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Zanzabum
4 years ago
7

What group of elements on the periodic table are already chemically stable?

Chemistry
1 answer:
AlladinOne [14]4 years ago
7 0
Anything in group 18. These are the noble gases, known for having the stable octet in the outermost electron shell (with the exception of Helium). These elements include Neon, Xenon, and Radon, to name a few.
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An iron bar at 200c is placed in thermal contact with an identical iron bar at 120c in an isolated system
Step2247 [10]

Conduction: In the conduction, the heat is transferred from the hotter body to the colder body until the temperature on both bodies are equal.

In thermal equilibrium, there is no heat transfer as the heat is transferred till the temperature on the bodies are not same.

In the given problem, an iron bar at 200°C is placed in thermal contact with an identical iron bar at 120°C in an isolated system. After 30 minutes, the thermal equilibrium is attained. Then, the temperature on both iron bars are equal.Both iron bars are at 160°C in an isolated system.

But in an open system, the temperatures of the iron bars after 30 minutes would be less than 160°C. There will be heat lost to the surrounding. The room temperature is 25°C. There will be exchange of the heat occur between the iron bars and the surrounding. But It would take more than 30 minutes for both iron bars to reach 160°C because heat would be transferred less efficiently.

7 0
3 years ago
Read 2 more answers
N2+3H2=>>2NH3. If 6.72dm³ of each of Nitrogen and hydrogen (measured at S.T.P.) are made to produced Ammonium. Which of th
torisob [31]

Answer:

Hydrogen, 0,2 mol

Explanation:

V_{M} = 22, 4 l/mol

V (N_{2}) = 6,72 dm^{2} = 6,72 l

V (H_{2}) = 6,72 dm^{2} = 6,72 l

The formula is:

n = \frac{V}{V_{M} }

n  (N_{2}) = 6,72l /22,4 l/mol = 0,3 mol

n  (H_{2}) = 6,72l /22,4 l/mol = 0,3 mol (the same)

But from the equation we get a proportion that 1 mol  (N_{2}) - 3 mol (H_{2})

\frac{0,3}{1} > \frac{0,3}{3} (we have more hydrogen than is need for the reaction)

So there is excess of H_{2}

We only need  \frac{0,3}{3} = 0,1 mol (H_{2})

0,3 mol (hydrogen quantity we have) - 0,1 mol (hydrogen quantity we need for the reaction) = 0,2 mol (excess)

make me the brainliest

3 0
4 years ago
What is the percentage by mass of oxygen in a formula unit of sodium hydrogen carbonate (NaHCO3)?
nadezda [96]

Percent by mass of an element can be calculated as

(Total mass of the element in the formula / formula weight ) * 100

Step 1 : Find formula weight of NaHCO₃

To find formula weight of NaHCO₃, we have to add atomic masses of the elements Na, H,C and O

Formula weight of NaHCO₃ = Na+H+C+3 (O)

Formula weight of NaHCO₃ = 22.99+1.01+12.01+3(16)

Formula weight of NaHCO₃ is 84.01 g

Step 2 : Use percent by mass formula

Percent by mass of oxygen = (total mass of oxygen / formula weight) * 100

Percent by mass of oxygen = \frac{3(16)}{84.01} * 100

Percent by mass of oxygen = 57.14%

Answer : percentage by mass of oxygen in a formula unit of sodium hydrogen carbonate (NaHCO3) is 57.14%

8 0
3 years ago
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In a model of an atom, where would I find a valence electron<br><br><br> Help
Snowcat [4.5K]
The valence electron is the outermost shell
5 0
3 years ago
How are the hydrogen atom and lithium atom similar?
Juli2301 [7.4K]
They both have 1 electron in their valence shell.....
5 0
3 years ago
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