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Romashka-Z-Leto [24]
3 years ago
10

Of the reactions below, only ________ is not spontaneous. A. Zn (s) + 2HI (aq) → ZnI2(aq) + H2 (g) B. 2Ag (s) + 2HNO3 (aq) → 2Ag

NO3 (aq) + H2 (g) C. Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2(g) D. 2Ni (s) + H2SO4 (aq) → Ni2SO4 (aq) + H2 (g) E. 2Al (s) + 6HBr (aq) → 2AlBr3 (aq) + 3H2 (g)
Chemistry
1 answer:
Fynjy0 [20]3 years ago
3 0

Answer:

2Ag (s) + 2HNO3 (aq) → 2AgNO3 (aq) + H2(g)

Explanation:

Usually, metals displace hydrogen from dilute acids to form the corresponding salt of the metal and hydrogen. This in turn depends on the position of the metal in the electrochemical series. Metals below hydrogen in the electrochemical series cannot spontaneously displace it from dilute acids. Silver is below hydrogen in the electrochemical series, hence it cannot spontaneously displace hydrogen from dilute nitric acid as shown in the answer.

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When the particles of two objects have the same average kinetic energy, what do we know about those objects?
Dmitriy789 [7]
That they both will be the same average kinetic energy
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3 years ago
Pyridine (c5h5n) is a particularly foul-smelling substance used in manufacturing pesticides and plastic resins. calculate the ph
zubka84 [21]

Answer is: ph value of pyridine solution is 9.1.

Chemical reaction: C₅H₅N + H₂O → C₅H₅NH⁺ + OH⁻.<span>
c(pyridine - C</span>₅H₅N) = 0.115M.<span>
Kb(C</span>₅H₅N) = 1.4·10⁻⁹.
[C₅H₅NH⁺] = [OH⁻] = x; equilibrium concentration.<span>
[</span>C₅H₅N] = 0.115 M - x.

Kb = [C₅H₅NH⁺] · [OH⁻] / [C₅H₅N].

1.4·10⁻⁹ = x² / (0.115 M -x)

Solve quadratic equation: x = [OH⁻] = 0.0000127 M.<span>
pOH = -log(0.0000127 M) = 4.9</span>

<span>pH = 14 - 4.9 = 9.1.</span>

7 0
3 years ago
In order to gain one pound of body weight, the average person must consume 3500 more calories
Fed [463]

Answer:

<em>17500 calories</em> of chocolate bars are needed to eat to gain 5 pounds.

Explanation:

We can use ratios to calculate the answer using the information given in the question.

1 pound : 3500 grams

5 pounds : x grams

As it is given that the individual is burning no calories, we do not have to factor in any additional numbers.

<u><em>Method</em><em> </em><em>A</em><em>:</em></u>

To go from 1 in the first ratio to 5 in the second ratio, they multipled 1 by 5. Hence, to go from 3500 in the first ratio to x in the second ratio, we must multiply by 5.

x = 3500 × 5

x = 17500

<em><u>Method B</u>:</em>

To solve for the answer x, we can convert the ratios into fractions.

1 / 5 = 3500 / x

3500 / x = 1 / 5

To make x the subject, multiply the denominator of the left fraction with the numerator of the right fraction and place it on the left side. Then multiply the numerator of the left fraction with the denominator of the right fraction and place it on the right side.

x = 5 × 3500

x = 17500

4 0
3 years ago
A characteristic of an organism; can be genetic or acquired
Fittoniya [83]
Genetic: an organism uses energy for cellular respiration, and has all the genes necessary to make the proteins to accomplish it.
8 0
3 years ago
If 45.00 g of precipitate is formed from the reaction of 0.100 mol/L
Tom [10]

Answer:

Approximately 2.53\; \rm L (rounded to three significant figures) assuming that {\rm HCl}\, (aq) is in excess.

Explanation:

When {\rm HCl} \, (aq) and {\rm AgNO_3}\, (aq) precipitate, {\rm AgCl} \, (s) (the said precipitate) and \rm HNO_3\, (aq) are produced:

{\rm HCl}\, (aq) + {\rm AgNO_3}\, (aq) \to {\rm AgCl}\, (s) + {\rm HNO_3}\, (aq) (verify that this equation is indeed balanced.)

Look up the relative atomic mass of \rm Ag and \rm Cl on a modern periodic table:

  • \rm Ag: 107.868.
  • \rm Cl: 35.45.

Calculate the formula mass of the precipitate, \rm AgCl:

\begin{aligned}& M({\rm AgCl})\\ &= (107.868 + 35.45)\; \rm g \cdot mol^{-1} \\\ &\approx 143.318 \; \rm  g\cdot mol^{-1}\end{aligned}.

Calculate the number of moles of \rm AgCl formula units in 45.00\; \rm g of this compound:

\begin{aligned}n({\rm AgCl}) &= \frac{m({\rm AgCl})}{M({\rm AgCl})} \\ &\approx \frac{45.00\; \rm g}{143.318\; \rm g \cdot mol^{-1}}\approx 0.313987\; \rm mol \end{aligned}.

Notice that in the balanced equation for this reaction, the coefficients of {\rm AgNO_3} \, (aq) and {\rm AgCl}\, (s) are both one.

In other words, if {\rm HCl}\, (aq) (the other reactant) is in excess, it would take exactly 1\; \rm mol of {\rm AgNO_3} \, (aq)\! formula units to produce 1\; \rm mol \! of {\rm AgCl}\, (s)\! formula units.

Hence, it would take 0.313987\; \rm mol of {\rm AgNO_3} \, (aq)\! formula units to produce 0.313987\; \rm mol\! of {\rm AgCl}\, (s)\! formula units.

Calculate the volume of the {\rm AgNO_3} \, (aq)\! solution given that the concentration of the solution is 0.124\; \rm mol \cdot L^{-1}:

\begin{aligned}V({\rm AgNO_3}) &= \frac{n({\rm AgNO_3})}{c({\rm AgNO_3})} \\ &\approx \frac{0.313987\; \rm mol}{0.124\; \rm mol \cdot L^{-1}}\approx 2.53\; \rm L\end{aligned}.

(The answer was rounded to three significant figures so as to match the number of significant figures in the concentration of {\rm AgNO_3} \, (aq)\!.)

In other words, approximately 2.53\; \rm L of that {\rm AgNO_3} \, (aq)\! solution would be required.

4 0
3 years ago
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