The region outside the nucleus where an electron can be found

The catabolic process of converting carbohydrates to CO2 requires ________ of carbon. The catabolic process of converting carboh

mars1129 [50]

Answer:

The catabolic process of converting carbohydrates to CO2 requires oxidation of carbon.

Explanation:

There are multiple definitions of reduction-oxidation. There is one that explains it with respect to oxygen, the other with respect to hydrogen and another with respect to electrons. The relevant one here is the one that explains it in terms of hydrogen. Oxidation is the removal of hydrogen while reduction is the addition of hydrogen. During repiration, the carbon loses the hydrogens attached to it and is therefore oxidized. These hydrogens attach themselves to oxygen which means oxygen is reduced.

4 0

3 years ago

How many moles of hcl are required to neutralize aqueous solutions of these bases:

Lemur [1.5K]

The answers are a.) 0.03 mol KOH requires 0.03 mol HCl, b.) 2 mol NH3 requires 2 mol HCl and c.) 0.1 mol Ca(OH)2 requires 0.2 mol HCl.
Solution:
We need to write the balanced equations for each reactions to find out the stoichiometry for each reactants.
a.) HCl (aq) + KOH (aq) → KCl (aq) + H2O(ℓ)
From the balanced equations, we can see that 1 HCl reacts with 1 KOH, therefore if 0.03 mol KOH is reacted then 0.03 mol HCl must also be present.

b.) HCl(aq) + NH3(aq) ) → NH4Cl(aq)
If 2 moles of NH3 are reacted then 2 moles of HCl must also be present since 1 HCl reacts with 1 NH3 from the balanced reaction.

c.) 2HCl(aq) + Ca(OH)2(s) → CaCl2(aq) + 2H2O(ℓ)
We can see that 2 HCl react with 1 Ca(OH)2, hence if 0.1 mol of Ca(OH)2 is reacted then 0.2 mol HCl must also be present.

8 0

3 years ago

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The most ideal way to perform a two-point turn is by

Stells [14]

Is by turning a few seconds before

5 0

4 years ago

Hydrogen sulfide,H2S, is a very toxic gas with a smell of rotten eggs. using the following: H2S+3/2 O2=SO2+H2O H2+1/2O2=H2O S+O2

Serga [27]

Answer:

ΔH = -20kJ

Explanation:

The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:

H₂(g) + S(g) → H₂S(g)

Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:

(1) H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ

(2) H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

(3) S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

The sum of -(1) + (2) + (3) gives:

-(1) SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ

(2) H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

(3) S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

-(1) + (2) + (3): H₂(g) + S(g) → H₂S(g)

ΔH = +519kJ - 242kJ - 297kJ = -20 kJ

I hope it helps!

5 0

3 years ago

Please help!!!!! I need the correct answer quickly!!!

xxMikexx [17]

Answer :

The oxidation state of oxygen (O) in $OF_2$ is, (+2)

The oxidation state of carbon (C) in $CO$ is, (+2)

The oxidation state of nitrogen (N) in $K_3N$ is, (-3)

Explanation :

Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

The oxidation number of a free element is always zero.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of oxygen (O) in compounds is usually -2, but it is -1 in peroxides.

The oxidation number of a Group 1 element in a compound is +1.

The oxidation number of a Group 2 element in a compound is +2.

The oxidation number of a Group 17 element in a binary compound is -1.

The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

(a) The given compound is, $OF_2$

Let the oxidation state of 'O' be, 'x'

$x+2(-1)=0\\\\x-2=0\\\\x=+2$

The oxidation state of oxygen (O) in $OF_2$ is, (+2)

(b) The given compound is, $CO$

Let the oxidation state of 'C' be, 'x'

$x+(-2)=0\\\\x-2=0\\\\x=+2$

The oxidation state of carbon (C) in $CO$ is, (+2)

(c) The given compound is, $K_3N$

Let the oxidation state of 'N' be, 'x'

$3(+1)+x=0\\\\3+x=0\\\\x=-3$

The oxidation state of nitrogen (N) in $K_3N$ is, (-3)

7 0

3 years ago

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