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eimsori [14]
3 years ago
9

Polysaccharides always:a. are a string of 3 or more sugar molecules.b. contain lipids.c. are polymers.d. The first and second ch

oices are correct.e. The first and third choices are correct.
Chemistry
1 answer:
tamaranim1 [39]3 years ago
7 0

Answer:e. The first and third choices are correct.

Explanation:

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Aqueous sodium hydroxide forms a light blue precipitate. What is the formula of the blue precipitate?
lisabon 2012 [21]

Answer:

Cu2+(aq) + 2OH-(aq) => Cu(OH)2(s)

Explanation:

Use of aqueous sodium hydroxide is a precipitation reaction to test for anions or cations. Aqueous sodium hydroxide in a precipitate test forms a insoluble precipitates along with some colors characteristics.

Aqueous sodium hydroxide (NaOH) when mixed with copper(II) (Cu2+) forms a blue precipitate. The formula is as follows:

Cu2+(aq) + 2OH-(aq) => Cu(OH)2(s)

8 0
3 years ago
2 Points
djyliett [7]
A. Camouflage Fur is the correct answer.
7 0
2 years ago
A 25.00 mL sample of the ammonia solution
musickatia [10]

Answer:

1.634 molL-1

Explanation:

The mol ration between NH3 and HCl is 1 : 1

Using Ca Va / Cb Vb = Na / Nb   where a = acid and b = base

Na = 1

Nb = 1

Ca = 0.208 molL-1

Cb = ?

Va = 19.64 mL

Vb = 25.00mL

Solving for Cb

Cb = Ca Va / Vb

Cb = 0.208 * 19.64 / 25.0

Cb = 0.1634 molL-1 (Concentration of diluted ammonia solution)

Using the dilution equation;

C1V1 = C2V2

Initial Concentration, C1 = ?

Initial Volume, V1 = 25.00 mL

Final Volume, V2 =  250 mL

Final Concentration, C2 = 0.1634 molL-1

Solving for C1;

C1 = C2 * V2 / V1

C1 = 0.1634 * 250 / 25.00

C1 = 1.634 molL-1

3 0
3 years ago
Trans-2-butene does not exhibit a signal in the double-bond region of the spectrum (1600–1850 cm−1); however, ir spectroscopy is
Elza [17]
However <em>trans</em>-2-Butene does not give a characteristic peak in 1620-1680 cm⁻¹ region but still the presence of carbon double bond carbon can be detected by detecting following peaks in IR Spectrum.

1)  3010-3100 cm⁻¹:
                               As in trans-2-Butene a hydrogen atoms ate attached to sp² hybridized carbon, therefore the stretching of =C-H (C-H) bond will give a peak of medium intensity in the range of 3010-3100 cm⁻¹.

2)  675-1000 cm⁻¹:
                             Another peak which is given by the bending of =C-H (C-H) bond with strong intensity will appear in the range of 675-1000 cm⁻¹.
3 0
3 years ago
Calculate the percentage of each element in acetic acid, hc2h3o2, and glucose, c6h12o6.
My name is Ann [436]
<em>Acetic acid, HC2H3O2</em>

First, calculate for the molar mass of acetic acid as shown below.
    M = 1 + 2(12) + 3(1) + 2(16) = 60 g

Then, calculating for the percentages of each element.
<em> Hydrogen:</em>
    P1 = ((4)(1)/60)(100%) = <em>6.67%</em>

<em> Carbon:</em>
   P2 = ((2)(12)/60)(100%) = <em>40%</em>

<em>Oxygen</em>
  P3 =((2)(16) / 60)(100%) = <em>53.33%</em>

<em>Glucose, C6H12O6</em>

The molar mass of glucose is as calculated below,
   6(12) + 12(1) + 6(16) = 180

The percentages of the elements are as follow,
 <em> Hydrogen:</em>
   P1 = (12/180)(100%) = <em>6.67%</em>

<em>Carbon:</em>
  P2 = ((6)(12) / 180)(100%) = <em>40%</em>

<em>Oxygen:</em>
  P3 = ((6)(16) / 180)(100%) = <em>53.33%</em>

b. Since the empirical formula of the given substances are just the same and can be written as CH2O then, the percentages of each element composing them will just be equal. 
6 0
3 years ago
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