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3 years ago

The major difference between a 1s orbital and a 2s orbital is that

Chemistry

1 answer:

Neko [114]3 years ago

5 0

Answer:

The 2s orbital is at a higher energy level.

Explanation:

1s and 2s are the sub-orbitals that are located in an atom. They are nearest to the nucleus and are found on the s sub-orbital. The difference between 1s and 2s is the difference in their level of energy. 1s has low energy as compared to 2s. 1s orbital has the lowest energy because it is located closed to the nucleus. 2s orbital has higher energy than 1s because it's orbit is larger than 1s.

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How many moles are in 1.2x10^3 grams of ammonia(NH3) ?

miskamm [114]

Answer : The number of moles present in ammonia is, 70.459 moles.

Solution : Given,

Mass of ammonia = $1.2\times 10^3g$

Molar mass of ammonia = 17.031 g/mole

Formula used :

$\text{Moles of }NH_3=\frac{\text{ given mass of }NH_3}{\text{ molar mass of }NH_3}$

$\text{Moles of }NH_3=\frac{1.2\times 10^3g}{17.031g/mole}=70.459moles$

Therefore, the number of moles present in ammonia is, 70.459 moles.

5 0

3 years ago

Read 2 more answers

A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c

Zarrin [17]

Answer:

$C_{alloy}=0.497\frac{J}{g\°C}$

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

$Q_{allow}=-(Q_{water}+Q_{Al})$

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

$m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})$

Then, we solve for specific heat of the metallic alloy to obtain:

$C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}$

Thereby, we plug in the given data to obtain:

$C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}$

Regards!

3 0

2 years ago

Based on the activity series of metal, which reaction with water will not happen?

laila [671]

Answer:

Explanation:

B, H2O + Na The elements toward the bottom left corner of the periodic table are the metals that are the most active in the sense of being the most reactive. Lithium, sodium, and potassium all react with water,

5 0

2 years ago

What happens if the same zero error occurs in every reading

nadya68 [22]

Zero error occur in every reading means that the measuring instrument needs to recalibrated and adjusted.

<h3>What is Zero error?</h3>

This error occurs when a measuring instrument reflects a digit which isn't zero despite the real value being zero.

When this occurs, only the zero screw on the device should be adjusted so as to correct this technical error. This is why recalibration should be the most appropriate solution.

Read more about Zero error here brainly.com/question/4704005

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3 0

1 year ago

The following equation is an example of a ______________ reaction. 2 NaCl + F2 → 2 NaF + Cl2 ???

olga55 [171]

Answer:

double replacement is the answer

3 0

2 years ago