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kumpel [21]
3 years ago
11

Amount of liquid measured in L or Lm

Chemistry
1 answer:
devlian [24]3 years ago
5 0

The amount of liquid is measured in L. As the unit Lm is only used when measuring the mass liquid fraction.


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How do valence electrons interact between non polar covalent, polar covalent, and ionic bonds?​
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Answer:

In non-polar covalent bonds, the electrons are equally shared between the two atoms. For atoms with differing electronegativity, the bond will be a polar covalent interaction, where the electrons will not be shared equally.

Explanation:

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Carbon-14 has a half-life of 5,730 years. How long will it take for 112. 5 g of a 120. 0-g sample to decay radioactively? 5,730
vichka [17]

The time taken by Carbon-14 to decay radioactively from 120g to 112.5g is 22,920 years.

<h3>How do we calculate the total time of decay?</h3>

Time required for the whole radioactive decay of any substance will be calculated by using the below link:

T = (n)(t), where

  • t = half life time = 5730 years
  • n = number of half life required for the decay

Initial mass of Carbon-14 = 120g

Final mass of Carbon-14 = 112.5g

Left mass = 120 - 112 = 7.5g

Number of required half life for this will be:

  • 1: 120 → 60
  • 2: 60 → 30
  • 3: 30 → 15
  • 4: 15 → 7.5

4 half lives are required, now on putting values we get

T = (4)(5730) = 22,920 years

Hence required time for the decay is 22,920 years.

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2 years ago
In atmospheric chemistry, the following chemical reaction converts SO2, the predominant oxide of sulfur that comes from combusti
Misha Larkins [42]

Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }  ⇄ \mathtt{3SO_{2(l)}}

Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:

i.e

K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}

However, since we are dealing with liquids solutions;

K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}   since the activity of a_{so_3} is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

K = \dfrac{1}{Pso_2Po_2^{1/2}}

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\  \simeq -68 \ kJ/mol

Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol

(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

So;

\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}

\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K }  \\ \\  K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\  K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }

K =7.98390356\times 10^{11} \\ \\  \mathbf{K = 7.98 \times 10^{11}}

(d)

The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).

This is because;

If  Q < K, then the reaction will proceed in the right direction towards the products.

However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.

So;

Q= \dfrac{1}{Pso_2Po_2^{1/2}}

Since we are dealing with liquids;

Q= \dfrac{1}{1 \times 1^{1/2}}

Q = 1

Since Q < K; Then, the reaction proceeds in the right direction.

Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.

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