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Akimi4 [234]
4 years ago
6

Compare and contrast the burning of wood and the metabolism of glucose in your cells. How are they similar, and how are they dif

ferent?
Chemistry
1 answer:
Alisiya [41]4 years ago
8 0

Burning of wood is a combustion reaction and the metabolism of glucose in your cells is cellular respiratory reaction.

Cellular respiration releases stored energy in glucose molecules and converts it into a form of energy that can be used by cells.

Cellular respiration is the process by which organisms combine oxygen with foodstuff molecules, diverting the chemical energy in these substances into life-sustaining activities and discarding, as waste products, carbon dioxide and water.

Wood as well as many common items that combust are organic (i.e., they are made up of carbon, hydrogen and oxygen). When organic molecules combust the reaction products are carbon dioxide and water (as well as heat)

Similarities:

1. Combustion reaction and metabolism of glucose both require oxygen.

2. Combustion and metabolism of glucose both product carbon dioxide and water

3. Both produces by-products: After cellular respiration and combustion have gotten what they needed from the wood, there will be byproducts from the conversion. In the case of combustion, they are noxious gases like carbon monoxide. In the case of respiration, the sugar molecule is broken into two molecules of pyruvic acid.

4. Catalyst: While breaking apart the bonds to release the stored energy either combustion or sugars for respiration the bonds will not broken by themselves. In each case, a catalyst is required to start the reaction that will break the bonds apart. In the case of combustion, the catalyst is a spark. Wood are flammable, so the spark will ignite the burning, breaking apart the bonds and releasing the energy. For respiration, enzymes are used to break the sugar molecule apart.

Differences

1. Glucose metabolism produces a chemical energy, while combustion produces light and heat energy.

2. Glucose metabolism is an endothermic reaction while combustion is an exothermic reaction (produces heat)

You might be interested in
The prefix mili-means a thousand
OLEGan [10]
<h2>Answer:</h2>

<em>Given data:</em>

1 kilogram = 1,000 grams

how many kilograms is 1,216 grams?

<em>Solution:</em>

1000 grams = 1 kilogram

1 gram = 1/1000 kilograms

1216 grams = 1/1000 * 1216 = 1.216 kilograms.

                                                    <em> Hence 1216 grams =  1.216 kilograms.</em>

3 0
3 years ago
Predict the product(s) of the following reaction. (Separate substances in a list with a comma. Include states-of-matter under th
kiruha [24]

Answer:

Cu₃(PO₄)₂(s), RbNO₃(aq)

Explanation:

This reaction looks like a possible <em>double replacement reaction</em>, in which the metal ions have exchanged partners.

3Cu(NO₃)₂(aq) + 2Rb₃PO₄(aq) ⟶ Cu₃(PO₄)₂ + 6RbNO₃

You must recall the pertinent <em>solubility rules</em>:

1. Salts of Group 1 elements (e.g., Rb⁺) are soluble.

2. Salts containing nitrate ions (NO₃⁻) are soluble.

3. Most salts containing phosphate ions (PO₄³⁻) are insoluble

According to Rules 1 and 1, RbNO₃ is soluble.

According to Rule 3, Cu₃(PO₄)₂ is insoluble.

∴ 3Cu(NO₃)₂(aq) + 2Rb₃PO₄(aq) ⟶ Cu₃(PO₄)₂(s) + 6RbNO₃(aq)

6 0
4 years ago
A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.25 grams of aluminum foil in a solut
mrs_skeptik [129]

Answer: The correct answer is about 0.20 grams of copper (II) is formed, and some aluminum is left in the reaction mixture.

Explanation:

To calculate the number of moles, we use the formula:

Moles=\frac{\text{Given mass}}{\text{Molar mass}}

  • <u>Moles of Aluminium:</u>

Molar mass of aluminium = 27 g/mol

Given mass of aluminium =  0.25g

Putting values in above equation, we get:

Moles=\frac{0.25g}{27g/mol}=0.00925mol

  • <u>Moles of Copper chloride:</u>

Molar mass of copper chloride = 134.45 g/mol

Given mass of copper chloride=  0.40g

Putting values in above equation, we get:

Moles=\frac{0.40g}{134.45g/mol}=0.00297mol

For the given chemical equation:  

4Al(s)+3CuCl_2(aq.)\rightarrow 2Al_2Cl_3(aq.)++3Cu(s)

By Stoichiometry,

3 moles of Copper chloride reacts with 4 moles of aluminium

So, 0.00297 moles of copper chloride will react with = \frac{4}{3}\times 0.00297 = 0.00396 moles of Aluminium.

As, the required moles of aluminium is less than the given moles of aluminium. Hence, it is considered as the excess reagent.

Copper chloride is the limiting reagent in the given chemical reaction because it limits the formation of products.

By Stoichiometry of the reaction:

3 moles of copper chloride produces 3 moles of copper metal.

So, 0.00297 moles of copper chloride will produce = \frac{3}{3}\times 0.00297 = 0.00297 moles of copper metal.

Now, to calculate the given mass of copper metal, we use the equation required to calculate moles:

Molar mass of copper = 63.5 g/mol

Putting values in that equation, we get:

0.00297mol=\frac{\text{given mass}}{63.5g/mol}

Given mas of copper metal = 0.188grams (approx 0.20grams)

For the given reaction, some of the aluminum is left behind because it is the excess reagent in the reaction.

Hence, the correct answer is about 0.20 grams of copper (II) is formed, and some aluminum is left in the reaction mixture.

6 0
3 years ago
Read 2 more answers
Select correct examples of linear molecules for five electron groups. Select correct examples of linear molecules for five elect
andrew-mc [135]

Explanation:

Formula to calculate hybridization is as follows.

                Hybridization = \frac{1}{2}[V+N-C+A]

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

So, hybridization of BeCl_{2} is as follows.

              Hybridization = \frac{1}{2}[V + N - C + A]

                                    = \frac{1}{2}[2 + 2]

                                    = 2

Hybridization of BeCl_{2} is sp. Therefore, BeCl_{2} is a linear molecule. There will be only two electron groups through which Be is attached.

Similarly, hybridization of XeF_{2} is calculated as follows.

         Hybridization = \frac{1}{2}[V + N - C + A]

                                    = \frac{1}{2}[8 + 2]

                                    = 5

Therefore, hybridization of XeF_{2} is sp^{3}d. Therefore, [tex]XeF_{2} is also a linear molecule. Though there are three lone pair of electrons present on a xenon atom and it is further attached with fluorine atoms through two electron pairs. Hence, there are in total five electron groups.

Thus, we can conclude that out of the given options XeF_{2} is the correct examples of linear molecules for five electron groups.

8 0
3 years ago
Electron sharing can be depicted by a Lewis dot structure, in which element symbols are surrounded by dots that represent the va
Nina [5.8K]

Answer:

Outermost

Covalent

Two

One

Two

Two

Covalent

One

Explanation:

A covalent bond is formed when an atom shares two electrons with another atom. These shared electrons could be contributed by each of the bonding atoms or by only one of the bonding atoms.

Hydrogen has the electronic configuration of 1s1. This implies that it has only one electron in its valence shell although the 1s shell can accommodate two electrons. When the atomic orbitals of carbon and hydrogen overlap, they share two electrons and hydrogen is now associated with two electrons in a covalent bond.

Since hydrogen possesses only one valence electron, it can not be bonded to two atoms.

4 0
3 years ago
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