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Slav-nsk [51]
3 years ago
8

The average air tempature in an area over a long period of time is called

Chemistry
2 answers:
Julli [10]3 years ago
5 0
Climate Change Hope I Helped
labwork [276]3 years ago
3 0
Climate change or it could just be Climate,
Hopefully that helps ❤
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Which statements about a sodium atom (22/11 Na ), are correct?
inysia [295]

Answer:D


Explanation:
Number of electrons=Number of protons
Sodium atom has 1 valence electron
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2 years ago
Given the reaction agi(s) ↔ ag+(aq) + i-(aq) solution equilibrium is reached in the system when
Julli [10]
Missing question:
1) the rate of dissolving reaches zero 
<span>2) the rate of crystallization reaches zero </span>
3) the rate of dissolving is zero and the rate of crystallization is greater than zero.
<span>4) both the rate of dissolving and the rate of crystallization are equal and greater than zero.
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Answer is: 4) both the rate of dissolving and the rate of crystallization are equal and greater than zero.
Silver chloride (AgCl) dissolves and form silver and chlorine ions, in the same time silver and chlorine ions crystallizate and form solid salt silver chloride.
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3 years ago
Need some help please?
VladimirAG [237]
26. B. 28. B. 20. A.
4 0
2 years ago
Given:
bulgar [2K]

Answer:

The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

Explanation:

Given;

CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol

From the combustion reaction above, it can be observed that;

1 mole of methane (CH₄) released 890 kilojoules of energy.

Now, we convert 59.7 grams of methane to moles

CH₄ = 12 + (1x4) = 16 g/mol

59.7 g of CH₄ = \frac{59.7}{16} = 3.73125 \ moles

1 mole of methane (CH₄) released 890 kilojoules of energy

3.73125 moles of methane (CH₄) will release ?

= 3.73125 moles x  -890 kJ/mol

= -3320.81 kJ

Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

5 0
3 years ago
In chemical reactions, the number of atoms in the product is ________________.
irina1246 [14]

Answer:

A

Explanation:

7 0
3 years ago
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