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Masteriza [31]
3 years ago
14

Atomic #

Chemistry
1 answer:
Anon25 [30]3 years ago
7 0

<em><u>In the modern periodic table, the elements are listed in order of increasing atomic number. The atomic number is the number of protons in the nucleus of an atom. The number of protons define the identity of an element (i.e., an element with 6 protons is a carbon atom, no matter how many neutrons may be present).</u></em>

<em><u /></em>

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To form a salt compound, the acid contributes the __ ion, and the base contributes the ___ ion
cupoosta [38]
To form a salt compound, the acid contribute a NON METAL ION  and the base contributes a METAL ION.
In chemistry, acid and base react together to form salt and water only. For the salt formed, the positive metal ion comes from the base while the negative non metal ion comes from the acid.
8 0
3 years ago
For the reaction N2O4(g) ⇋ 2NO2(g), Kc = 0.25 at 98°C. At a point during the reaction, the concentration of N2O4 = 0.50 M and th
Scilla [17]

Answer:

Q = 0.50

No

Left

Explanation:

At a generic reversible equation

aA + bB ⇄ cC + dD

The reaction coefficient (Q) is the ratio of the substances concentrations:

Q = \frac{[C]^c*[D]^d}{[A]^a*[B]^b}

Solids and liquid water are not considered in this calculus.

When the reaction achieves equilibrium (concentrations are constant), the Q value is named as Kc, which is the equilibrium constant of the reaction. If Q > Kc, it indicates that the concentration of the products is higher, so, the reaction must progress to the left and form more reactants; if Q < Kc, than the concentrations of the reactants, are higher, so, the reaction progress to the right.

In this case:

Q = \frac{[NO_2]^2}{[N_2O_4]}

Q = \frac{0.50^2}{0.50}

Q = 0.50

So, Q > Kc, the reaction is not at equilibrium and it progresses to the left.

6 0
3 years ago
Is a burning log an exothermic or endothermic event if the log is the system?
denpristay [2]
It would be endothermic because the log is in the system.
6 0
2 years ago
In a percentage composition investigation a compound was decomposed into its elements: 20.0 g of calcium, 6.0 g of carbon, and 2
inysia [295]

The percentage composition of this compound : 40%Ca, 12%C and 48%O

<h3>Further explanation</h3>

Given

20.0 g of calcium,

6.0 g of carbon,

and 24.0 g of oxygen.

Required

The percentage composition

Solution

Total mass of compound :

=mass calcium + mass carbon + mass oxygen

=20 g + 6 g + 24 g

=50 g

Percentage composition :

  • Ca-calcium

\tt \dfrac{20}{50}\times 100\%=40\%

  • C-carbon

\tt \dfrac{6}{50}\times 100\%=12\%

  • O-oxygen

\tt \dfrac{24}{50}\times 100\%=48\%

3 0
2 years ago
Use the following half-reactions to construct a voltaic cell:
velikii [3]

<u>Answer:</u> The correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

<u>Explanation:</u>

We are given:

Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V

Reduction half reaction: Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V       ( × 3)

Net equation:  3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.80-(-0.73)=1.53V

Hence, the correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

4 0
3 years ago
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