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Bingel [31]
4 years ago
5

Is adding hot cocoa to hot water a chemical or physical change?

Chemistry
1 answer:
saul85 [17]4 years ago
6 0
Adding hot cocoa to hot water would be a chemical change
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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]}.

8 0
3 years ago
Balance the single replacement chemical reaction.
OLEGan [10]

Answer:

Option c. 2Kl + Cl2 —> 2KCl + l2

Explanation:

A balanced equation must have the same number of atoms on the reactants side and the products side. This do the fact of the postulation by the mass conservation principle that says: Atoms can neither be broken nor destroyed but can be arranged. The atoms on the left side for each element are equal. Thus, C presents a balanced equation.

3 0
3 years ago
The early ancestors of the horse (shown in the picture) were adapted to life in tropical forests. Gradually the forests disappea
Gre4nikov [31]

So here you go are taller with long legs and are capable of great speed to escape predators

8 0
3 years ago
Read 2 more answers
Find the mass of 3.27 x 10^23 molecules of H2SO4. Use 3 significant digits<br> and put the units.
marta [7]

Answer:

Approximately 53.3\; \rm g.

Explanation:

Lookup Avogadro's Number: N_{\rm A} = 6.02\times 10^{23}\; \rm mol^{-1} (three significant figures.)

Lookup the relative atomic mass of \rm H, \rm S, and \rm O on a modern periodic table:

  • \rm H: 1.008.
  • \rm S: 32.06.
  • \rm O: 15.999.

(For example, the relative atomic mass of \rm H is 1.008 means that the mass of one mole of \rm H\! atoms would be approximately 1.008\! grams on average.)

The question counted the number of \rm H_2SO_4 molecules without using any unit. Avogadro's Number N_{\rm A} helps convert the unit of that count to moles.

Each mole of \rm H_2SO_4 molecules includes exactly (1\; {\rm mol} \times N_\text{A}) \approx 6.02\times 10^{23} of these \rm H_2SO_4 \! molecules.

3.27 \times 10^{23} \rm H_2SO_4 molecules would correspond to \displaystyle n = \frac{N}{N_{\rm A}} \approx \frac{3.27 \times 10^{23}}{6.02 \times 10^{23}\; \rm mol^{-1}} \approx 0.541389\; \rm mol of such molecules.

(Keep more significant figures than required during intermediary steps.)

The formula mass of \rm H_2SO_4 gives the mass of each mole of \rm H_2SO_4\! molecules. The value of the formula mass could be calculated using the relative atomic mass of each element:

\begin{aligned}& M({\rm H_2SO_4}) \\ &= (2 \times 1.008 + 32.06 + 4 \times 15.999)\; \rm g \cdot mol^{-1} \\ &= 98.702\; \rm g \cdot mol^{-1}\end{aligned}.

Calculate the mass of approximately 0.541389\; \rm mol of \rm H_2SO_4:

\begin{aligned}m &= n \cdot M \\ &\approx 0.541389\; \rm mol \times 98.702\; \rm g \cdot mol^{-1}\\ &\approx 53.3\; \rm g\end{aligned}.

(Rounded to three significant figures.)

6 0
3 years ago
Help ASAP pleaseeee thank you
Artyom0805 [142]
The correct answer is B. balance
8 0
3 years ago
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