Answer the question PLS 100 points!!!

G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu

Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ $Ag^{+}$ + $Br^{-1}$ .

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains $\frac{0.6485}{45}$ ×1000 = 14.411 gm-mole. Thus the solubility product $K_{sp}$ of AgBr = [ $Ag^{+}$ ]× $Br^{-}$ .

Or, 5.0× $10^{-13}$ = $S^{2}$ (the given value of solubility product of AgBr is 5.0× $10^{-13}$ and the charge of the both ions are same).

Thus S = (5.00× $10^{-13}$ ) $^{1/2}$ = 7.071× $10^{-7}$ g/mL.

Thus the concentration of Br $^{-1}$ or HBr is 7.071× $10^{-7}$ g/mL.

4 0

3 years ago

What is a living thing that can carry out life process by itself

9966 [12]

Answer:

All Living things

Explanation:

All living things are capable of carrying out all life processes. Some examples of such organisms are plants, animals, microorganisms (such as bacteria)

And organism too

6 0

3 years ago

Read 2 more answers

If you have a 150-gram sample of CrO3: • How many moles of CrO3 do you have? • How many oxygen atoms do you have? • How many gra

docker41 [41]

Answer :

(a) The moles of $CrO_3$ is, 1.5 moles.

(b) The number of oxygen atoms are, $27.099\times 10^{23}$

(c) The mass of oxygen is, 72 grams.

Explanation : Given,

Mass of $CrO_3$ = 150 g

Molar mass of $CrO_3$ = 100 g/mole

Molar mass of oxygen = 16 g/mole

(a) First we have to calculate the moles of $CrO_3$ .

$\text{Moles of }CrO_3=\frac{\text{Mass of }CrO_3}{\text{Molar mass of }CrO_3}=\frac{150g}{100g/mole}=1.5moles$

The moles of $CrO_3$ is, 1.5 moles.

(b) Now we have to calculate the number of oxygen atoms.

In $CrO_3$ , there are 1 atom of chromium and 3 atoms of oxygen.

According to the mole concept,

1 mole of $CrO_3$ contains $3\times 6.022\times 10^{23}$ number of oxygen atoms.

So, 1.5 mole of $CrO_3$ contains $1.5\times 3\times 6.022\times 10^{23}=27.099\times 10^{23}$ number of oxygen atoms.

The number of oxygen atoms are, $27.099\times 10^{23}$

(c) Now we have to calculate the mass of oxygen.

As, 1 mole of $CrO_3$ contains 3 moles of oxygen

So, 1.5 mole of $CrO_3$ contains $1.5\times 3=4.5$ moles of oxygen

$\text{Mass of oxygen}=\text{Moles of oxygen}\times \text{Molar mass of oxygen}$

$\text{Mass of oxygen}=(4.5mole)\times (16g/mole)=72g$

The mass of oxygen is, 72 grams.

3 0

3 years ago

Green plants absorbs sunlight to power photosynthesis, the chemical synthesis of food from water and carbon dioxide. The compoun

Serggg [28]

Answer: The frequency of this light is $6.62\times 10^{14}s^{-1}$

Explanation:

To calculate the wavelength of light, we use the equation:

$\lambda=\frac{c}{\nu}$

where,

$\lambda$ = wavelength of the light = $453nm=453\times 10^{-9}m$

c = speed of light = $3.0\times 10^8m/s$

$\nu$ = frequency of light = ?

$\nu=\frac{3.0\times 10^8m/s}{453\times 10^{-9}m}=6.62\times 10^{14}s^{-1}$

The frequency of this light is $6.62\times 10^{14}s^{-1}$

3 0

3 years ago

A brick has a mass of 4.0 kg and the Earth has a mass of 6.0 × 1027 g. Use this information to answer the questions below. Be su

Vedmedyk [2.9K]

Answer:

a) $2.4\times 10^{24} kg$ is the mass of 1 mole of bricks.

b) $2.5\times 10^3$ moles of bricks have a mass equal to the mass of the Earth.

Explanation:

a) Mass of brick = 4.0 kg

1 mole = $N_A=6.022\times 10^{23}$ particles/ atoms/molecules

Mass of $N_A$ bricks :

$=6.022\times 10^{23}\times 4.0 kg=2.4088\times 10^{24} kg\approx 2.4\times 10^{24} kg$

$2.4\times 10^{24} kg$ is the mass of 1 mole of bricks.

b)

Mass of the Earth = M = $6.0\times 10^{27} kg$

Mass of 1 mole of brick = m= $2.4\times 10^{24} kg$

Let the moles of brick with equal mass of the Earth be x.

$m\times x=M$

$x=\frac{M}{m}=\frac{6.0\times 10^{27}kg}{2.4\times 10^{24} kg}=2.5\times 10^3$

$2.5\times 10^3$ moles of bricks have a mass equal to the mass of the Earth.

7 0

3 years ago