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ANEK [815]
3 years ago
11

What type of compound is fe3n2?

Chemistry
1 answer:
Kobotan [32]3 years ago
3 0
I believe the type of compound fe3n2 is ionic. 
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allsm [11]

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north is the word that best completes the sentence

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3 years ago
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When comparing differences in an amino acid residue at one position between the different proteins in the multiple alignment, wh
bogdanovich [222]

Answer:

To understand the utility in sequence comparison and in the search for proteins that have a common evolutionary origin, you need to be clear about some concepts about how to evolve proteins. The idea that is accepted is that throughout the evolution some species are giving rise to new ones. Behind this is the genetic variation of organisms, that is, the evolution of genomes and their genes, as well as the proteins encoded by them.

Explanation:

Three ways can be distinguished by which genes evolve, and by proteins: mutation, duplication and shuffling of domains. When differences between homologous protein sequences are observed, these differences change to do with the way of life of the organism, an example of this, bacteria that live in hot springs at very high temperatures have proteins with a very high denaturation temperature, and these proteins are usually richer in cysteines. On the other hand, the fact that in positions of the sequences they remain unchanged (preserved positions), means that these have a special importance for the maintenance of the structure or function of the protein and its modification has not been tolerated throughout of evolution

8 0
2 years ago
Which gas has approximately the same density as c2h6 at stp no nh3 h20 so2?
zmey [24]
1 mole of any gas occupy 22.4 L at STP (standard temperature and pressure, 0°C and 1 atm).

Let given gases be 1 mole. So their volumes will be the same, 22.4 liters.

Density is the ratio of mass to volume.

By formula; density= mass/volume; d=m/V

To find out masses of gases, do the mole calculation.

By formula; mole= mass/molar mass; n= m/M; m= n*M

Molar masses are calculated as
1. C₂H₆ (ethane)                 = 2*12 g/mol + 6*1 g/mol= 30 g/mol
2. NO (nitrogen monoxide) = 1*14 g/mol + 1*16 g/mol= 30 g/mol
3. NH₃ (ammonia)              = 1*14 g/mol + 3*1 g/mol= 17 g/mol
4. H₂O (water)                    = 2*1 g/mol + 1*16 g/mol= 18 g/mol
5. SO₂ (sulfur dioxide)        = 1*32 g/mol + 2*16 g/mol= 64 g/mol
Use Periodic Table to get atomic mass of elements.

Since their volumes are equal, compounds having the same molar mass will have the same density. 
Recall the formula d= m/V.

Ethane and nitrogen monoxide have the same density.

The answer is C₂H₆ and NO.
6 0
2 years ago
A sample of a compound contains 60.0g C and 5.05 H it’s molar mass is 78.12g what is the compounds molecular formula
Svetach [21]
The compound's molecular formula is C6H6
6 0
3 years ago
Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241
anygoal [31]

Answer : The enthalpy of the given reaction will be, -1048.6 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is:

XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)    \Delta H=?

The intermediate balanced chemical reactions are:

(1) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)     \Delta H_1=-241.8kJ

(2) X(s)+2Cl_2(g)\rightarrow XCl_4(s)    \Delta H_2=+461.9kJ

(3) \frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)    \Delta H_3=-92.3kJ

(4) X(s)+O_2(g)\rightarrow XO_2(s)    \Delta H_4=-789.1kJ

(5) H_2O(g)\rightarrow H_2O(l)    \Delta H_5=-44.0kJ

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) 2H_2O(g)\rightarrow 2H_2(g)+O_2(g)     \Delta H_1=2\times 241.8kJ=483.6kJ

(2) XCl_4(s)\rightarrow X(s)+2Cl_2(g)    \Delta H_2=-461.9kJ

(3) 2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)    \Delta H_3=4\times -92.3kJ=-369.2kJ

(4) X(s)+O_2(g)\rightarrow XO_2(s)    \Delta H_4=-789.1kJ

(5) 2H_2O(l)\rightarrow 2H_2O(g)    \Delta H_5=2\times 44.0kJ=88.0kJ

The expression for enthalpy of main reaction will be:

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5

\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)

\Delta H=-1048.6kJ

Therefore, the enthalpy of the given reaction will be, -1048.6 kJ

4 0
3 years ago
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