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2 years ago

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If f(x) is an even function and (6, 8) is one the points on the graph of f(x), which reason explains why (–6, 8) must also be a

point on the graph?
Since the function is even, the outputs of a negative x-value and a positive x-value are the same.

Since the function is even, the outputs of a negative y-value and a positive y-value are the same.

The graph has rotational symmetry, so the point will be reflected across the y-axis.

The graph has rotational symmetry, so the point will be rotated 90 degrees about the origin.

PLEASE HELP IM IN THE MIDDLE OF A TEST

Guest

1 year ago

I'm taking that test rn lol

1 answer:

Vikki [24]2 years ago

4 0

If $f(x)$ is an even function, then $f(x)=f(-x)$ . In this case, $f(6)=8=f(-6)$ .

This is because an even function is symmetric about the $y-axis$ . Hence the outputs of a negative x-value and a positive x-value are the same.

The correct option is the first option.

Since the graph is even, the outputs of a negative x-value and a positive x-value are the same.

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Step-by-step explanation:

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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

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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.

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1 year ago