All categories

2 years ago

What is the amount in moles of a 52 g sample of aluminium fluoride, AIF3

Chemistry

1 answer:

Karo-lina-s [1.5K]2 years ago

6 0

Answer:

Try looking for some kind of answer that has to do with .61904

Explanation:

52 g x 1 mol/ (molar mass of aluminum fluoride) 84 g = .619047619

Sorry if this is wrong but that's my best guess

You might be interested in

Define Evaporation.<br>Thanks

Kay [80]

Answer:

evapouration is a type of vaporization that occur on the surface of liquid as it change into the gas phase.

hope it helps.

5 0

3 years ago

Read 2 more answers

What is the pH of a 4.08*10-'M solution of H30+?

Mariana [72]

Answer:

8.4

Explanation:

-log(4.08x10^-9) = 8.4

- Hope that helped! Please let me know if you need further explanation.

4 0

3 years ago

Which statement describes what happens to the DNA in the process shown? A. Some of the DNA is destroyed, so each daughter cell h

kolbaska11 [484]

hhjjjbvhjbvb

hhmkn

hjkk

hhjj

7 0

3 years ago

A pycnometer is a precisely weighted vessel that is used for highly accurate density determinations. Suppose that a pycnometer h

dexar [7]

Answer:

5.758 is the density of the metal ingot in grams per cubic centimeter.

Explanation:

1) Mass of pycnometer = M = 27.60 g

Mass of pycnometer with water ,m= 45.65 g

Density of water at 20 °C = d = $998.2 kg/m^3$

1 kg = 1000 g

$1 m^3=10^6 cm^3$

$998.2 kg/m^3=\frac{998.2 \times 1000 g}{10^6 cm^3}=0.9982 g/cm^3$

Mass of water ,m'= m - M = 45.65 g - 27.60 g =18.05 g

Volume of pycnometer = Volume of water present in it = V

$Density=\frac{Mass}{Volume}$

$V=\frac{m'}{d}=\frac{18.05 g}{0.9982 g/cm^3}=18.08 cm^3$

2) Mass of metal , water and pycnometer = 56.83 g

Mass of metal,M' = 9.5 g

Mass of water when metal and water are together ,m''= 56.83 g - M'- M

56.83 g - 9.5 g - 27.60 g = 19.7 g

Volume of water when metal and water are together = v

$v=\frac{m''}{d}=\frac{19.7 g}{0.9982 g/cm^3}=19.73 cm^3$

Density of metal = d'

Volume of metal = v' = $\frac{M'}{d'}$

Difference in volume will give volume of metal ingot.

v' = v - V

$v'=19.73 cm^3-18.08 cm^3=$

$v'=1.65 cm^3$

Since volume cannot be in negative .

Density of the metal =d'

= $d'=\frac{M'}{v'}=\frac{9.5 g}{1.65 cm^3}=5.758 g/cm^3$

5 0

3 years ago

"acid is responsible for the odor in rancid butter. a solution of 0.25 m butyric acid has a ph of 2.71. what is the ka for"

Salsk061 [2.6K]

Answer:- The Ka for the acid is $1.53*10^-^5$ .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

$Ka = [H^+][A^-]\frac{1}{HA}$

Where, Ka is the acid ionization constant. Let's plug in the values.

$Ka = \frac{X^2}{0.25-X}$

Let's calculate the value of X first using the equation:

$pH = -log[H^+]$ [/tex]

on taking antilog ob above equation we get:

$[H^+]=10^-^p^H$

$[H^+]=10^-^2^.^7^1$

$[H^+]$ = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

$Ka=\frac{(0.00195)^2}{0.25-0.00195}$

$Ka=1.53*10^-^5$

So, the value of Ka for butyric acid is $1.53*10^-^5$ .

8 0

3 years ago

Read 2 more answers