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Anuta_ua [19.1K]
3 years ago
9

In any chemical compound, the elements are always combined in the same proportion by ___.?

Chemistry
2 answers:
irga5000 [103]3 years ago
6 0

Answer is: elements are always combined in the same proportion by mass.

Law of multiple proportions or Dalton's Law said that the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers.

For example, nitrogen(I) oxide N₂O; m(N) : m(O) = 2·14 : 16 = 7 : 4.

Another example, water (H₂O) is made of two hydrogen atoms and one oxygen atom:  

m(H) : m(O) = 2·1 : 16 = 1: 8.

Andru [333]3 years ago
6 0

Answer:

mass

Explanation:

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The molarity of a solution prepared by dissolving 4.11 g of NaI in enough water to prepare 312 mL of solution is
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Answer:

The correct answer is 8.79 × 10⁻² M.

Explanation:

Based on the given information, the mass of NaI given is 4.11 grams. The molecular mass of NaI is 149.89 gram per mole. The moles of NaI can be determined by using the formula,

No. of moles of NaI = Weight of NaI/ Molecular mass

= 4.11 / 149.89

= 0.027420

The vol. of the solution given is 312 ml or 0.312 L

The molarity can be determined by using the formula,

Molarity = No. of moles/ Volume of the solution in L

= 0.027420/0.312

= 0.0879 M or 8.79 × 10⁻² M

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A mixture of water and graphite is heated to 600 k in a 1 l container. when the system comes to equilibrium it contains 0.15 mol
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(missing in Q) : Calculate the concentration of CO & H2 & H2O when the system returns the equilibrium???

when the reaction equation is:

C(s) + H2O(g)  ↔ H2(g) + CO(g) 

∴ Kc = [H2] [CO] / [H2O]

and we have Kc = 0.0393 (given missing in the question)

when the O2 is added so, the reaction will be:

2H2(g) + O2(g) → 2H2O(g)

that means that 0.15 mol H2 gives 0.15 mol of H2O

∴ by using ICE table:

            [H2O]          [H2]        [CO]

initial 0.57 + 0.15      0               0.15

change  -X                +X              +X

Equ   (0.72-X)             X            (0.15+X)

by substitution:

0.0393 = X (0.15+X) / (0.72-X)  by solving for X

∴ X = 0.098 

∴[H2] = X = 0.098 M

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∴[H2O] = 0.72 - X

             = 0.72 - 0.098

             = 0.622 M


8 0
2 years ago
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