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andrew11 [14]
2 years ago
15

Consider the following data for five hypothetical elements: Q, W, X, Y, and Z. Rank the elements from most reactive to least rea

ctive. Combination Observation of reaction Q+W+ Reaction occurs X+Z+ No reaction W+Z+ Reaction occurs Q++Y Reaction occurs
Chemistry
2 answers:
yanalaym [24]2 years ago
4 0

Answer:

Y>Q>W>Z>X

Explanation:

Each more reactive element on the list replaces from a compound any of the elements below it (less reactive).

Q>W

Z>X

W>Z

Y>Q

Thus,

Y>Q>W>Z>X

pochemuha2 years ago
4 0

Answer:

Y > Q > W > Z > X

Explanation:

Reactivity series gives information on the reactive nature of element. Only more reactive element will replace a less reactive element in it compound. More reactive element usually displace less reactive element in it compound.

From the reaction above Y displace element Q from it compound. Element Q displace W from it compound.  W displace Z in it compound. The element X cannot displace Z from it compound because it is less reactive .Their is no reaction.

Base on the reactivity element  Y is the most reactive.

Y > Q > W > Z > X

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Write the balanced chemical equation for the reaction shown.
son4ous [18]

Explanation:

        Carbon                  black sphere

        Nitrogen                blue sphere

        Chlorine                green sphere

        Hydrogen              white sphere

 Reactants:

  three molecules that contain one black sphere and four white spheres

           3CH₄

three molecules that contain two blue spheres and four green spheres:

           3N₂Cl₄

Products:

 three molecules that contain one black sphere and four green spheres:

           3CCl₄

 three molecules that contain two blue spheres:

            3N₂

   six molecules that contain two white spheres

            6H₂

The reaction proper:

                    3CH₄ +      3N₂Cl₄  →   3CCl₄ + 3N₂ + 6H₂

Learn more:

Chemical reactions  brainly.com/question/4216541

#learnwithBrainly

7 0
2 years ago
Calculate the number of grams of solute in 500.0 mL of 0.189 M KOH.
KIM [24]

Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

Solution : Given,

Volume of solution = 500 ml

Molarity of KOH solution = 0.189 M

Molar mass of KOH = 56 g/mole

Formula used :

Molarity=\frac{\text{Mass of KOH}\times 1000}{\text{Molar mass of KOH}\times \text{Volume of solution in ml}}

Now put all the given values in this formula, we get the mass of solute KOH.

0.189M=\frac{\text{Mass of KOH}\times 1000}{(56g/mole)\times (500ml)}

\text{Mass of KOH}=5.292g

Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

7 0
2 years ago
Read 2 more answers
A carpet sells for 27.99 a square yard. What is the price of the carpet square per square meter ?
klio [65]
There are 9 square meters in one square yard so divide 27.99 by 9

27.99/9=3.11

one square meter is $3.11

I hope I've helped!
3 0
2 years ago
What is the empirical formula of C6H18O3?
Aloiza [94]

Answer:

The answer to your question is C₂H₆O

Explanation:

Data

Molecular formula = C₆H₁₈O₃

Empirical formula = ?

Empirical formula is defined as the simplest ratio of the elements that form part of a molecule.

Process

To find the empirical formula find the greatest common factor of the subscripts.

                             6    18   3   2

                             3      9  3   3

                              1     3    1   3

                                     1

The GCF is 3, so factor 3 of the molecular formula

                         3 ( C₂H₆O)  

The result is the empirical formula C₂H₆O

7 0
2 years ago
Choose chlorous acid, hclo2, from the list above and press the equilibrate button. when the 0.10 m solution of hclo2 is allowed
Arturiano [62]
Chlorous Acid Ionizes as,

                                        HClO₂   ⇆   H⁺  +   ClO₂⁻

                                        Ka  =  [H⁺] [ClO₂⁻] / [HClO₂]

 Ka of Chlorous Acid  =  1.1 × 10⁻²

The concentration of H⁺ ions at equilibrium are calculated as,

Initial                               0.1 M     ⇆    0          0
At Equilibrium                 0.1-X      ⇆    X         X
So,
                                       Ka  =  [X] [X] / [0.1-X]
Putting value of Ka,
                                1.1 × 10⁻²  =  X² / 0.1-X
Solving for X,
(www.cymath.com)
                                 X  =  0.028 M

Hence at equilibrium concentration of H⁺ and ClO₂⁻ is 0.028 M.

Percentage Ionization is calculated as,

                                 =  [H⁺] / [HA] × 100

                                 =  (0.028 / 0.1) × 100

                                 =  0.28 × 100

                                 =  28 %     (Percentage Ionization)
6 0
2 years ago
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