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3 years ago

(easy)+(20 points)

Chemistry

2 answers:

Semmy [17]3 years ago

6 0

I am not sure but cold to hot

Vitek1552 [10]3 years ago

4 0

Liquid to solid and solid to liquid

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A 60 and N net force is applied to a 12 kg box which is the acceleration of the box?

GarryVolchara [31]

The relation between force and mass and acceleration is

$Force = mass X acceleration$

The SI unit will be

Force = Newton

mass =kg

acceleration= $\frac{m}{s^{2} }$

thus

$acceleration = \frac{Force}{mass}$

putting values

$acceleration=\frac{60 N}{12kg}=5\frac{m}{s^{2} }$

Thus the acceleration will be

$5\frac{m}{s^{2} }$

6 0

3 years ago

How many moles of HCl are in 30.00mL of a 0.1000M HCl solution? A. 0.003000mol. B. 300.0mol C. 0.03000mol D. 3.000mol

Katyanochek1 [597]

Given the volume of HCl solution = 30.00 mL

Molarity of HCl solution = 0.1000 M

Molarity, moles and volume are related by the equation:

Molarity = $\frac{Moles of solute}{Volume of solution (L)}$

Converting volume of HCl from mL to L:

$30.00 mL * \frac{1 L}{1000mL}=0.030000 L$

Calculating moles of HCl from volume in L and molarity:

$0.03000 L * \frac{0.1000mol}{L}= 0.003000 mol HCl$

The final moles would be reported to 4 sig figs. So the correct answer will be 0.03000 mol HCl

Correct option: C. 0.03000mol

7 0

3 years ago

Which of the following is equal to 1 mole?

nadya68 [22]

They are all equal to one mole (all of the above).

7 0

3 years ago

Please help Please help

alexdok [17]

Follow Avogadro’s Number
1 mole = 6.02 x 10^23
So we can do it
4.77x10^25/6.02x10^23 = 79.2 mole

5 0

1 year ago

Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

3 years ago