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3 years ago

Chemistry skeletal structure

Chemistry

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yan [13]3 years ago

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Answer:

Your answer is attached DAME!!!

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Calculate the equilibrium concentration of c2o42− in a 0.20 m solution of oxalic acid.

spin [16.1K]

Answer:

[C2O2−4] =1.5⋅10−4⋅mol⋅dm−3

Explanation:

For the datasheet found at Chemistry Libretext,

Ka1=5.6⋅10−2 and Ka2=1.5⋅10−4 [1]

for the separation of the primary and second nucleon once oxalic corrosive C2H2O4 breaks up in water at 25oC (298⋅K).

Build the RICE table (in moles per l, mol⋅dm−3, or identically M) for the separation of the primary oxalic nucleon. offer the growth access H+(aq) fixation be x⋅mol⋅dm−3.

R C2H2O4(aq)⇌C2HO−4(aq)+H+(aq)

I 0.20

C −x +x

E 0.20−x x

By definition,

Ka1=[C2HO−4(aq)][H+(aq)][C2H2O4(aq)]=5.6⋅10−2

Improving the articulation can provides a quadratic condition regarding x, sinking for x provides [C2HO−4]=x=8.13⋅10−2⋅mol⋅dm−3

(dispose of the negative arrangement since fixations can faithfully be additional noteworthy or appreciate zero).

Presently build a second RICE table, for the separation of the second oxalic nucleon from the amphoteric C2HO−4(aq) particle. offer the modification access C2O2−4(aq) be +y⋅mol⋅dm−3. None of those species was out there within the underlying arrangement. (Kw is neglectable) therefore the underlying centralization of each C2HO−4 and H+ are going to be appreciate that at the harmony position of the first ionization response.

R C2HO−4(aq)⇌C2O2−4(aq)+H+(aq)

I 8.13⋅10−2 zero eight.13⋅10−2

C −y +y

E 8.13⋅10−2−y y eight.13⋅10−2+y

It is smart to just accept that

a. 8.13⋅10−2−y≈8.13⋅10−2,

b. 8.13⋅10−2+y≈8.13⋅10−2 , and

c. The separation of C2HO−4(aq)

Accordingly

Ka2=[C2O2−4(aq)][H+(aq)][C2HO−4(aq)]=1.5⋅10−4≈(8.13⋅10−2 )⋅y8.13⋅10−2

Consequently [C2O2−4(aq)]=y≈Ka2=1.5⋅10−4]⋅mol⋅dm−3

3 0

4 years ago

A vessel of volume 22.4 dm3 contains 20 mol h2 and 1 mol n2 ad 273.15 k initially. All of the nitrogen reacted with sufficient h

NikAS [45]

Nitrogen combine with hydrogen to produce ammonia $\text{NH}_3$ at a $1:3:2$ ratio:

$\text{N}_2 \; (g) + 3 \; \text{H}_2 \; (g) \leftrightharpoons 2\; \text{NH}_3 \; (g)$

Assuming that the reaction has indeed proceeded to completion- with all nitrogen used up as the question has indicated. $3 \; \text{mol}$ of hydrogen gas would have been consumed while $2 \; \text{mol}$ of ammonia would have been produced. The final mixture would therefore contain

$17 \; \text{mol}$ of $\text{H}_2 \; (g)$ and
$2 \; \text{mol}$ of $\text{NH}_3 \; (g)$

Apply the ideal gas law to find the total pressure inside the container and the respective partial pressure of hydrogen and ammonia:

$\begin{array}{lll} P(\text{container}) &= & n \cdot R \cdot T / V \\ & = & (17 + 2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 1.926 \times 10^{3} \; \text{kPa} \end{array}$
$\begin{array}{lll} P(\text{H}_2) &= & n \cdot R \cdot T / V \\ & = & (17) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 1.723 \times 10^{3} \; \text{kPa} \end{array}$
$\begin{array}{lll} P(\text{NH}_3) &= & n \cdot R \cdot T / V \\ & = & (2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 2.037 \times 10^{2} \; \text{kPa} \end{array}$

6 0

3 years ago

Many hospitals, and some doctors\' offices, use radioisotopes for diagnosis and treatment, or in palliative care (relief of symp

jekas [21]

Answer:

For a: The isotopic symbol of the above atom will be $_{53}^{131}\textrm{I}$

For b: The isotopic symbol of the above atom will be $_{77}^{192}\textrm{Ir}$

For c: the isotopic symbol of the above atom will be $_{62}^{153}\textrm{Sm}$

Explanation:

The isotopic representation of an atom is: $_Z^A\textrm{X}$

where,

Z = Atomic number of the atom

A = Mass number of the atom

X = Symbol of the atom

For a: Iodine-131

Atomic number of iodine = 53

Mass number of iodine = 131

Symbol of iodine = I

The isotopic symbol of the above atom will be $_{53}^{131}\textrm{I}$

For b: Iridium-192

Atomic number of iridium = 77

Mass number of iridium = 192

Symbol of iridium = Ir

The isotopic symbol of the above atom will be $_{77}^{192}\textrm{Ir}$

For c: Samarium-153

Atomic number of samarium = 62

Mass number of samarium = 153

Symbol of samarium = Sm

The isotopic symbol of the above atom will be $_{62}^{153}\textrm{Sm}$

7 0

3 years ago

Which equation agrees with the ideal gas law?

photoshop1234 [79]

$\frac{(P1)(V1)}{T1}=\frac{(P2)(V2)}{T2}$ is directly related to the ideal gas law.

The ideal gas law states that: $PV=nRT$ . Rewriting this gives: $\frac{PV}{T}=nR$ . R is the universal gas constant, so its value is fixed. For a given sample, n is the number of moles of the gas, so its value would be fixed too. This means that the value of $\frac{PV}{T}$ would be a fixed constant as well. Therefore, whatever the initial value $\frac{(P1)(V1)}{T1}$ is, it should be equal to the final value $\frac{(P2)(V2)}{T2}$ .

6 0

3 years ago

Read 2 more answers

Which images show a container holding a heterogeneous mixture?

marshall27 [118]

Answer:

Explanation:

a heterogeneous mixture is one that is not uniform in composition.It is made up of non-uniform mix of smaller constituent parts that could be separated from one another using various means.

4 0

3 years ago

Read 2 more answers