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lina2011 [118]
2 years ago
10

30. how many grams of boric acid, b(oh)3 (fm 61.83), should be used to make 2.00 l of 0.050 0 m solution? what kind of fl ask is

used to prepare this solution?

Chemistry
1 answer:
grigory [225]2 years ago
7 0

Answer:

             Mass  =  6.183 g

Solution:

Step 1: Calculate number of moles of Boric acid using following formula,

                                        Molarity  =  Moles ÷ Volume

Solving for Moles,

                                        Moles  =  Molarity × Volume

Putting Values,

                                        Moles  =  0.05 mol.L⁻¹ × 2.0 L

                                        Moles  =  0.1 mol

Step 2: Calculate Mass of Boric Acid using following formula,

                                        Moles  =  Mass ÷ M.mass

Solving for Mass,

                                        Mass  =  Moles × M.mass

Putting values,

                                        Mass  =  0.1 mol × 61.83 g.mol⁻¹

                                        Mass  =  6.183 g

Flask used to prepare this solution is called as Volumetric flask. Take 2 L volumetric flask, add 6.183 g of Boric acid and fill it to the mark with distilled water.

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Answer:

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Explanation:

Forbes gives somewhat of an explanation if you are curious.

(Ethan Siegal, "The Universe Would Be Very Different Without Dark Matter", Forbes)

3 0
1 year ago
Three mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,
8_murik_8 [283]

Answer:

(a). 46,666.7 g/mol; 78,571.4 g/mol

(b). 86950g/mol; 46,666.7 g/mol.

(c). 86950g/mol; 43,333.33 g/mol

Explanation:

So, we are given the molar masses for the three samples as: 10,000, 30,000 and 100,000 g mol−1.

Thus, the equal number of molecule in each sample = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

The average molar mass = [ ( 10,000)^2 + (30,000)^2 + 100,000)^2] ÷ 10,000 + 30,000 + 100,000 = 78,571. 4 g/mol.

(b). The equal masses of each sample = 3/[ ( 1/ 10,000) + (1/30,000 ) + (1/100,000) ] = 20930.23 g/mol.

Average molar mass = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

(c). Equal masses of the two samples = (0.145 × 10,000) + (0.855 × 100,000)/ 0.145 + 0.855 = 86950g/mol.

The weight average molar mass = 1.7 + 10,000 + 100,000/ 1.7 + 1 = 43,333.33 g/mol.

6 0
2 years ago
A gas occupies a volume of 72 ml at 400 k and 800 torr. if the temperature drops to 200 k and the pressure changes to 400 torr,
satela [25.4K]
We are tasked to solve for the volume of the gas that occupies when pressure and temperature changes to 400 Torr and 200 Kelvin from Torr and 400 Kelvin. We can use ideal gas law assuming constant gas composition and close system. The solution is shown below:
P1V1 / T1 = P2V2 / T2
V2 = P1V1T2 / T1P2
V2 = 800*72*200 / 400*400
V2 = 72 ml

The answer for the volume is 72 ml.
7 0
2 years ago
6) A 0.20 ml CO2 bubble in a cake batter is at 27°C. In the oven it gets
Nata [24]

Answer: The new volume of cake is 1.31 mL.

Explanation:

Given: V_{1} = 0.20 mL,         T_{1} = 27^{o}C

V_{2} = ?,                    T_{2} = 177^{o}C

Formula used to calculate new volume is as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{0.20 mL}{27^{o}C} = \frac{V_{2}}{177^{o}C}\\V_{2} = 1.31 mL

Thus, we can conclude that the new volume of cake is 1.31 mL.

5 0
2 years ago
By which process is a precipitate most easily separated from the liquid in which it is suspended
Gelneren [198K]
<span>Filtration, if its a precipitate that means its insoluble. </span>
7 0
2 years ago
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