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3 years ago

When water H2O freezes into ice some of the properties have changed but the blank of the H2O is the same.

1 answer:

6 0

I would see chemical constitution and physical components are not changed.

Because physical reaction won't change the component itself

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Complete the sentences about heme. Some terms will not be used. The prosthetic group of hemoglobin and myoglobin is . The organi

nevsk [136]

Explanation:

Haemoglobin consists of heme unit which is comprised of an $Fe^{2+}$ and porphyrin ring. The ring has four pyrrole molecules which are linked to the iron ion. In oxyhaemoglobin, the iron has coordinates with four nitrogen atoms and one to the F8 histidine residue and the sixth one to the oxygen. In deoxyhaemoglobin, the ion is displaced out of the ring by 0.4 Å.

The prosthetic group of hemoglobin and myoglobin is - Heme

The organic ring component of heme is - Porphyrin

Under normal conditions, the central atom of heme is - $Fe^{2+}$

In deoxyhemoglobin , the central iron atom is displaced 0.4 Å out of the plane of the porphyrin ring system.

The central atom has six bonds: four to nitrogen atoms in the porphyrin, one to a histidine residue, and one to oxygen.

7 0

3 years ago

Balance the following reaction:

kvv77 [185]

C_5H_8+13/2O_2—»5CO_2+4H_2O

Balanced one

2C_5H_8+13O_2—»10CO_2+8H_2O

Moles of Pentyne

Given mass/Molarmass
34/68
0.5mol

Moles of H_2O

8/2(0.5)
4(0.5)
2mol

1mol releases 241.8KJ

2mol releases 241.8(2)=483.6KJ

8 0

2 years ago

How many particles are in 7.07x10^-6 moles of a substance?

Butoxors [25]

Answer:

4.25*10^18

Explanation:

1 mole =6.023*10^23 particles

so 7.07*10^-6 mole=6.023*10^23*7.07*10^-6 particles=4.25*10^18 particles

5 0

3 years ago

Represent the decomposition of aluminum oxide using the same number of atoms of molecules as are

pantera1 [17]

Answer:

Decomposition of aluminium oxide forms aluminium atoms and oxygen atoms.

Explanation:

Decomposition reaction:

When a single compound break down into two or more simpler products.

For example "AB" reactant undergoes decomposition to form "A" and "B" products.

The chemical reaction is as follows.

$AB\rightarrow A+B$

The given compound is aluminium oxide.

The decomposition reaction of aluminium oxide is a follows.

$Al_{2}O_{3}\rightarrow Al+O_{2}$

The balanced equation is as follows.

$2Al_{2}O_{3}\rightarrow 4Al+3O_{2}$

Therefore, Decomposition of aluminium oxide forms aluminium atoms and oxygen atoms.

4 0

3 years ago

Be sure to answer all parts. Consider the reaction N2(g) + 3H2(g) → 2NH3(g)ΔH o rxn = −92.6 kJ/mol If 4.0 moles of N2 react with

Sphinxa [80]

Answer : The work done is, $1.98\times 10^4J$

Explanation :

The given balanced chemical reaction is:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

When 4 moles of $N_2$ react with 12 moles of $H_2$ then it gives 8 moles of $NH_3$

First we have to calculate the change in moles of gas.

Moles on reactant side = Moles of $N_2$ + Moles of $H_2$

Moles on reactant side = 4 + 12 = 16 moles

Moles on product side = Moles of $NH_3$

Moles on reactant side = 8 moles

Change in moles of gas = 16 - 8 = 8 moles

Now we have to calculate the change in volume of gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of gas = 1.0 atm

V = Volume of gas = ?

n = number of moles of gas = 8 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas = $25^oC=273+25=298K$

Putting values in above equation, we get:

$1.0atm\times V=8mole\times (0.0821L.atm/mol.K)\times 298K$

$V=195.7L$

As the number of moles of gas decreased. So, the volume also deceased. Thus, the volume of gas will be, -195.7 L

Now we have to calculate the work done.

Formula used :

$w=-p\Delta V$

where,

w = work done

p = pressure of the gas = 1.0 atm

$\Delta V$ = change in volume = -195.7 L

Now put all the given values in the above formula, we get:

$w=-p\Delta V$

$w=-(1.0atm)\times (-195.7L)$

$w=195.7L.artm=195.7\times 101.3J=19824.41J=1.98\times 10^4J$

conversion used : (1 L.atm = 101.3 J)

Thus, the work done is, $1.98\times 10^4J$

6 0

3 years ago