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Anika [276]
4 years ago
13

What is not within the scope of chemistry

Chemistry
1 answer:
ankoles [38]4 years ago
5 0

hello!

your answer will be

The origin of the matter does not usually fall into chemistry

have a goed day

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A system gains 687 kJ of heat, resulting in a change in internal energy of the system equal to 156 kJ. How much work is done?
Maslowich

Answer:

w = -531 kJ

1. Work was done by the system.

Explanation:

Step 1: Given data

  • Heat gained by the system (q): 687 kJ (By convention, when the system absorbs heat, q > 0).
  • Change in the internal energy of the system (ΔU°): 156 kJ

Step 2: Calculate the work done (w)

We will use the following expression.

ΔU° = q + w

w = ΔU° - q

w = 156 kJ - 687 kJ

w = -531 kJ

By convention, when w < 0, work is done by the system on the surroundings.

4 0
3 years ago
4. Write the electronic configuration of first 20 elements in the periodic table.
DIA [1.3K]

Answer:

Name Atomic Number Electron Configuration Period 1 Hydrogen 1 1s1 Helium 2 1s2 Period 2 Lithium 3 1s2 2s1 Beryllium 4 1s2 2s2 Boron 5 1s2 2s22p1 Carbon 6 1s2 2s22p2 Nitrogen 7 1s2 2s22p3 Oxygen 8 1s2 2s22p4 Fluorine 9 1s2 2s22p5 Neon 10 1s2 2s22p6 Period 3 Sodium 11 1s2 2s22p63s1 Magnesium 12 1s2 2s22p63s2 Aluminum 13 

7 0
3 years ago
Which of the following alkyl halides will react fastest with CH3OH in an SN1 mechanism?
yaroslaw [1]

Answer:

IV

Explanation:

The complete question is shown in the image attached.

Let us call to mind the fact  that the SN1 mechanism involves the formation of carbocation in the rate determining step. The order of stability of cabocations is; tertiary > secondary > primary > methyl.

Hence, a tertiary alkyl halide is more likely to undergo nucleophilic substitution reaction by SN1 mechanism since it forms a more stable cabocation in the rate determining step.

Structure IV is a tertiary alkyl halide, hence it is more likely to undergo nucleophilic  substitution reaction by SN1 mechanism.

5 0
3 years ago
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r
weqwewe [10]

Answer:

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by \bigtriangledown H^\circ_{rxn}

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}=?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1)       \bigtriangledown H^\circ_{rxn}= +157KJ

P₄(g)+6Cl₂(g)→  4PCl₃(g).............(2)     \bigtriangledown H^\circ_{rxn}= -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3)          \bigtriangledown H^\circ_{rxn}= -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4)          \bigtriangledown H^\circ_{rxn}=4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s)    \bigtriangledown H^\circ_{rxn}=( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s)      \bigtriangledown H^\circ_{rxn}= - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}= -1835 KJ

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

5 0
3 years ago
If we wish to prepare 250. mL of 0.200 M BaCl2 solution, how much solid BaCl2 is needed?
n200080 [17]

Answer:

10.4g

Explanation are in the Picture

8 0
3 years ago
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