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Anna35 [415]
3 years ago
10

What is the mass of 0.25 mol of sulfur atom

Chemistry
1 answer:
tatiyna3 years ago
6 0

Answer:

8.0 g

Explanation:

To convert moles to grams, you need to use the molar mass.  The molar mass of sulfur is 32.06.  The molar mass is the atomic mass of the atom.

(0.25 mol) × (32.06 g/mol) = 8.015 g

Round for significant figures.

8.015 g ≈ 8.0 g

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I think you want to ask about Keq. At equilibrium, we can know [SO2Cl2] is 2.2*10-2 M -1.3*10-2M=9*10^-3 M. And [SO2]=[Cl2]. So the Keq=1.88*10^-2.
5 0
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What will happen to the flask?
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Which statement best describes how two hydrogen atoms combine to form molecular hydrogen (H2)?
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8 0
3 years ago
The empirical formula of glucose is CH2O the molecular formula of glucose is 6 times than the empirical formula. what is the mol
jeka57 [31]

Answer:

The molecular formula of glucose is C₆H₁₂O₆

Explanation:

Empirical formula:

It is the simplest formula gives the ratio of smallest whole number of atoms.

Molecular formula:

It gives the total number of atoms in a molecule of compound.

The molecular formula and empirical formula can be related as follow:

Molecular formula = n × empirical formula

Given data:

Empirical formula = CH₂O

Molecular formula = ?

It is stated in given problem that molecular formula is the 6 times of the empirical formula.

Molecular formula = n × empirical formula

Molecular formula = 6 × CH₂O

Molecular formula = C₆H₁₂O₆

The molecular formula of glucose is C₆H₁₂O₆.

5 0
3 years ago
A chemist fills a reaction vessel with 0.750 M lead (II) (Pb2+) aqueous solution, 0.232 M bromide (Br) aqueous solution, and 0.9
Ronch [10]

Answer:

The free energy = -20.46 KJ

Explanation:

given Data:

Pb²⁺ = 0.750 M

Br⁻ = 0.232 M

R = 8.314 Jk⁻¹mol⁻¹

T = 298K

The Gibb's free energy is calculated using the formula;

ΔG = ΔG° + RTlnQ -------------------------1

Where;

ΔG° = standard Gibb's freeenergy

R = Gas constant

Q = reaction quotient

T = temperature

The chemical reaction is given as;

Pb²⁺(aq) + 2Br⁻(aq) ⇄PbBr₂(s)

The ΔG°f are given as:

ΔG°f (PbBr₂)  = -260.75 kj.mol⁻¹

ΔG°f (Pb²⁺)   = -24.4 kj.mol⁻¹

ΔG°f (2Br⁻)    = -103.97 kj.mol⁻¹

Calculating the standard gibb's free energy using the formula;

ΔG° = ξnpΔG°(product) - ξnrΔG°(reactant)

Substituting, we have;

ΔG° =[1mol*ΔG°f (PbBr₂)] - [1 mol *ΔG°f (Pb²⁺) +2mol *ΔG°f (2Br⁻)]

ΔG° =(1 *-260.75 kj.mol⁻¹) - (1* -24.4 kj.mol⁻¹) +(2*-103.97 kj.mol⁻¹)

      = -260.75 + 232.34

     = -28.41 kj

Calculating the reaction quotient Q using the formula;

Q = 1/[Pb²⁺ *(Br⁻)²]

   = 1/(0.750 * 0.232²)

  = 24.77

Substituting all the calculated values into equation 1, we have

ΔG = ΔG° + RTlnQ

ΔG = -28.41 + (8.414*10⁻³ * 298 * In 24.77)

     = -28.41 +7.95

    = -20. 46 kJ

Therefore, the free energy of reaction = -20.46 kJ

8 0
3 years ago
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