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Maksim231197 [3]
2 years ago
6

The electron configuration 1sÆ 2sÆ2pÒ 3sÆ3pÒ represents which noble gas?

Chemistry
1 answer:
ch4aika [34]2 years ago
4 0
This noble gas is argon (Ar), because its third energy level is full.
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How do you view failure? Is failure the end, or is failure an important lesson and motivator?
Katyanochek1 [597]

Answer:

Failure is a lesson

Explanation:

Without failure, we'd be less capable of compassion, empathy, kindness, and great achievement. It's through failure that we learn the greatest lessons that life could teach us.

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How did the different reform movements we've discussed change and shape America?<br> Pls help
tigry1 [53]

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6 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

<u>Initial temperature of gold= Unknow</u>

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 <u>heat equation</u>:

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 <u>all heat of gold is transfer to the water</u>, so:

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

Now we can <u>put the values into the equation</u>:

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 <u>solve for the initial temperature of gold</u>, 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
How many moles are in 50 g of CO2
murzikaleks [220]

Answer:

1.1 mol

Explanation:

n=m/M, where n is moles, m is mass, and M is molar mass.

M of CO2 = 12.01+16.00+16.00 = 44.01g/mol

n=50g/44.01g/mol

n = 1.13610543 mol

n ≈ 1.1 mol

Hope that helps

8 0
2 years ago
Equal masses (in grams) of hydrogen gas and oxygen gas are reacted to form water. Which substance is limiting?
Sonbull [250]

Answer:

a. Oxygen gas is limiting

Explanation:

hydrogen gas and oxygen gas are reacted to form water

2H₂ + O₂  →  2H₂O

the above balanced equation shows that 2 moles of H₂ is required for 1 mole of O₂

Given equal masses of H₂ and O₂

assuming 'x' gm for each, no. of moles of each gas  =

no. of moles of H₂ = x/2 = 0.5x moles

no.of moles of O₂ = x/32 = 0.031x moles

This shows that no. of moles of O₂ is very less so O₂ will become the limiting reagent.

3 0
3 years ago
Read 2 more answers
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