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vladimir2022 [97]
3 years ago
8

Negative charge is placed in an eletric

Chemistry
1 answer:
leonid [27]3 years ago
7 0
I would i have think it is either a) east or c) west b/c couldn’t go by it self
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The Great Lakes were formed by a retreating glacier that dug out large basins. Then as the glacier melted, the water filled in those basins

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How many Noble gases are there?
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six noble gases

Here are five of the six noble gases: helium, neon, argon, kypton and xeon. They're all colourless and transparent. Krypton and xeon form compounds only with difficulty. Helium, neon and argon don't form compounds at all.

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How many significant figures are in 4.800x10-3?​
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4 after solving =0.004800 where 4,800 is the significant number
5 0
2 years ago
When N2(g) reacts with H2(g) to form NH3(g), 92.2 kJ of energy are evolved for each mole of N2(g) that reacts. Write a balanced
Naddika [18.5K]

Answer:

N₂ + 3H₂ → 2NH₃   ΔH = - 92.2KJ

Explanation:

Let's write out the chemical equation between Nitrogen and Hydrogen to Form Ammonia.

Nitrogen + Hydrogen = Ammonia

N₂ + H₂ → NH₃

A Thermochemical Equation is a balanced stoichiometric chemical equation that includes the enthalpy change, ΔH.

The balanced stoichiometric chemical equation is given as;

N₂ + 3H₂ → 2NH₃

92.2 kJ of energy are evolved for each mole of N2(g) that reacts. And from the equation, 1 mole of N2 reacts.

The enthalpy change, ΔH = - 92.2KJ. The negative sign is because heat is being evolved.

The balanced thermochemical equation;

N₂ + 3H₂ → 2NH₃   ΔH = - 92.2KJ

6 0
3 years ago
A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture co
DaniilM [7]

Answer:

Explanation:

The air 9% mole% methane have an average molecular weight of:

9%×16,04g/mol + 91%×29g/mol = 27,8g/mol

And a flow of 700000g/h÷27,8g/mol = 25180 mol/h

In the reactor where methane solution and air are mixed:

In = Out

Air balance:

91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)

Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>

<em></em>

The air in the product gas is

95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>

43058 mol air×29g/mol <em>1249 kg air</em>

Percent of oxygen is: \frac{289kg}{1249 kg} =<em>0,231 kg O₂/ kg air</em>

<em></em>

I hope it helps!

4 0
2 years ago
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