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puteri [66]
3 years ago
10

How many moles of oxygen are in 1 mole of fe(no3)3?

Chemistry
1 answer:
Naddika [18.5K]3 years ago
8 0
One mole of Fe(NO3)3, or iron(III) nitrate, has three moles of nitrate molecules, which have three moles of oxygen atoms each. We can show this mathematically:
1 mole Fe(NO3)3 * (3 moles NO3)/(1 mole Fe(NO3)3) = 3 moles NO3
3 moles NO3 * (3 moles Oxygen)/(1 mole NO3) = 9 moles Oxygen
9 moles of Oxygen in one mole Fe(NO3)3
You might be interested in
Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe
leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

c = \frac{n}{V}

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}

M_{KOH} = 56.106 \frac{g}{mol}

<u>Calculate the moles of potassium hydroxide.</u>

n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}

n_{KOH} = 0.25941(9)mol

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

V = \frac{500.0mL}{1}*\frac{1L}{1000mL}

V = 0.5000L

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

c = \frac{n}{V}

c = \frac{0.25941(9)mol}{0.5000 L}              

c = 0.5188(3) \frac{mol}{L}         <= Keep an insignificant figure for rounding

c = 0.5188 \frac{mol}{L}              <= Rounded up

c = 0.5188M               <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0
3 years ago
The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal
spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

5 0
3 years ago
Instructions:
never [62]

Answer:

See explanation

Explanation:

A. Constitutional or structural isomers have the same molecular formula but different structural formulas.

B. Conformational isomers are compounds having the same atom to atom connectivity but differ by rotation about one or more single bonds.

C. Stereo isomers are compounds having the same molecular mass and atom to atom connectivity but different arrangement of atoms and groups in space.

I. Enantiomers are stereo isomers (optical isomers particularly) that are non-superimposable mirror images of each other.

II. Diasteromers are optical isomers that are not mirror images of each other.

Both diasteromers and enantiomers are types of optical isomers which in turn is one of the types of stereo isomers.

Stereo isomers differ from conformational isomers in that the arrangement of atoms in stereo isomers is permanent while conformational isomers results from free rotations in molecules about single bonds.

7 0
2 years ago
The following thermochemical equation is for the reaction of Fe 3 O 4 (s) with hydrogen (g) to form iron and water vapor Fe 3 O
Mila [183]

Answer:

41.3kJ of heat is absorbed

Explanation:

Based in the reaction:

Fe₃O₄(s) + 4H₂(g) → 3Fe(s) + 4H₂O(g) ΔH = 151kJ

<em>1 mole of Fe3O4 reacts with 4 moles of H₂, 151kJ are absorbed.</em>

63.4g of Fe₃O₄ (Molar mass: 231.533g/mol) are:

63.4g Fe₃O₄ × (1mol / 231.533g) = <em>0.274moles of Fe₃O₄</em>

These are the moles of Fe₃O₄ that react. As 1 mole of Fe₃O₄ in reaction absorb 151kJ, 0.274moles absorb:

0.274moles of Fe₃O₄ × (151kJ / 1 mole Fe₃O₄) =

<h3>41.3kJ of heat is absorbed</h3>

<em />

6 0
3 years ago
Determine the value of the equilibrium constant, KgoalKgoalK_goal, for the reaction CO2(g)⇌C(s)+O2(g)CO2(g)⇌C(s)+O2(g), Kgoal=?
NISA [10]

Answer:

The value of the equilibrium constant for reaction asked is 1.24\times 10^{-7}.

Explanation:

CO_2(g)\rightleftharpoons C(s)+O_2(g)

K_{goal}=?

K_{goal}=\frac{[C][O_2]}{[CO_2]}

2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2(g)..[1]

K_1=\frac{[CH_3COOH][O_2]^2}{[CO_2]^2[H_2O]^2}

2H_2(g)+O2(g)\rightleftharpoons 2H_2O(l)..[2]

K_2=\frac{[H_2O]^2}{[H_2]^2[O_2]}

CH_3COOH(l)\rightleftharpoons 2C(s)+2H_2(g)+O_2(g)..[3]

K_3=\frac{[C]^2[H_2]^2[O_2]}{[CH_3COOH]}

[1] +  [2] + [3]

2CO_2(g)\rightleftharpoons 2C(s)+2O_2(g)

 ( on adding the equilibrium constant will get multiplied with each other)

K=K_1\times K_2\times K_3

K=5.40\times 10^{-16}\times 1.06\times 10^{10}\times 2.68\times 10^{-9}

K=1.53\times 10^{-14}

K=\frac{[C]^2[O_2]^2}{[CO_2]^2}

On comparing the K and K_{goal}:

K^2=K_{goal}

K_{goal}=\sqrt{K}=\sqrt{1.53\times 10^{-14}}=1.24\times 10^{-7}

The value of the equilibrium constant for reaction asked is 1.24\times 10^{-7}.

4 0
3 years ago
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