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Marysya12 [62]
3 years ago
10

A theater director reinterprets a play by replacing its classical score with

Chemistry
2 answers:
cricket20 [7]3 years ago
4 0

Answer:

The play will be more  appealing to a younger audience.

Explanation:

A younger audience will more likely appreciate current pop hits rather than classical score.

Sunny_sXe [5.5K]3 years ago
4 0

Answer:

B. The play will be more appealing to a younger audience.

Explanation:

If the audience is younger, then this would be more appealing to them because pop has always been one of the fan-favorties of younger generations

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What volume is equivalent to 12.0 m^3?
anzhelika [568]

1.2 x 10^10 mm^3 is equivalent to 12.0 m^3

8 0
3 years ago
7. There are 7. 0 ml of 0.175 M H2C2O4 , 1 ml of water , 4 ml of 3.5M KMnO4 what is the molar concentration ofH2C2O4 ?
Illusion [34]

Answer:

7. 0.1021 M

8. 1.167 M

10. Increase in volume of water would lower the rate of reaction

Explanation:

7. What is the molar concentration of H₂C₂O₄ ?

Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.

Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of H₂C₂O₄, M = number of moles of H₂C₂O₄/volume = n/V

= 1.225 × 10⁻³ mol/12 × 10⁻³ L

= 0.1021 mol/L

= 0.1021 M

8. Using the data from question 7 what is the molar concentration of KMnO₄ ?

Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.

Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V

= 14 × 10⁻³ mol/12 × 10⁻³ L

= 1.167 mol/L

= 1.167 M

10. From question number 7, what effect increasing the volume of water has on the reaction rate?

Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.

3 0
3 years ago
If 28 grams of N reacts completely with 12 grams of H2, then how many
Bogdan [553]

Answer:

Mass of NH₃ produced = 34 g

Explanation:

Given data:

Mass of nitrogen = 28 g

Mass of Hydrogen = 12 g

Mass of NH₃ produced = ?

Solution:

Chemical equation:

N₂ +  3H₂    →   2NH₃

Moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 28 g/ 28 g/mol

Number of moles = 1 mol

Moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 12 g/ 2 g/mol

Number of moles = 6 mol

Now we will compare the moles of hydrogen and nitrogen with ammonia.

                            H₂              :               NH₃

                            3                :                2

                            6                :             2/3×6 = 4 mol

                           N₂              :                NH₃

                            1                :                 2

Number of moles of ammonia produced by nitrogen are less thus it will act as limiting reactant.

Mass of ammonia produced:

Mass = number of moles × molar mass

Mass =  2 mol  ×  17 g/mol

Mass = 34 g

                     

5 0
3 years ago
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PLEASE HELP ASAP
Dahasolnce [82]

Answer:0,25 g/cm3 creo

Explanation:densidad es masa entre volumen

5 0
3 years ago
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At each step, a dichotomous key involves how many choices
kvasek [131]
Two, hope this helps
8 0
3 years ago
Read 2 more answers
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