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kakasveta [241]
3 years ago
8

What is easier to observe about an object,it’s physical properties or it’s chemical properties. Explain why.

Chemistry
1 answer:
Sladkaya [172]3 years ago
7 0

Answer:

Physical Properties

Explanation:

You are able to see physical properties but are unable to see chemical properties.

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<span>orbital shell is the</span><span> the circular paths around the nucleus of an atom along which the electrons traverse.</span>
4 0
3 years ago
What is electrons configuration for arsenic ?
Anna35 [415]

Answer:

[Ar] 3d¹⁰ 4s² 4p³

Explanation:

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7 0
2 years ago
Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,
WITCHER [35]

Answer: Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]

We are given:

\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ

Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

6 0
2 years ago
Below are the steps of the engineering process (not necessarily in order): 1) Know the background 2) Do the work 3) Make a plan
LuckyWell [14K]

Answer:

4 - 1 - 3 - 2 - 6 - 5

Explanation:

During an engineering process, first, we need to identify the problem, or the need because the process only will occur because of some need. Then, it's necessary to know as much as possible about the problem and the things that already exist or already were tested to solve it. Knowing the background will make the work easy.

After that, it's necessary to plan the things we'll do, knowing the costs, the time needed for activities, how many people will be necessary for each step, etc. It's really important to make a plan. Then, do the work, following the plan. Thus, the process must be tested. During the test of the results, some problems must be found, so it's time to evaluate and redesign the process, to solve these problems found.

3 0
3 years ago
7. Convert 5.2 cm of magnesium (Mg) ribbon to mm of Mg ribbon.
Effectus [21]
<h3>Answer:</h3>

52 mm

<h3>Explanation:</h3>

We are given;

  • 5.2 cm of magnesium

Required to convert it to cm

We are going to use the appropriate conversion factor;

  • The units used to measure length include;

Kilometer(km)

10

Hectometer (Hm)

10

Decameter (dkm)

10

Meter(m)

10

Decimeter (dm)

10

Centimeter (cm)

10

Millimeter (mm)

Therefore; the appropriate conversion factor is 10mm/cm

Thus;

5.2 cm will be equivalent to;

= 5.2 cm × 10 mm/cm

= 52 mm

Therefore, the length of magnesium ribbon is 52 mm

7 0
2 years ago
Read 2 more answers
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