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emmainna [20.7K]
3 years ago
7

Which statements about reducing sugars are true? D‑Glucose (an aldose) is a reducing sugar. The oxidation of a reducing sugar fo

rms a carboxylic acid sugar. A disaccharide with its anomeric carbons joined by the glycosidic linkage cannot be a reducing sugar. Reducing sugars contain ketone groups instead of aldehyde groups. A reducing sugar will not react with the Cu2+ in Benedict's reagent.
Chemistry
1 answer:
qaws [65]3 years ago
4 0

Answer:

The true statements are given below.

Explanation:

1 D glucose is a reducing sugar

2 The oxidation of reducing sugar forms a carboxylic acid sugar.

D glucose is a reducing sugar because glucose contain a free hydroxyl group (-OH)in its anomeric carbon.

 The oxidation of reducing sugar result in the conversion of -CHO group in case of aldose sugar and -CH2OH group in case of ketose sugar into carboxylic acid(-COOH).

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Clasifica los siguientes elementos en metales, no metales, metaloides , o gases , nobles : flúor , azufre , nitrogeno , cloro ,
professor190 [17]

Answer:

<u>  Translation:</u>

Classify the following elements into metals, nonmetals, metalloids, or gases, noble : fluorine, sulfur, nitrogen, chlorine, magnesium, helium, sodium, bromine, silicon, tellurium.

  • Metals: magnesium and sodium.
  • Nonmetals: fluorine, chlorine, bromine, sulfur and nitrogen.
  • Metalloids: silicon and tellurium.
  • Noble gases: helium.

Explanation:

Metals: They occupy the left and central areas of the Periodic Table; therefore, they constitute a majority group of the elements.

Nonmetals: They are located in the upper right region of the Periodic Table.

Within the Periodic Table, the metalloids lie diagonally from boron to polonium. Items above on the right are nonmetals, and items below on the left are metals.

The noble gases are located in group 18 of the Periodic Table.

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3 years ago
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A student dissolves 4 g of salt into 10 g of water. The total weight of the solution is 14 g. The student then heats the solutio
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C. or B. ok :) !! ! Hopes this hepls
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3 years ago
A civil engineering student is considering buying an electric car. However, the conscious student wants to make sure her carbon
svetlana [45]

Answer:

\text{An gasoline-powered vehicle has }\\\boxed{\text{five times the carbon footprint}}\text{ of an electric vehicle.}

Explanation:

1. Miles travelled in an average month  

\text{Miles} = \text{1 mo} \times \dfrac{\text{52 wk}}{\text{12 mo}} \times \dfrac{\text{5 da}}{\text{1 wk}} \times \dfrac{\text{16 mi}}{\text{1 da}} = \text{347 mi}

2. Using a gasoline powered vehicle  

(a) Moles of heptane used  

n = \text{347 mi} \times \dfrac{\text{1 gal}}{\text{36.5 mi}}\times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{679.5 g}}{\text{1L}} \times \dfrac{ \text{1 mol}}{\text{100.20 g}}= \text{244 mol}

(b) Equation for combustion  

C₇H₁₆ + O₂ ⟶ 7CO₂ + 8H₂O  

(c) Moles of CO₂ formed  

n = \text{244 mol heptane} \times \dfrac{\text{7 mol CO$_{2}$}}{\text{1 mol heptane}} = \text{1710 mol CO$_{2}$}

(d) Volume of CO₂ formed  

At 20 °C and 1 atm, the molar volume of a gas is 24.0 L.  

V = \text{1710 mol } \times \dfrac{\text{24.0 L}}{\text{1 mol}} \times \dfrac{\text{1 gal}}{\text{3.785 L}} = \textbf{10 800 gal}

3. Using an electric vehicle  

(a) Theoretical energy used  

\text{Theor. Energy} = \text{347 mi} \times \dfrac{\text{1 kWh}}{\text{5.2 mi}} = \text{66.7 kWh theor.}

(b) Actual energy used  

The power station is only 85 % efficient.  

\text{Actual energy used} = \text{66.7 kWh theor.} \times \dfrac{\text{100 kWh actual}}{\text{ 85 kWh theor.}}\times \dfrac{\text{3600 kJ}}{\text{1 kWh}}\\\\ = 2.82\times 10^{5} \text{ kJ}\

(c) Combustion of CH₄

CH₄ + 2O₂ ⟶ CO₂ +2 H₂O

(d) Equivalent volume of CO₂

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4. Comparison  

\dfrac{V_{\text{gasoline}}}{V_{\text{electric}}} = \dfrac{10800}{2180} = 5.0\\\\ \text{An gasoline-powered vehicle has }\\ \boxed{\textbf{ five times the carbon footprint}}\text{ of an electric vehicle.}

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Use a Punnett square to predict the cross of a homozygous tall parent with a homozygous short parent, if tall is dominant over s
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Answer:

All offspring are tall when a homozygous tall parent with homozygous short parent.

Explanation:

When we crossed  homozygous tall parent with homozygous short parent, we conclude that all offspring are tall, because homozygous short parent are supressed under the homozygous tall parent, due to law of dominance.

Law of dominance states that, recessive alleles are suppressed by dominant alleles but they can appear in F2 generation.

Using a punett square, we can predict the cross between homozygous tall and homozygous short parent.      

  The phenotypes are: All are tall plants (4:0).

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3 years ago
Calculate the average weight of the dogs listed in the chart below.
pishuonlain [190]

Answer:

19

Explanation:

Once you add all the weights together, you end up with 114 then you divide it by how many dogs there are, 6. 114/6 = 19. The average is 19 lbs.

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3 years ago
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