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Mice21 [21]
3 years ago
5

so I am working on this question for chemestry and can't figure out the format it wants since I am blind. The question states Wr

ite the identity of the missing particle for the following nuclear decay reaction: 4925Mn→4924Cr+? Express your answer as a particle. and my answer is currently 24 Cr49 + e+24 Cr49 + e+. O also I can't read mathmatical graphics/equation editor since majority of those aren't accessible for blind students so could you explain this in just html text? Thanks in advanced
Chemistry
1 answer:
skelet666 [1.2K]3 years ago
6 0

Answer:

Positron\rightarrow _{1} ^{0}e

Explanation:

lets take the particle is X, the given decay reaction can be written as follows

_{25} ^{49}Mn\rightarrow _{24}^{49}Cr+X

Now by using mass balancing:

The sum of the mass of right hand side=The sum of the mass of left hand side

49=49+M

M=0

Now by using atomic number balancing:

Atomic number on right side=Atomic number on left side

25 = 24 + Z

Z=1

Therefore we can say that , the mass of that unknown particle is 0 unit and atomic number is 1.

Thus that particle will be positron.

Positron\rightarrow _{1} ^{0}e

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Mkey [24]

This requires familiarity with the different theories (or concepts) of acids and bases.

On the Arrhenius concept, an acid is a substance that produces an H⁺ ion in water such that the H⁺ concentration increases, and a base is a substance that produces an OH⁻ ion in water such that the OH⁻ concentration increases.

On the Brønsted–Lowry concept, an acid is a substance that donates a proton (which is basically an H⁺ ion) in a solvent, and a base is a substance that accepts a proton in a solvent.

On the Lewis concept, an acid is a substance that accepts an electron pair in a solvent, and a base is a substance that donates an electron pair in a solvent.

The concepts become progressively broader, i.e., the Arrhenius concept is the most restrictive and the Lewis concept is the least restrictive. As a corollary, an Arrhenius acid or base is also both a Brønsted–Lowry acid or base and a Lewis acid or base, respectively; a Brønsted–Lowry acid or base is not necessarily an Arrhenius acid or base, but an Arrhenius acid or base is also a Lewis acid or base, respectively. And finally, a Lewis acid or base may not necessarily be either an Arrhenius or a Brønsted–Lowry acid or base.

So, with the above concepts in mind, we can match the statements in column A with the type of acid or base in column B:

\begin{center}\begin{tabular}{ c c } 1 & Bronsted Lowry acid \\  2 & Bronsted Lowry base \\   3 & Arrhenius acid \\ 4 & Arrhenius base \\ 5 & Lewis base \\ 6 & Lewis acid\end{tabular}\end{center}

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3 years ago
What is the specific heat capacity of a 70 g sample of an unknown metal that releases 6700J of heat when it cools from 90 to 25
swat32
Q = mCpΔT —> Solve for Cp —> Cp = q/mΔT

q = -6700 J (negative because it released heat
m = 70 g
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Cp = -6700 J / (70 g)(-65°C) = 1.47 J/g°C
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3 years ago
Write the equilibrium constant expression for this reaction: 2H+(aq)+CO−23(aq) → H2CO3(aq)
MrRissso [65]

Answer:

Equilibrium constant expression for \rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq) \rightleftharpoons H_2CO_3\, (aq):

\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{[\mathrm{H_2CO_3}]}{\left[\mathrm{H^{+}\, (aq)}\right]^{2} \, \left[\mathrm{CO_3}^{2-}\right]}.

Where

  • a_{\mathrm{H_2CO_3}}, a_{\mathrm{H^{+}}}, and a_{\mathrm{CO_3}^{2-}} denote the activities of the three species, and
  • [\mathrm{H_2CO_3}], \left[\mathrm{H^{+}}\right], and \left[\mathrm{CO_3}^{2-}\right] denote the concentrations of the three species.

Explanation:

<h3>Equilibrium Constant Expression</h3>

The equilibrium constant expression of a (reversible) reaction takes the form a fraction.

Multiply the activity of each product of this reaction to get the numerator.\rm H_2CO_3\; (aq) is the only product of this reaction. Besides, its coefficient in the balanced reaction is one. Therefore, the numerator would simply be \left(a_{\mathrm{H_2CO_3\, (aq)}}\right).

Similarly, multiply the activity of each reactant of this reaction to obtain the denominator. Note the coefficient "2" on the product side of this reaction. \rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq) is equivalent to \rm H^{+}\, (aq) + H^{+}\, (aq) + {CO_3}^{2-}\, (aq). The species \rm H^{+}\, (aq) appeared twice among the reactants. Therefore, its activity should also appear twice in the denominator:

\left(a_{\mathrm{H^{+}}}\right)\cdot \left(a_{\mathrm{H^{+}}}\right)\cdot \, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right = \left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right.

That's where the exponent "2" in this equilibrium constant expression came from.

Combine these two parts to obtain the equilibrium constant expression:

\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \quad\begin{matrix}\leftarrow \text{from products} \\[0.5em] \leftarrow \text{from reactants}\end{matrix}.

<h3 /><h3>Equilibrium Constant of Concentration</h3>

In dilute solutions, the equilibrium constant expression can be approximated with the concentrations of the aqueous "(\rm aq)" species. Note that all the three species here are indeed aqueous. Hence, this equilibrium constant expression can be approximated as:

\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{\left[\mathrm{H_2CO_3\, (aq)}\right]}{\left[\mathrm{H^{+}\, (aq)}\right]^2\cdot \left[\mathrm{{CO_3}^{2-}\, (aq)}\right]}.

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