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Sergio039 [100]
2 years ago
10

15) Bob has 12 black pen, 14 red pen, 15 green pens, 24 blue pens and 3 boxes of

Mathematics
1 answer:
cricket20 [7]2 years ago
8 0

Answer:

13, 16, 18

Step-by-step explanation:

Sabemos que podemos calcular el total de plumas amarillos, por medio de la media.

16 = (12 + 14 + 15 + 24 + x + y + z) / 7

16 * 7 = 12 + 14 + 15 + 24 + x + y + z

112 = 65 + x + y + z

x + y + z = 112-65

x + y + z = 47

sabemos que la mediana es el valor medio, y que ese 46 debe dividirse entre 3 valores porque son 3 cajas, entonces debemos buscar los números adecuados para el valor 15 esté en la mitad, es decir de cuarto.

Sin meter los valores nuevos, el ranking sería así:

12

14

15

24

el 15 esta de tercero, para que quede de cuarto, debe tener dos valores por debajo y un por arriba, así:

X

12

14

15

y

z

24

Por lo tanto, valores mayores a 15 serían por ejemplo y = 16 y z = 18:

16 + 18 = 34

x = 47-34 = 13

quedaría:

12

13

14

15

dieciséis

18 años

24

Si recalculamos la media:

m = (12 + 14 + 15 + 24 + 13 + 16 + 18) / 7 = 16

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How much heat is produced when 100 mL of 0.250 M HCl (density, 1.00 g/mL) and 200 mL of 0.150 M NaOH (density, 1.00 g/mL) are mi
alexira [117]

Answer:

1.45 kilo joules of heat is produced when 100 mL of 0.250 M HCl and 200 mL of 0.150 M NaOH.

1.15°C is the temperature increase.

Step-by-step explanation:

HCl+NaOH\rightarrow NaCl+H_2O ,ΔH°=-58 kJ/mol

Molarity=\frac{Moles}{Volume (L)}

Molarity of HCl = 0.250 M

Volume of HCl = 100 ml = 0.1 L

Moles of HCl = n

n=0.250\times 0.1 L=0.0250 mol

Molarity of NaOH= 0.150 M

Volume of NaOH= 200 ml = 0.2 L

Moles of NaOH= n'

n'=0.150\times 0.2 L=0.030 mol

According to reaction, 1 mol of HCl reacts with 1 mol of NaOH. Then 0.0250 mole of HCl will reacts with 0.0250 mol of NaOH.

\frac{1}{1}\times 0.0250 mol=0.0250 mol of NaOH

As we can see that moles of NaOH are in excess.Hence, excessive agent.

The enthalpy of the reaction = ΔH°=-58 kJ/mol

Energy released when 0.0250 moles of HCl reacted with 0.0250 moles of NaOH:

Q=\Delta H^o\times 0.0250 mol=-58 kJ/mol\times 0.0250 mol=-1.45 kJ

(Negative sign indicates that heat is liberated.)

Mass of the HCL solution = m

Volume of HCl ,v= 100 ml

Density of HCl solution = d = 1.00 g/mL

m=d\times v=1.00 g/mL\times 100 mL=100 g

Mass of NaOH solution = m'

Volume of NaOH ,v' = 200 ml

Density of NaOH solution = d' = 1.00 g/mL

m'=d'\times v'=1.00 g/mL\times 200 mL=200 g

Mass of the solution after mixing,M = m + m' = 100  g + 200 g = 300 g

Heat absorbed by the final solution formed after mixing = Q'

Heat absorbed by the final solution formed after mixing = Heat released during reaction

Q' = -Q = -(-1.45 kJ)= 1.45 kJ=1450 J  (1kJ = 1000 J)

Q' = mcΔT

generally, m = mass of the substance

c  = specific heat of the substance

ΔT = Change in temperature

Specific heat capacity of the product formed after mixing = c = 4.19 J/g°C

Mass of the resulting  mix = 300 g

ΔT = ?

1450 J=300g\times 4.19 J/g^oC\times \Delta T

\Delta T=\frac{1450 J}{300g\times 4.19 J/g^oC}

\Delta T=1.15^oC

1.15°C is the temperature increase.

8 0
1 year ago
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