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boyakko [2]
3 years ago
6

Write the electron configuration (ground and excited states) for element with atomic number 16 ?

Chemistry
1 answer:
Talja [164]3 years ago
3 0
2, 8,6 because it has to be in a configuration of 2,8,8
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True or false 12,785.000 has five significant digits.
Yuki888 [10]

Answer:

True

Explanation:

Significant digits include non-zero digits (unless the zero is between two non-zero numbers)

In 12,785.000, there are 5 non-zero digits:

1 2 7 8 5

4 0
3 years ago
Read 2 more answers
This stadium can hold 100,000, or 1 x 10^5, people. The number of atoms in a grain of iron is about 1 x 10^18. Would you need 1 
polet [3.4K]
<h3>Answer:</h3>

1 x 10^13 stadiums

<h3>Explanation:</h3>

From the question;

1 x 10^5 people can fill 1 stadium

We are given,  1 x 10^18 atoms of iron

We are required to determine the number of stadiums that  1 x 10^18 atoms of iron would occupy.

We are going to assume that a stadium would occupy a number of atoms equivalent to the number of people.

Therefore;

One stadium =  1 x 10^5 atoms

Then, to find the number of stadiums that will be occupied by  1 x 10^18 atoms;

No. of stadiums = Total number of atoms ÷ Atoms in a single stadium

                           =  1 x 10^18 atoms ÷  1 x 10^5 atoms

                          =  1 x 10^13 stadiums

Therefore,  1 x 10^18 atoms of iron would occupy  1 x 10^13 stadiums

7 0
3 years ago
Which pieces of equipment would be used to measure the density of a solution?
Inessa05 [86]

Answer:

c I've taken chem 3 times

Explanation:

5 0
3 years ago
Read 2 more answers
Describe the electron sharing between 2 oxygen atoms
erastova [34]
When two oxygen atoms get close to each other, the attractions from the nucleus of both atoms attract the outer electrons.

(BRAINLIEST PLEASE!!!)
3 0
3 years ago
What is the density of CHCL3 vapor at 1.00atm and 298K?
Advocard [28]

Answer:

4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.

Explanation:

By ideal gas equation:

PV=nRT

Number of moles (n)

can be written as: n=\frac{m}{M}

where, m = given mass

M = molar mass

PV=\frac{m}{M}RT\\\\PM=\frac{m}{V}RT

where,

\frac{m}{V}=d which is known as density of the gas

The relation becomes:

PM=dRT    .....(1)

We are given:

M = molar mass of chloroform= 119.5 g/mol

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the gas = 298K

P = pressure of the gas = 1.00 atm

Putting values in equation 1, we get:

1.00atm\times 119.5g/mol=d\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\d=4.88g/L

4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.

7 0
3 years ago
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