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iragen [17]
3 years ago
11

An eccentric chemistry professor (not mentioning any names) stops in every day and orders 200ml of Sumatran coffee at precisely

65.0°C. You then need to add enough milk at 7.00°C to drop the temperature of the coffee, initially at 85.0°C, to the ordered temperature. Calculate the amount of milk (in ml) you must add to reach this temperature. In order to simplify the calculations, you will start by assuming that milk and coffee have the specific heat and density as if water. In the following parts, you will remove these simplifications. Solve now this problem assuming the density is 1.000 g/ml for milk and coffee and their specific heat capacity is 4.184 J/(g ºC). How much milk will you add?
Chemistry
1 answer:
Damm [24]3 years ago
3 0

Answer:

68.97g of milk must be added

Explanation:

To solve this problem we need to use the equation:

Q = C*m*ΔT -Coffee cup calorimeter equation-

This equation relates mass and change in temperature with heat of solution, Q.

The energy that decreases in the milk is the same that increases in the milk. That is:

Q(milk) = Q(tea)

Using the equation:

4.184J/molK* Mass milk * (65.0°C - 7.00°C) = 4.184J /molK * 200g (Tea) * (85.0°C - 65.0°C)

Mass milk * 58.0°C = 200g * 20.0°C

Mass milk = 4000g°C / 58.0°C

Mass milk =

<h3>68.97g of milk must be added</h3>
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Read 2 more answers
What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )
yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution :  Given,

Concentration (c) = 0.150 M

Acid dissociation constant = k_a=4.5\times 10^{-4}

The equilibrium reaction for dissociation of HNO_2 (weak acid) is,

                           HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+

initially conc.         c                       0         0

At eqm.              c(1-\alpha)                c\alpha        c\alpha

First we have to calculate the concentration of value of dissociation constant (\alpha ).

Formula used :

k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}

Now put all the given values in this formula ,we get the value of dissociation constant (\alpha ).

4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}

4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2

0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0

By solving the terms, we get

\alpha=0.0533

No we have to calculate the concentration of hydronium ion or hydrogen ion.

[H^+]=c\alpha=0.150\times 0.0533=0.007995 M

Now we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.007995 M)

pH=2.097\approx 2.1

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

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3 years ago
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