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Misha Larkins [42]
3 years ago
6

3. Why do elements in group 1-17 react to form bonds?

Chemistry
1 answer:
iren [92.7K]3 years ago
6 0

Answer:

Elements in the same group of the periodic table have the same number of valence electrons. These are the electrons in their outer energy level that can be involved in chemical reactions. ... All the elements in group 1 have just one valence electron. This makes them very reactive.

Explanation:

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What element is located in group 2, period 2?
telo118 [61]

This element is beryllium.

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How is phosphorylation of glyceraldehyde 3-phosphate in the payment phase of glycolysis different from phosphorylation of glucos
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In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place.  During this phase, the conversion of glyceraldehyde-3-phosphate to pyruvate and the coupled formation of ATP take place. Because Glucose is split to yield two molecules of D-Glyceraldehyde-3-phosphate, each step in the payoff phase occurs twice per molecule of glucose.

Glyceraldehyde 3-phosphate dehydrogenase Simultaneous oxidation and phosphorylation of G3P produce 1,3-bisphosphoglycerate (1,3-BPG) and nicotine adenine dinucleotide (NADH).

The divalent cation also affected the response of the enzyme from the endosperm and shoots to adenine nucleotides and inorganic pyrophosphate.

This phase is also called the glucose activation phase. In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place. Steps 1, 2, 3, 4, and 5 together are called the preparatory phase.

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7 0
1 year ago
HELPPPPPP<br>Calculate the frequency of light with a wavelength 559nm​
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5 0
3 years ago
How many moles of oxygen are necessary to generate 28 moles of water, according to the following equation: 2H2+O2→2H2O
valentinak56 [21]

The number of moles of oxygen required to generate 28 moles of water from the reaction is 14 moles

<h3>Balanced equation </h3>

2H₂ + O₂ —> 2H₂O

From the balanced equation above,

2 moles of water were obtained from 1 mole of oxygen

<h3>How to determine the mole of oxygen needed </h3>

From the balanced equation above,

2 moles of water were obtained from 1 mole of oxygen

Therefore,

28 moles of water will be obtained from = 28 / 2 = 14 moles of oxygen

Thus, 14 moles of oxygen are needed for the reaction

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5 0
2 years ago
Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following
bearhunter [10]

Answer:

210.7~g~MgCO_3

Explanation:

We have to start with the <u>reaction</u>:

MgCO_3~->~MgO~+~CO_2

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the <u>number of moles</u> that we have in the 110.0 g of carbon dioxide, to this, we have to know the <u>atomic mass of each atom</u>:

C: 12 g/mol

O: 16 g/mol

Mg: 23.3 g/mol

If we take into account the number of atoms in the formula, we can calculate the <u>molar mass</u> of carbon dioxide:

(12*1)+(16*2)=44~g/mol

In other words: 1~mol~CO_2=~44~g~CO_2. With this in mind, we can calculate the moles:

110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2

Now, the <u>molar ratio</u> between carbon dioxide and magnesium carbonate is 1:1, so:

2.5~mol~CO_2=2.5~mol~MgCO_3

With the molar mass of MgCO_3 ((23.3*1)+(12*1)+(16*3)=84.3~g/mol. With this in mind, we can calculate the <u>grams of magnesium carbonate</u>:

2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3

I hope it helps!

8 0
3 years ago
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