</span>
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All the mixture are harmful for us. is this statement true? justify with two examples​
Taya2010 [7]

Answer:

no.it is false.some are useful.like mixture of h2 and O 2 to form water.

3 0
2 years ago
Read 2 more answers
Cars run on gasoline, where octane (C8H18) is the principle component. This combustion reaction is responsible for generating en
Bezzdna [24]

Answer:

  • 10.19 g CO₂
  • 4.69 g H₂O

Explanation:

The combustion reaction of Octane is:

  • C₈H₁₈ → 8CO₂ + 9H₂O

To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.

We calculate the mass of Octane from the given volume and density, using the following <em>conversion factors</em>:

  • 1 gallon = 3.785 L
  • 1 L = 1000 mL

Now we<u> convert 1.24 gallons to mL</u>:

  • 1.24 gallon * \frac{3.785L}{1gallon} *\frac{1000mL}{1L} = 4693.4 mL

We <u>calculate the mass of Octane</u>:

  • 4693.4 mL * 0.703 g/mL = 3.30 g Octane

Now we use the <em>stoichiometric ratios</em> and <em>molecular weights</em> to <u>calculate the mass of CO₂ and H₂O</u>:

  • CO₂ ⇒ 3.30 g Octane ÷ 114g/mol * \frac{8molCO_{2}}{1molOctane} * 44 g/mol =  10.19 g CO₂
  • H₂O ⇒ 3.30 g Octane ÷ 114g/mol * \frac{9molH_{2}O}{1molOctane} * 18 g/mol = 4.69 g H₂O

7 0
3 years ago
Which of the following statements is true?
rjkz [21]

its A. Genes make up chromosomes. it did it on study island..... i did the other answer the other person got and i got it wrong....and then it told me it was A. Genes make up chromosomes.

HOPE I COULD HELP

3 0
3 years ago
Read 2 more answers
How many bromine atoms are present in 39.0 g of ch2br2?
Mandarinka [93]
Answer is: there is  2,69·10²³ atoms of bromine.
m(CH₂Br₂) = 39,0 g.
n(CH₂Br₂) = m(CH₂Br₂) ÷ M(CH₂Br₂).
n(CH₂Br₂) = 39 g ÷ 173,83 g/mol.
n(CH₂Br₂) = 0,224 mol.
In one molecule of CH₂Br₂, there is two bromine atoms, so:
n(CH₂Br₂) : n(Br) = 1 : 2.
n(Br) = 0,448 mol.
N(Br) = n(Br) · Na.
N(Br) = 0,448 mol · 6,022·10²³ 1/mol.
n(Br) = 2,69·10²³.
7 0
3 years ago
If 20.0 mL of a 0.0800 M HNO3, 35.0 mL of a 0.0800 M KSCN, and 40.0 mL of a 0.0800 M Fe(NO3)3 are combined, what is the initial
Tems11 [23]

Answer:

0.0295M

Explanation:

As you can see, in the mixture you have KSCN and other compounds. The KSCN in solution is dissolved in K⁺ ions and SCN⁻ ions. That means initial concentration of SCN⁻ ions is the same of KSCN, 0.0800M.

You are adding 35.0mL of this solution and the total volume of the mixture is 20.0mL + 35.0mL + 40.0mL = 95.0mL.

That means you are diluting your solution 95.0mL / 35.0mL = 2.714 times.

And the concentration of SCN⁻ is:

0.0800M / 2.714 =

<h3>0.0295M </h3>

4 0
3 years ago
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