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3 years ago

A. calculate the quantity of heat released when 1.00 g of sulfur is burned in oxygen.

Chemistry

1 answer:

s344n2d4d5 [400]3 years ago

4 0

Answer: A) 9.25 KJ.

B) - 148.296 KJ.

C) 296 KJ/mol.

Explanation:

For the reaction: S(s) + O2(g) → SO2(g), ΔH = -296 KJ/mol

A) The number of moles in 1.00 g of S is

n = mass/atomic mass = (1.00 g) / (32.00 g/mol) = 0.03125 mole

So, the quantity of heat when 1.00 g of sulfur burned in oxygen = ΔH of the reaction for 1.0 mole x no. of moles = (-296 KJ/mol) x (0.03125 mole) = 9.25 KJ.

B) The quantity of heat released when 0.501 mole of sulfur is burned in air = ΔH of the reaction for 1.0 mole x no. of moles = (-296 KJ/mol) x (0.501 mol) = - 148.296 KJ.

C) The quantity of energy is required to break up 1.0 mole of SO2(g) into its constituent elements = 296 KJ/mol.

It is the same that the amount of energy released when 1.0 mole of S is burned in oxygen.

The process of formation (burning of S) is exothermic.

On the other hand, the reverse operation (breakdown) must be endothermic (and therefore a positive energy change) = 296 KJ/mol.

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For a gaseous carbon atom in its' ground state which of the following are possible sets of quantum numbers for two different ele

MAXImum [283]

Answer:

Option e=>

electron 1; n l ml ms = 2 1 0 +1/2.

electron 2;n l ml ms= 2 1 -1 +1/2.

Explanation:

The electronic configuration of carbon in its ground state is 1s2 2s2 2p2. The excited electronic configuration of carbon is given as; 1s2 2s1 2p3 because one of the electrons moves to the 2p sub- orbital because of the absorption of the atom.

Recall that there are four different types of quantum numbers and they are the principal quantum number,n; the Azimuthal quantum number, l; the magnetic quantum number, ml and the spin quantum number, Ms.

Our principal quantum number,n is TWO(2) which is the highest energy level from the electronic configuration written above.

The Azimuthal quantum number is 1 from l= n - 1.

The magnetic quantum number is -1 and 0 because we have two electrons filling the 3 orbitals of 2p sublevels. That is;

2px = -1 , 2py = 0 and 2pz = +1.

Only 2px and 2py will be considered because we have 2 electrons on the 2p in the ground state electronic configuration.

Therefore, the first electron will have ml= 0 and the second electron will have ml= -1.

Spin quantum number, ms; in both electrons the spins is an up spin so we have a +1/2 for both electron.

Hence; electron 1; n l ml ms = 2 1 0 +1/2.

Electron 2;n l ml ms= 2 1 -1 +1/2.

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3 years ago

How atoms form chemical bond discuss in detail

motikmotik

Answer:

Atoms form chemical bonds to make their outer electron shells more stable. An ionic bond, where one atom essentially donates an electron to another, forms when one atom becomes stable by losing its outer electrons and the other atoms become stable (usually by filling its valence shell) by gaining the electrons.

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uranmaximum [27]

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Explanation:

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3 years ago

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1. 51.2 g of NaClO (pKa of HClO is 7.50) and 25.3 g of KF (pKa of of HF is 3.17) were combined in enough water to make 0.250 L o

Andrews [41]

Answer:

* $pH=11.0$

* $[HClO]_{eq}=9.32x10^{-4}M$

* $[HF]_{eq}=5.07x10^{-6}M$

Explanation:

Hello,

At first, the molarities of NaClO and KF are computed as shown below:

$M_{NaClO}=\frac{51.2g*\frac{1mol}{74.45g} }{0.250L}=2.75M \\\\M_{KF}=\frac{25.3g*\frac{1mol}{58.1g} }{0.250L} =1.74M$

Now, since both NaClO and KF are ionic, one proposes the dissociation reactions as:

$NaClO\rightarrow Na^++ClO^-\\KF\rightarrow K^++F^-$

In such a way, by writing the formation of HClO and HF respectively, due to the action of water, we obtain:

$ClO^-+H_2O\rightleftharpoons HClO+OH^-\\F^-+H_2O\rightleftharpoons HF+OH^-\\$

Now, the presence of OH⁻ immediately implies that Kb for both NaClO and KF must be known as follows:

$Kb=\frac{Kw}{Ka} \\Kb_{NaClO}=\frac{1x10^{-14}}{10^{-7.50}}=3.16x10^{-7}\\Kb_{KF}=\frac{1x10^{-14}}{10^{-3.17}}=1.48x10^{-11}$

In such a way, the law of mass action for each case is:

$Kb_{NaClO}=3.16x10^{-7}=\frac{[HClO][OH]^-}{[ClO^-]}=\frac{(x_{ClO})(x_{ClO})}{2.75-x_{ClO}}\\Kb_{KF}=1.48x10^{-11}=\frac{[HF][OH]^-}{[F^-]}=\frac{(x_{F})(x_{F})}{1.74-x_{F}}$