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Tasya [4]
3 years ago
7

The Henry's Law constant of methyl bromide, CH3Br, is KH = 0.159 mol/(L ∙ atm) at 25°C. What is the solubility of methyl bromide

in water at 25°C and at a partial pressure of 270. Mm Hg?
Please show steps
Chemistry
1 answer:
Tanzania [10]3 years ago
4 0

Answer:

Henry's law constant for a gas ch3br

28 The Henry's Law constant of methyl bromide, CH3Br, is k = 0.159 mol/(L⋅atm) at 25°C.

Explanation:

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What is creation in terms of chemical reactions?
shtirl [24]

Answer:

Products

Explanation:

In a chemical reaction, the atoms and molecules produced by the reaction are called products

7 0
3 years ago
In the following reaction, how many grams of NaBr will produce 244 grams of NaNO3? Pb(NO3)2(aq) 2 NaBr(aq) PbBr2(s) 2 NaNO3(aq)
lyudmila [28]

Answer : The mass of NaBr is, 295.323 grams

Solution :

First we have to calculate the moles of NaNO_3.

\text{Moles of }NaNO_3=\frac{\text{Mass of }NaNO_3}{\text{Molar mass of }NaNO_3}=\frac{244g}{85g/mole}=2.87moles

Now we have to calculate the moles of NaBr.

The balanced chemical reaction is,

Pb(NO_3)_2(aq)+2NaBr(aq)\rightarrow PbBr_2(s)+2NaNO_3(aq)

From the balanced reaction we conclude that

As, 2 moles of NaNO_3 react with 2 moles of NaBr

So, 2.87 moles of NaNO_3 react with 2.87 moles of NaBr

Now we have to calculate the mass of NaBr.

\text{Mass of }NaBr=\text{Moles of }NaBr\times \text{Molar mass of }NaBr

\text{Mass of }NaBr=(2.87mole)\times (102.9g/mole)=295.323g

Therefore, the mass mass of NaBr is, 295.323 grams

7 0
3 years ago
1. Show that heat flows spontaneously from high temperature to low temperature in any isolated system (hint: use entropy change
Inga [223]

Answer:

1 ) Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

                 = Q/tc - Q/th

2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

Explanation:

<u>1) To show that heat flows spontaneously from high temperature to low temperature </u>

example :

Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )

Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment

Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )

Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

                 = Q/tc - Q/th

<u>2) Entropy change for Decomposition of mercuric oxide </u>

2HgO (s) → 2Hg(l) + O₂ (g)

Δs = positive

there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C

hence ΔSdecomposition = S⁻ Hg  -  S⁻ HgO =

Δh of reaction = 181.6 KJ

Temp = 500 + 273 = 773 k

hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

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icang [17]

Answer:

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