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3 years ago

Explain how you would make 450 mL of a .250M NaOH solution?

Chemistry

1 answer:

Neporo4naja [7]3 years ago

4 0

Answer:

Weigh 4.5 grams of sodium hydroxide and add it to the dry volumetric flask of 450 mL followed by small amount of water to dissolve all the NaOH .After this add the water upto tye mark of 450 mL.

Explanation:

Molarity of the solution is the moles of compound in 1 Liter solutions.

$Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}$

Mass of NaOH = x

Molar mass of NaOH = 40 g/mol

Volume of the NaOH solution = 450 mL =- 0.450 L ( 1 ml = 0.450 L)

Molarity of the solution of NaOH = 0.250 M

$Molarity=\frac{1.248 g}{26 g/mol\times 0.9102254 L}=0.528 mol/L$

$0.250 M=\frac{x}{40 g/mol\times 0.450 L}$

Solving for x:

x = 4.5 g

Weigh 4.5 grams of sodium hydroxide and add it to the dry volumetric flask of 450 mL followed by small amount of water to dissolve all the NaOH .After this add the water upto tye mark of 450 mL.

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When a sample of a compound in the vitamin D family was burned in a combustion analysis, 5.983 mg of the compound gave 18.490 mg

Jet001 [13]

We calculate it as follows:

Moles CO2 = 0.01849 g / 44 = 0.000420
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Moles H = 2 x 0.006232 / 18 = 0.000692 
Mass H = 0.000692 g 
Mass O = 0.005982 - ( 0.00504 + 0.000692) = 0.00025 
Moles O = 0.00025 / 16 = 0.0000156 
C 0.000420
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divide each by the smallest value, giving you the chemical formula as:

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5 0

3 years ago

Read 2 more answers

What is the associated deBroglie wavelength of a H2 molecule moving on one direction with kinetic energy of (3/2 kT) at 30 K

andrey2020 [161]

Answer: The de-Broglie's wavelength of a hydrogen molecule is $3.26\AA$

Explanation:

Kinetic energy is the measure of temperature of the system.

The equation used to calculate kinetic energy of a particle follows:

$E=\frac{3}{2}kT$

where,

E = kinetic energy of the particles = ?

k = Boltzmann constant = $1.38\times 10^{-23}J/K$

T = temperature of the particle = 30 K

Putting values in above equation, we get:

$E=\frac{3}{2}\times 1.38\times 10^{-23}J/K\times 30K\\\\E=6.21\times 10^{-22}J$

Calculating the mass of 1 molecule of hydrogen gas:

Conversion factor used: 1 kg = 1000 g

1 mole of hydrogen gas has a mass of 2 grams or $2\times 10^{-3}kg$

According to mole concept:

$6.022\times 10^{23}$ number of molecules occupy 1 mole of a gas.

As, $6.022\times 10^{23}$ number of hydrogen molecules has a mass of $2\times 10^{-3}kg$

So, 1 molecule of hydrogen will have a mass of = $\frac{2\times 10^{-3}kg}{6.022\times 10^{23}}\times 1=3.32\times 10^{-27}kg$

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

$\lambda=\frac{h}{\sqrt{2mE_k}}$

where,

$\lambda$ = De-Broglie's wavelength = ?

h = Planck's constant = $6.624\times 10^{-34}Js$

m = mass of 1 hydrogen molecule = $3.32\times 10^{-27}kg$

$E_k$ = kinetic energy of the particle = $6.21\times 10^{-22}J$

Putting values in above equation, we get:

$\lambda=\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 3.32\times 10^{-27}kg\times 6.21\times 10^{-22}J}}$

$\lambda=3.26\times 10^{-10}m=3.26\AA$ (Conversion factor: $1\AA=10^{-10}m$ )

Hence, the de-Broglie's wavelength of a hydrogen molecule is $3.26\AA$

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3 years ago

Please help

KonstantinChe [14]

Answer: There are 0.5 grams of barium sulfate are present in 250 of 2.0 M $BaSO_{4}$ solution.

Explanation:

Given: Molarity of solution = 2.0 M

Volume of solution = 250 mL

Convert mL int L as follows.

$1 mL = 0.001 L\\250 mL = 250 mL \times \frac{0.001 L}{1 mL}\\= 0.25 L$

Molarity is the number of moles of solute present in liter of solution. Hence, molarity of the given $BaSO_{4}$ solution is as follows.

$Molarity = \frac{mass}{Volume (in L)}\\2.0 M = \frac{mass}{0.25 L}\\mass = 0.5 g$

Thus, we can conclude that there are 0.5 grams of barium sulfate are present in 250 of 2.0 M $BaSO_{4}$ solution.

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3 years ago

BEST GETS BRAINLIEST!!

tatuchka [14]

Gee. I'll have to guess at what's "commonly thought".

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Book ... It's getting late, Hillary's fading, and that's all I can think of.
I hope this much is some help.

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mart [117]

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