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2 years ago

Asuming all orbitals are in the same energy level which type of orbital has the lowest energy?

Chemistry

1 answer:

Virty [35]2 years ago

6 0

Answer: C. S orbital

Explanation: The s orbital is the lowest orbital shell which is much closer to the nucleus. This orbital can hold a maximum of 2 electrons, a perfect example for this is the 1s orbital which is the closest to the atomic nucleus.

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How many moles in 28.0 grams of Oxygen

anastassius [24]

Answer:

1.) 28.0 grams of oxygen28 grams (1 mole/16 grams per mole)=1.75 moles oxygen2.)5.0 moles of Iron5 moles(55.845 grams/1 mole)=279.225 grams Iron3.) 452 grams Argon452 grams(1 mole/39.948 grams)=11.315 moles Argon4.) 16.5 moles Hydrogen16.5 moles(1.01 grams/1 mole)=16.665 grams Hydrogen

Explanation:

3 0

2 years ago

What is the resultant displacement?

Lelu [443]

Answer:

The resultant is the vector sum of two or more vectors. It is the result of adding two or more vectors together. ... When displacement vectors are added, the result is a resultant displacement.

7 0

2 years ago

A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from

sweet-ann [11.9K]

Answer:

$T_i~=163.1$ ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

Initial temperature of gold= Unknow

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the heat equation:

$Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT$

$Q_A_u=m_A_u*Cp_A_u*deltaT$

We can relate these equations if we take into account that all heat of gold is transfer to the water, so:

$m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT$

Now we can put the values into the equation:

$60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)$

Now we can solve for the initial temperature of gold, so:

$T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20$

$T_i~=163.1$ ºC

I hope it helps!

5 0

2 years ago

If an automobile air bag has a volume of 11.5 l , what mass of nan3 (in

True [87]

Moles of NaN3 at STP = volume of gas / 22.4 = 11.5/22.4 = 0.5mole. Massof NaN3 = moles of NaN3 x molecular weight = 0.5 x 65 = 32.5 grams.

8 0

2 years ago

Read 2 more answers

Describe the temperature changes that occur in ice as energy is added, starting in the frozen state and ending in the vapor stat

Ket [755]

Explanation:

When water is frozen then it is known as ice and its state is solid. So, its molecules will be held closer to each other as they are held by strong intermolecular forces of attraction.

As a result, its temperature will be minimum as its molecules have least kinetic energy.

It is known that kinetic energy of a substance is directly proportional to temperature.

As, K.E = $\frac{3}{2}kT$

where K.E = kinetic energy

T = temperature

k = boltzmann constant

When solid changes into liquid state then it means molecules of a substance has gained kinetic energy due to which there occurs more collisions between the molecules.

Hence, temperature of substance also increases.

Whereas when liquid state of a substance changes intro vapor state then it means that more kinetic energy has gained by the molecules due to which there will be much more collisions between the molecules.

Hence, temperature will be maximum in vapor state.

8 0

2 years ago