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allochka39001 [22]
3 years ago
15

Glucose, one type of monosaccharide, is represented by the chemical formula, c6h12o6. if you were to put together 2 glucose mole

cules, why wouldn't the molecular formula for the newly made disaccharide be c12h24o12

Chemistry
2 answers:
andrew11 [14]3 years ago
7 0

Answer:

This is because of the removal of water molecule during condensation reaction between two glucose molecules.

Explanation:

The molecule of glucose has molecular formula ofC_6H_{12}O_6.

When two molecules of monosaccharide adds up together they form disaccharide molecule of carbohydrate. During the addition of two monosaccharides condensation reaction occurs in which removal of small molecule like water takes place.

C_6H_{12}O_6+C_6H_{12}O_6\rightarrow C_{12}H_{22}O_{11}+H_2O

ElenaW [278]3 years ago
4 0
To combine these you would need to perform Dehydration synthesis which removes a water molecule. So it would be c12h22o11
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What is the speed of a basketball that is thrown 18meters in 4 seconds?
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Explanation:

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speed = 18/4

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If the volume of two containers stays the same , and one containers temperature is lowered, what happens to the pressure in the
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3 years ago
Find the initial concentration of the weak acid or base in each of the following aqueous solutions: (a) a solution of HClO with
Luda [366]

Answer:

a) 0.021 M

b) 0.019 M

Explanation:

To do this, you need to calculate the concentration of ions in solution with the given value of pH for each solution, then, write the chemical equation for both solutions, Set an ICE chart, use the value of Ka and Kb reported for both solutions, and solve for the initial concentration.

This is the general procedure to do it, now let's do it by parts.

<em><u>a) Concentration of HClO pH = 4.6</u></em>

With the given pH, we use the following expression:

pH = -log[H₃O⁺]      From here, we solve for [H₃O⁺]

[H₃O⁺] = 10^(-pH)   (1)

Let's calculate first the hydronium concentration:

[H₃O⁺] = 10^(-4.6) = 2.51x10⁻⁵ M

This value indicates the equilibrium concentration of this ion in solution. Now, to know the initial concentration of the acid, we need to do an ICE chart and write the chemical equation. This is an acid - base reaction, so we need the value of Ka of the acid.

         HClO + H₂O <---------> H₃O⁺ + ClO⁻       Ka = 3x10⁻⁸

I:            Y                                 0          0

C:          -x                                +x         +x

E:           Y - x                            x          x

With this chart, we need to write the expression for Ka which is:

Ka = [H₃O⁺] * [ClO⁻] / [HClO] = x² / Y-x

But we already know the concentration of [H₃O⁺], which is the same for [ClO⁻], and the value of Ka, so all we have to do is replace the values in the above expression and solve for Y:

3x10⁻⁸ = (2.51x10⁻⁵)² / Y - 2.51x10⁻⁵

We can round to Y because "x" is a very small value as it's value of Ka so:

3x10⁻⁸ = (2.51x10⁻⁵)²/Y

Y = (2.51x10⁻⁵)²/3x10⁻⁸

<h2><em>Y = [HClO] = 0.021 M</em></h2>

<em>And this is the initial concentration of the acid.</em>

<u><em>b) Solution of hidrazine pH = 10.2</em></u>

We do the same procedure as part a) with the difference that instead of using Ka , we use Kb and concentration of [OH⁻]. The Kb for hydrazine is 1.3x10⁻⁶

Let's calculate the [OH⁻]:

pOH = 14 - pH

pOH = 14 - 10.2 = 3.8

[OH⁻] = 10^(-3.8) = 1.58x10⁻⁴ M

The chemical equation:

          N₂H₄ + H₂O <---------> N₂H₅⁺ + OH⁻    Kb = 1.3x10⁻⁶

I:            Y                                  0           0

C:          -x                                +x           +x

E:         Y-x                                 x           x

Kb = x²/(Y-x)

1.3x10⁻⁶ = (1.58x10⁻⁴)²/Y

Y = (1.58x10⁻⁴)²/1.3x10⁻⁶

<h2><em><u>Y = [OH⁻] = 0.019 M</u></em></h2>

And this is the initial concentration of hydrazine

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