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Juliette [100K]

3 years ago

15

33. An object with a mass of 2300g has a force of 6.2N applied to it. What is the resulting acceleration

1 answer:

slamgirl [31]3 years ago

5 0

Answer:

The formula

Explanation:

acceleration=force(N)÷mass

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You need to produce a buffer solution that has a pH of 5.31. You already have a solution that contains 10. mmol (millimoles) of

Karolina [17]

Answer:

37 mmol of acetate need to add to this solution.

Explanation:

Acetic acid is an weak acid. According to Henderson-Hasselbalch equation for a buffer consist of weak acid (acetic acid) and its conjugate base (acetate)-

$pH=pK_{a}(acetic acid)+log[\frac{mmol of CH_{3}COO^{-}}{mmol of CH_{3}COOH }]$

Here pH is 5.31, $pK_{a}$ (acetic acid) is 4.74 and number of mmol of acetic acid is 10 mmol.

Plug in all the values in the above equation:

$5.31=4.74+log[\frac{mmol of CH_{3}COO^{-}}{10}]$

or, mmol of $CH_{3}COO^{-}$ = 37

So 37 mmol of acetate need to add to this solution.

3 0

3 years ago

Which has a higher kinetic energy an object at 20° C or an object at 50°C

Simora [160]

Answer:

The object at 50°C will have a higher kinetic energy.

Explanation:

Temperature is a measure of the average kinetic energy of the particles in an object. As you introduce more energy into the system (e.g. heat the object), the particles on average move faster because they have more kinetic energy.

5 0

3 years ago

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

3 years ago

How much heat is required to change temperature of 10 g of water from 4 °C to 8 °C? (Water has a specific heat of 4.18 )?

Andrews [41]

Answer: 167.2 J

Explanation:

5 0

2 years ago

I'm in a rush please help

Alexeev081 [22]

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4 0

4 years ago