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Maru [420]
3 years ago
15

What are the three elements found in all sugars? What is their ratio?

Chemistry
2 answers:
Vikki [24]3 years ago
3 0

Answer: Carbon Hydrogen and water

1:2:1

Explanation:

Carbohydrates are made of Carbon Hydrogen and water. Hence, they are represented by a general formula Cx(H2O)y.

Carbohydrates have carbon, hydrogen and oxygen in ratio 1:2:1, as is found in glucose (C6H12O6), Glycerose (C3H6O3), Ribose (C5H10O5).

Thus, in all sugars carbon, hydrogen and oxygen are found in ratio 1:2:1

Alborosie3 years ago
3 0

Answer:

C12H22O11

12:22:11

Explanation:

Sugar is composed from:

12 atoms of carbon

22 atoms of hydrogen

11 atoms of oxygen

So their ratio in this order is 12:22:11

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Will ethane ionize in a solution
Jet001 [13]

Answer:

Ethane can be ionized

Explanation:

Ionization or ionisation is the process by which an atom or a molecule acquires a negative or positive charge by gaining or losing electrons, often in conjunction with other chemical changes. The resulting electrically charged atom or molecule is called an ion.

hope this helps out!

<em>~</em><em>A</em><em>d</em><em>r</em><em>i</em><em>a</em><em>n</em><em>n</em><em>a</em>

5 0
3 years ago
What does an atomic number represent in an atom? 1. Number of protons 2. Number of electrons. 3. Number of neutrons. 4. Number o
WINSTONCH [101]

4: The mass amount of protons and neutrons in an atom

We just covered this in my chem class

5 0
4 years ago
Read 2 more answers
The equilibrium constant Kp for the reaction I2(g) + Br2(g) ⇀↽ 2 IBr(g) + 11.7 kJ is 280 at 150◦C. Suppose that a quantity of IB
mixas84 [53]

Answer:

\large \boxed{\text{0.0120 atm }}

Explanation:

The balanced equation is

I₂(g) + Br₂(g) ⇌ 2IBr(g)

Data:

      Kp = 280

p(IBr) = 0.200 atm

1. Set up an ICE table.

Let p = the initial pressure of IBr. Then

\begin{array}{ccccccc}\rm \text{I}_{2}& + & \text{Br}_{2} & \, \rightleftharpoons \, & \text{2IBr} &  &  \\0 & & 0 & & p & & \\+x &   & +x  & & -2x &   &\\x &   & x} &   & 280 & & \\\end{array}

2. Calculate p(I₂)

\begin{array}{rcl}K_{\text{p}}&=&\dfrac{p_{\text{IBr}}^{2}} {p_{\text{I}_2}^{2}}\\\\280&=&{\dfrac{0.200^{2}}{x^{2}}&&\\\\280x^{2} & = &0.0400\\x^{2} & = &\dfrac{0.0400}{280 }\\\\& = & 1.429 \times 10^{-4}\\x & = & \textbf{0.0120 atm}\\\end{array}\\\text{The partial pressure of iodine is $\large \boxed{\textbf{0.0120 atm }}$}}

Check:

\begin{array}{rcl}{\dfrac{0.200^{2}}{0.0120^{2}}}&=&280\\\\280& =& 280\\\end{array}

8 0
3 years ago
Label the signals due to ha, hb, and hc in the 1h nmr spectra.
trapecia [35]
Structure along with ¹H-NMR is shown below.

Signal for Hₐ;
                     Based on multiplicity of of the peak, a Singlet peak (the only singlet peak present in spectrum) at 2.2 ppm was assigned to Hₐ. As the methyl group in not coupling with any other proton so it gave a singlet peak.

Signal for Hb;
                     Based on multiplicity of of the peak, a Quartet peak (the only quartet peak present in spectrum) at 2.4 ppm was assigned to Hb. As the methylene group in coupling with methyl group having three protons so it gave a Quartet peak.

Signal for Hc;
                     Based on multiplicity of of the peak, a Triplet peak (the only triplet peak present in spectrum) at 1.1 ppm was assigned to Hc. As the methyle group in coupling with methylene group having two protons so it gave a Triplet peak.

3 0
4 years ago
Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring
ludmilkaskok [199]

Answer:

1.20 V

Explanation:

Pb(s) + Br_2(l)\rightarrow Pb^{2+}(aq) + 2Br^-(aq)

Here Pb undergoes oxidation by loss of electrons, thus act as anode. Bromine undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials.

Given,

Pb^{2+}(aq) + 2 e^-\rightarrow Pb(s)

E^0_{[Pb^{2+}/Pb]}= -0.13\ V

Br_2(l) + 2 e^-\rightarrow 2 Br(aq)

E^0_{[Br_2/Br^{-}]}=+1.07\ V

E^0=E^0_{[Br_2/Br^{-}]}- E^0_{[Pb^{2+}/Pb]}

E^0=+1.07- (-0.13)\ V=1.20\ V

6 0
4 years ago
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