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zepelin [54]
3 years ago
10

Hey sample of helium occupies 235 mL at 788 tour and 25°C. If the sample is condensed into a 0.115 L flask what will be the new

pressure assuming constant temperature?
Chemistry
1 answer:
7nadin3 [17]3 years ago
3 0

the system is in constant temperature

so it follows Boyle's law

i.e.P1V1=P2P2

where P1=initial pressure=788 tour

P2=final pressure= ?

V1=initial volume=235 ml

V2=final volume=0.115 ml

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The process by which wind removes surface materials is called abrasion.

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3 years ago
If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th
fredd [130]

Answer:

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

We assume that s to be  the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

 [Pb²⁺] =   s

Then [I⁻] = 2s

K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} =  4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of  }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

B)

The Concentration of Pb²⁺  in water is calculated as :

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}

\mathbf{s} =\sqrt[3]{1.75*10^{-9}}

\mathbf{s} =\mathbf{1.21*10^{-3}  \ mol/L }

The Concentration of Pb²⁺  in 1.0 mol·L⁻¹ NaI

\mathbf{PbCl{_2}}  \leftrightharpoons    \ \ \ \ \ \ \  \mathbf{Pb^{2+}}   \ \ \ \  \ +   \ \  \ \ \ \ \ \mathbf{2 I^-}

                             \ \ \ \ \ \ \  \ \   \ \  \ \ \ \ \ \ \  \mathbf0}   \ \ \ \  \ \ \ \ \ \   \ \ \ \ \ \mathbf{1.0}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{+2x}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{1.0+2x}

The equilibrium constant:

K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \  m/L

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the  common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

3 0
3 years ago
This is a chemistry question that i don't know how to do. Need some help
koban [17]
C₀=8.10M
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v₁-?

n=c₀v₀=c₁v₁

v₁=c₀v₀/c₁

v₁=8.10*1.58/5.28=2.42 L 
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Answer:

Eutrophication is the enrichment of a body of water with excessive nutrients (nitrogen and phosphorus), which causes algal growth and subsequent decline of dissolved oxygen after decomposition.

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3 years ago
Calculate the Ka of your acetic acid solution. Discuss this calculation. Based on the value of Ka, is acetic acid a strong acid
lord [1]

Based on our knowledge of strong and weak acids, we can confirm that the Ka value for acetic acid will be relatively low since it is a weak acid.

Acids can be strong or weak. This is determined by its <u><em>tendency to break apart into ions or stay together to form molecules.</em></u> Although somewhat counter-intuitive, strong acids are those that are most likely to break apart and therefore contain a <em><u>high number of </u></em><em><u>ions </u></em><em><u>within their solutions</u></em>.

Weak acids, on the other hand, are those that<em><u> tend to stay together in the form of </u></em><em><u>molecules </u></em><em><u>and therefore possess very </u></em><em><u>low ion counts </u></em><em><u>in their solutions.</u></em> The acid dissociation constant, Kₐ, is used to measure whether an acid is weak or strong and how much so. In the case of Acetic acid, the ka measurement will offer a low value, indicating a weak acid.

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brainly.com/question/4131966?referrer=searchResults

4 0
2 years ago
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