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3 years ago

How many amu are there in 8.3g?

Chemistry

1 answer:

jasenka [17]3 years ago

4 0

Answer:

4.998 × 10²⁴ amu

Explanation:

Given data:

Mass of substance = 8.3 g

Atomic mass unit = ?

Solution:

The mass of one mole of substance is equal the gram per mole of that substance.

atomic mass unit = g/mol

For 8.3 g:

8.3 × 6.022×10²³

4.998 × 10²⁴ amu

The number 6.022 × 10²³ is called Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

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How many valence electrons does B (boron) have?<br> A. 13<br> B. 5<br> C. 10<br> D. 3

pshichka [43]

Answer:

Boron has 3 valence electrons

8 0

3 years ago

Why does spda fizz when you put a paper straw in it?

Zanzabum

The paper in the straw makes the soda fizz.

3 0

3 years ago

How much<br>much hydrogen gas evolved<br>when 1.5 current is passed through water for 1.5 hours?

evablogger [386]

0.042 moles of Hydrogen evolved

<h3>Further explanation</h3>

Given

I = 1.5 A

t = 1.5 hr = 5400 s

Required

Number of Hydrogen evolved

Solution

Electrolysis of water ⇒ decomposition reaction of water into Oxygen and Hydrogen gas.

Cathode(reduction-negative pole) : 2H₂O(l)+2e⁻ ⇒ H₂(g)+2OH⁻(aq)

Anode(oxidation-positive pole) : 2H₂O(l)⇒O₂(g)+4H⁻(aq)+4e⁻

Total reaction : 2H₂O(l)⇒2H₂(g)+O₂(g)

So at the cathode H₂ gas is produced

Faraday : 1 mole of electrons (e⁻) contains a charge of 96,500 C

$\tt mol~e^-=\dfrac{Q}{96500}$

Q = i.t

Q = 1.5 x 5400

Q = 8100 C

mol e⁻ = 8100 : 96500 = 0.084

From equation at cathode , mol ratio e⁻ : H₂ = 2 : 1, so mol H₂ = 0.042

4 0

3 years ago

Ethanol (c2h5oh) melts at -114Â°c. the enthalpy of fusion is 5.02 kj/mol. the specific heats of solid and liquid ethanol are 0.9

maria [59]

You'll want to add three amounts of heat.

(1) Specific heat of lowering the temperature from -135°C to the melting point -114°C
(2) Latent heat of fusion/melting
(3) Specific heat of elevating the temperature from -114°C to -50°C

(1) E = mCΔT = (25 g)(0.97 J/g·°C)(1 kJ/1000 J)(-114 - -135) = 0.509 kJ
(2) E = mΔH = (25 g)(5.02 kJ/mol)(1 mol/46.07 g ethanol) = 2.724 kJ
(3) E = mCΔT = (25 g)(2.3 J/g·°C)(1 kJ/1000 J)(-50 - -114) = 3.68 kJ

<em>Summing up all energies, the answer is 6.913 kJ.</em>

7 0

4 years ago

High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen. (a) Determine t

Lemur [1.5K]

Answer:

Concentration of Flourine = 24.756%

Explanation:

Given that :

High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen.

the objective is to determine he concentration of Flourine (in wt%) that must be added if this substitution occurs for 12% of all the original hydrogen atoms.

At standard conditions , the atomic weight of the these compounds are as follows:

Carbon = 12.01 g/mol

Chlorine = 35.45 g/mol

Fluorine = 19.00 g/mol

Hydrogen = 1.008 g/mol

Oxygen = 16.00 g/mol

The chemical formula for polyethylene = (CH₂ - CH₂)ₙ

Therefore, for two carbons, there will be 4 hydrogens;

i.e

(CH₂ - CH₂)₂

( C₂H₄ - C₂H₄ )

Suppose the number of original hydrogen = 4moles

number of moles of Flourine F = 12% of 4

= 0.12 × 4

= 0.48 mol

∴ the number of remaining moles of Hydrogen is:

= 4 - 0.48

= 3.52 moles

number of moles of Carbon = 2 moles

∴ the mass of flourine F = number of moles of F × molar mass of F

= 0.48 × 19

= 9.12

The total mass of the compound now is = (0.48 × 19 ) + (3.52 × 1) + (2× 12)

= 9.12 + 3.52 + 24

= 36.64

Concentration of Flourine = (mass of flourine/total mass) × 100

Concentration of Flourine = (9.12/36.84 ) × 100

Concentration of Flourine = 0.24756 × 100

Concentration of Flourine = 24.756%

3 0

3 years ago