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Llana [10]
3 years ago
6

Calcium reacts with water. This reaction is represented by the balanced equation below. The aqueous product of this reaction can

be heated to evaporate the water, leaving a white solid, Ca(OH)2(s).
Ca(s) + H2O(l) ==> Ca(OH)2(aq) + H2(g)

Compare the electrical conductivity of the aqueous product in the reaction to the electrical conductivity of the white solid that remains after the water is evaporated from the solution.

Write the chemical name of the base produced in the reaction.

State one change in reaction conditions that will increase the rate of the reaction.
Chemistry
2 answers:
vodomira [7]3 years ago
7 0

Answer:

- Calcium is more conductive than calcium hydroxide.

- Calcium hydroxide.

- Addition of more either calcium or water to the reaction vessel

Explanation:

Hello,

In this case, based on the given reaction and information, calcium's electrical conductivity 0.298 106/(cm*Ω) which is a relatively high value compared to it its base has due to its large insolubility.

In addition, the name of the formed base is calcium hydroxide based on the IUPAC rules.

Moreover, the addition of more either calcium or water to the reaction vessel will generate more calcium hydroxide.

Best regards.

Harrizon [31]3 years ago
6 0
The aqueous product conducts electricity while the solid does not. Ionic solids do not conduct electricity, but ions dissolved in solution do. The chemical name of the base is calcium hydroxide. Increasing the temperature of the system would increase the rate of the reaction.
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A 508-g sample of sodium bicarbonate (NaHCO3) contains how many moles of sodium bicarbonate (NaHCO3)?
Blizzard [7]

molar mass = (22.99) + (1.01) + (12.01) + 3(16.00)

molar mass = 84.01 g/mol

//

(508g)(1 mol/84.01 g) = 6.0

There are 6.0 moles of sodium bicarbonate

3 0
3 years ago
La formula quimica de la molecula de agua H20 , si H = 1 gramos y O = 16 gramos . Cual es su composicion porcentual ? 88,88 % de
vesna_86 [32]

Answer:

88,88 % de O y 11,11 % de H

Explanation:

La composición porcentual se define como la masa que hay de cada mol de átomo en 100g. Las moles de agua en 100g son:

<em>Masa molar agua:</em>

2H = 2*1g/mol = 2g/mol

1O = 1*16g/mol = 16g/mol

Masa molar = 2 + 16 = 18g/mol

100g H2O * (1mol / 18g) = 5.556 moles H2O.

Moles de hidrógeno:

5.556 moles H2O * (2mol H / 1mol H2O) = 11.11 moles H

Moles Oxígeno = Moles H2O = 5.556 moles

La masa de hidrógeno es:

11.11mol * (1g/mol) 11.11g H

La masa de oxígeno es:

5.556 mol * (16g / 1mol) = 88.89g O

Así, el porcentaje de O es 88.89% y el de H es 11.11%. La opción correcta es:

<h3>88,88 % de O y 11,11 % de H</h3>
7 0
3 years ago
A voltaic cell consists of a Zn&gt;Zn2+ half-cell and a Ni&gt;Ni2+ half-cell at 25 °C. The initial concentrations of Ni2+ and Zn
nlexa [21]

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of Ni^{2+} has fallen to 0.500 M is, 0.52 V

(c) The concentrations of Ni^{2+} and Zn^{2+} when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M

Explanation :

The values of standard reduction electrode potential of the cell are:

E^0_{[Ni^{2+}/Ni]}=-0.23V

E^0_{[Zn^{2+}/Zn]}=-0.76V

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : Zn\rightarrow Zn^{2+}+2e^-     E^0_{[Zn^{2+}/Zn]}=-0.76V

Reaction at cathode (reduction) : Ni^{2+}+2e^-\rightarrow Ni     E^0_{[Ni^{2+}/Ni]}=-0.23V

The balanced cell reaction will be,  

Zn(s)+Ni^{2+}(aq)\rightarrow Zn^{2+}(aq)+Ni(s)

First we have to calculate the standard electrode potential of the cell.

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

E^o=E^o_{[Ni^{2+}/Ni]}-E^o_{[Zn^{2+}/Zn]}

E^o=(-0.23V)-(-0.76V)=0.53V

(a) Now we have to calculate the cell potential.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = emf of the cell = ?

Now put all the given values in the above equation, we get:

E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}

E_{cell}=0.49V

(b) Now we have to calculate the cell potential when the concentration of Ni^{2+} has fallen to 0.500 M.

New concentration of Ni^{2+} = 1.50 - x = 0.500

x = 1 M

New concentration of Zn^{2+} = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}

Now put all the given values in the above equation, we get:

E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}

E_{cell}=0.52V

(c) Now we have to calculate the concentrations of Ni^{2+} and Zn^{2+} when the cell potential falls to 0.45 V.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}

Now put all the given values in the above equation, we get:

0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}

x=1.49M

The concentration of Ni^{2+} = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of Zn^{2+} = 0.100 + x = 0.100 + 1.49 = 1.59 M

5 0
3 years ago
Which element is most common in sea water?
SVEN [57.7K]

Answer:

The answer is (B.) Chlorine

Explanation:

Its because i took the test

4 0
3 years ago
Which of the following situations indicates an alpha emission?
Angelina_Jolie [31]
Your answer would be C. Alpha decay involves the ejection of 2 protons and 2 neutrons from the nucleus for a total of 4 amu lost. This form of decay is most common in heavy elements.
4 0
3 years ago
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