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3 years ago

How many moles are 18.06*10^23 formula units of salt?

Chemistry

1 answer:

xxTIMURxx [149]3 years ago

4 0

Answer:

mole

Originally, a mole was the quantity of anything that has the same number of particles found in 12.000 grams of carbon-12. That number of particles is Avogadro's Number, which is roughly 6.02x1023. A mole of carbon atoms is 6.02x1023 carbon atoms.

Explanation:

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Suppose we have a compound that is 4.330 % Li, 22.10 % Cl, 39.89 % O, and 33.69 % H2O. What is the compounds formula?

amid [387]

The formula of compound is LiClO4.3H2O

<em><u>calculation</u></em>

<em><u> </u></em>find the mole of each element

that is moles for Li,Cl,O and that of H2O

moles = % composition/ molar mass

For Li = 4.330/ 6.94 g/mol= 0.624 moles

Cl=22.10/35.5=0.623 moles

39.89/16 g/mol =2.493 moles

H20= 33.69/18 g/mol= 1.872 moles

find the mole ratio by dividing each moles by smallest number of mole ( 0.624 moles)

that is for Li= 0.624/0.623= 1

Cl= 0.623/0.623=1

O = 2.493/0.623 =4

H2O= 1.872/0.623=3

<h3>Therefore the formula=LiClO4.3H2O</h3><h3 />

8 0

3 years ago

Molten gallium reacts with arsenic to form the semiconductor, gallium arsenide, GaAs, used in light-emitting diodes and solar ce

Vsevolod [243]

Answer:

a) 1.2g of arsenic b) 0.64g of arsenic c) 3.481g of gallium d) 3.806g of gallium e) 2.61g arsenic

Explanation:

The balanced equation is:

Ga + As = GaAs, 1:1 mole ratio

a) mass (gallium)/ molar mass of Ga = 4/ 69.723 = 0.0574mol

Mass (arsenic)/ molar mass of As = 5.5/74.9216 = 0.0734, subtracting the moles from each other (knowing already that the ratio is 1:1), arsenic is in excess by 1.2g

b) repeating the procedure (changing the values)

It will be 0.0574 to 0.06593

Arsenic is in excess by 0.00854

0.00854* mass of arsenic (such must be done for the first remaining mole) = 0.64g

c) the mole ratio is 0.0574: 0.00747

Gallium is in excess by 0.05

Mass of excess gallium = 0.05* 69.723 = 3.481g

d) using the mass given, the new ratio is

0.128: 0.0734

Gallium is in excess by 0.054mol

Mass of excess gallium = 0.054*69.723 = 3.806g

e) using the mass again, the new ratio is 0.0574: 0.02,

Gallium is in excess by 0.0374*69.723 = 2.61g

7 0

3 years ago

Using the Rydberg formula, calculate the initial energy level when an electron in a hydrogen atom transitions into n= 2 and emit

EleoNora [17]

Answer:

Explanation:

Using the Rydberg formula as:

$\frac {1}{\lambda}=R_H\times Z^2\times (\frac {1}{n_{1}^2}-\frac {1}{n_{2}^2})$

where,

λ is wavelength of photon

R = Rydberg's constant (1.097 × 10⁷ m⁻¹)

Z = atomic number of atom

n₁ is the initial final level and n₂ is the final energy level

For Hydrogen atom, Z= 1

n₂ = 2

Wavelength = 410.1 nm

Also,

1 nm = 10⁻⁹ m

So,

Wavelength = 410.1 × 10⁻⁹ m

Applying in the formula as:

$\frac {1}{410.1\times 10^{-9}}=1.097\times 10^7\times 1^2\times (\frac {1}{n_{1}^2}-\frac {1}{2^2})$

Solving for n₁ , we get

n₁ ≅ 1

3 0

4 years ago

Read 2 more answers

Which is the net result of the proton-proton chain? 6 protons = 2 heliums + 3 positrons + 3 neutrinos + gamma rays 4 protons = 1

saveliy_v [14]

Answer: 5.125 neutrinos

Explanation:

8 0

3 years ago

For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ

Komok [63]

Answer:

Check the explanation

Explanation:

Answer – Given, $H_3PO_4$ acid and there are three Ka values

$K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}$

The transformation of $H_2PO_4- (aq) to HPO_4^2-(aq)$ is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

= 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

= 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

= -log 6.2x10-8

= 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

= 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

6 0

3 years ago