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3 years ago

Why noble gas neon is an unreactive element?

2 answers:

6 0

This is because it has a full outer valence shell! so there are 8 electrons and that means it doesn't have the urge the gain anymore

katrin2010 [14]3 years ago

3 0

Noble gases already have 8 valence electrons, which means their outermost energy level is completely filled. It doesn't need other elements to react with it. The noble gases are in the last group of the periodic table. Group 1, alkali meatballs, are the most reactive since they carry only 1 valence electrons and need to react with other elements

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Can someone help me with these 3 questions <br> please and thank you

Novay_Z [31]

Answer:

1 i would find a way out of the unsafe situation and get away

2 not really to much

3 take care of each other and care for one another

Explanation:

there is my answers please be happy i am only in middle school so that is the best i can do for you.

6 0

3 years ago

Identify two tools in the lab classroom, one that is used to measure volume and another that is used to observe objects. How wou

vovangra [49]

Answer:

Ciditegkdkcifiyrkchkfyifoyr

3 0

3 years ago

Earth's oceans have an average depth of 3800 m, a total surface area of 3.63 x 108 km2, and an average concentration of dissolve

kakasveta [241]

Answer:

a) Mass of gold in ocean = 8.1*10¹² g

b) Volume of gold = 4.2*10¹¹ cm³

c) Value of gold = $ 4.1*10¹³

Explanation:

a) Depth of ocean = 3800 m

Surface area of ocean = 3.63*10⁸ km²

Now, 10⁶ m² = 1 km²

Therefore, the area in units of m² = 3.63*10¹⁴ m²

$Volume\ of\ ocean = Depth*Area = 3800m*3.63*10^{14} m2=1.4*10^{18} m3$

Average concentration of gold in ocean = 5.8*10⁻⁹ g/L

Now, 1 g/L = 1000 g/m³

Therefore, concentration of gold in g/m³ = 5.8*10⁻⁶ g/m³

$Mass\ of\ gold =1.4*10^{18} m^{3} *5.8*10^{-6} g/m3 = 8.1*10^{12} g$

b) Mass of gold present = 8.1*10¹² g

Density of gold = 19.3 g/cm³

$Volume\ of\ gold = \frac{Mass}{density} =\frac{8.1*10^{12} }{19.3} =4.2*10^{11} cm^{3}$

c) Price of gold = $1595/troy oz

Unit conversion

1 troy oz = 311 g

Therefore, 8.1*10¹² g of gold is equivalent to:

$\frac{8.1*10^{12} g* 1\ troy\ oz}{311g} =2.6*10^{10} \ troy\ oz$

The value of gold in the ocean is:

$\frac{2.6*10^{10} troy\ oz*1595\ dollars}{1\ troy\ oz} =4.1*10^{13} \ dollars$

5 0

4 years ago

WHEN IN FRICTION not useful

Paladinen [302]

Friction is not useful in Machinery sometimes, Since it leads to the wear and tear of machine parts.

6 0

3 years ago

Calculate the osmotic pressure of 5.0g of sucrose ssolution in 1L. Answer should be in Torr

Pani-rosa [81]

Answer: The osmotic pressure of 5.0g of sucrose solution in 1 L is 271.32 torr.

Explanation:

Given: Mass = 5.0 g

Volume = 1 L

Molar mass of sucrose = 342.3 g/mol

Moles are the mass of a substance divided by its molar mass. So, moles of sucrose are calculated as follows.

$Moles = \frac{mass}{molarmass}\\= \frac{5.0 g}{342.3 g/mol}\\= 0.0146 mol$

Hence, concentration of sucrose is calculated as follows.

$Concentration = \frac{moles}{Volume (in L)}\\= \frac{0.0146 mol}{1 L}\\= 0.0146 M$

Formula used to calculate osmotic pressure is as follows.

$\pi = CRT$

where,

$\pi$ = osmotic pressure

C = concentration

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

$\pi = CRT\\= 0.0146 \times 0.0821 L atm/mol K \times 298 K\\= 0.357 atm (1 atm = 760 torr)\\= 271.32 torr$

Thus, we can conclude that the osmotic pressure of 5.0g of sucrose solution in 1 L is 271.32 torr.

4 0

3 years ago