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andriy [413]
3 years ago
5

The hydroxide ion concentration in pure water at 25ºC is NOT equal to

Chemistry
1 answer:
Flura [38]3 years ago
6 0
Answer here is D.

All other answers are different ways of saying <span>1 X 10^-7 M.
</span>
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Incomplete contribution happens when there is a limited supply of air so only half is much oxygen adds to the carbon forming carbon monoxide
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Complete the analogy solid:Table:.......water​
prisoha [69]

Answer:

of

Explanation:

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3 years ago
Is a push or a pull.<br><br><br> A.FORCE<br> B.MASS <br><br> ???
Alex_Xolod [135]
A.Force because your adding pressure to what you are pushing or pulling.
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3 years ago
Manganese forms several oxides when combined with oxygen. One of the oxides (Oxide 1) contains 63.2% of Mn and another oxide (Ox
Nina [5.8K]

Explanation:

Defining law of definite proportions, it states that when two elements form more than one compound, the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers.

A. One of the oxides (Oxide 1) contains 63.2% of Mn.

Mass of the oxide = 100g

Mass of Mn = 63.2 g

Mass of O = 100 - 63.2

= 36.8 g

Ratio of Mn to O = 63.2/36.8

= 1.72

Another oxide (Oxide 2) contains 77.5% Mn.

Mass of oxide = 100 g

Mass of Mn = 77.5 g

Mass of O = 100 - 77.5

= 22.5 g

Ratio of Mn to O = 77.5/22.5

= 3.44

Therefore, the ratio of the masses of Mn and O in Oxide 1 and Oxide 2 is in the ratio 1.72 : 3.44, which is also 1 : 2. So the law of multiple proportions is obeyed.

B.

Oxide 1

Mass of Mn per 1 g of O = mass of Mn/mass of O

= 77.5/22.5

= 3.44 g/g of Oxygen.

Oxide 2

Mass of Mn per 1 g of O = mass of Mn/mass of O

= 77.5/22.5

= 3.44 g/g of Oxygen.

3 0
3 years ago
The equilibrium constant, Kc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium co
Anarel [89]

Answer:

Kp = 1.41 x 10⁻⁶

Explanation:

We have the chemical equation:

2 A(g) + 3 B(g)⇌ C(g)

In which A and B are the reactants and C is the product. We calculate first the change in the number of moles of gas (Δn or dn):

dn= (sum moles products - sum moles reactants)

   = (moles C - (moles A + moles B))

   = (1 - (2+3))

   = 1 - 5

   = -4

We have also the following data:

Kc = 63.2

T= 81∘C + 273 = 354 K

R = 0.082 L.atm/K.mol (it is a constant)

Thus, we introduce the data in the mathematical expression for the relation between Kp and Kc:

Kc = (RT)^{dn}= (0.082 L.atm/K.mol x 354 K)⁻⁴ = 1.41 x 10⁻⁶

3 0
3 years ago
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