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Lera25 [3.4K]
3 years ago
7

A sample of thionylchloride, SOCl2, contains 0.206 mol of the compound. What is the mass of the sample, in grams?

Chemistry
2 answers:
Ilia_Sergeevich [38]3 years ago
3 0

Explanation:

Moles=mass/molar mass

moles × molar mass = mass

0.206 x 119= mass

Mass= 24. 51grams

Galina-37 [17]3 years ago
3 0

Answer:

24.51 g to the nearest hundredth

Explanation:

1 mole of SOCl2 =  32 + 16 + 2 *35.5 = 119 g

0.206 mol = 0.206 * 119

=  24.51 g.

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Draw one representation that shows the intermolecular interactions between NH_3 and water and another that shows the intermolecu
Lera25 [3.4K]

Answer:

NH3 has greater water solubility due to intermoleculate interactions

Explanation:

Hi:

If we represent the structures of NH3 and SbH3 we can see that they are similar to the naked eye, this is because N and Sb belong to the same group of the periodic table (group 15).

However, the electronegativity of N is greater than that of Sb. The NH3 molecule is polar and can form an intermolecular interaction called hydrogen bridge with water.

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The zone with positive charge density of the water molecule (hydrogens) is oriented towards the zone with positive charge density of SBH3 (the pair of electrons not shared)

Stronger intermolecular junctions allow greater solubility of NH3 molecules.

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What is the highest point of the wave a transverse wave called
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This decomposition is first order with respect to phosphine, and has a half‑life of 35.0 s at 953 K. Calculate the partial press
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Answer:

0.57 atm

Explanation:

When a a reaction is first order, we have from calculus the following relation:

ln[A]t/[A]₀ = - kt

where [A]t is the concentration of A ( phosphine in this case ) after a time, t

           [A]₀ is the initial concentration of A

           k is the rate constant, and

           t is the time

We also know that for a first order reaction

           k = 0.693/ t 1/2

wnere t 1/2 is the half-life.

This equation is derived for the case when A]t/= 1/2 x [A]₀ which occurs at the half-life.

Thus, lets first find k from the half life time, and then solve for t = 70.5 s

k = 0.693 /  35.0 s = 0.0198 s⁻¹

ln [ PH₃ ]t / [ PH₃]₀ = - kt

from the ideal gas law we know pV = nRT, so the volumes cancel:

ln (pPH₃ )t / p(PH₃)₀ = - kt

taking inverse log to both sides of the equation:

(pPH₃ )t / p(PH₃)₀  = - kt

thus:

(pPH₃ )t  = 2.29 atm x e^(- 0.0198 s⁻¹ x 70.5 s ) = 0.57 atm

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