Use the words in the list to answer each question

Which of the following is true regarding a voltaic (or galvanic) cell? a) It can only be used with hydrogen. b) It produces elec

N76 [4]

Answer:

b) It produces electrical current spontaneously.

Explanation:

Cells capable of converting chemical energy to electrical energy and vice versa are termed Electrochemical cells. There are two types of electrochemical cells viz: Galvanic or Voltaic cells and Electrolytic cells. Voltaic cell is an elctrochemical cell capable of generating electrical energy from the chemical reaction occuring in it.

The voltaic cell uses spontaneous reduction-oxidation (redox) reactions to generate ions in a half cell that causes electric currents to flow. An half cell is a part of the galvanic cell where either oxidation or reduction reaction is taking place. Hence, the spontaneous production of electric currents is true about Voltaic/Galvanic cells.

3 0

3 years ago

Analysis of a gaseous chlorofluorocarbon, CClxFy, shows that it contains 11.79% C and 69.57% Cl. In another experiment, you find

uranmaximum [27]

Answer:

The molecular formula = $C_2Cl_{4}F_2$

Explanation:

$Moles =\frac {Given\ mass}{Molar\ mass}$

% of C = 11.79

Molar mass of C = 12.0107 g/mol

% moles of C = 11.79 / 12.0107 = 0.9816

% of Cl = 69.57

Molar mass of Cl = 35.453 g/mol

% moles of Cl = 69.57 / 35.453 = 1.9623

Given that the gaseous chlorofluorocarbon only contains chlorine, flourine and carbon. So,

% of F = 100% - % of C - % of C = 100 - 11.79 - 69.57 = 18.64

Molar mass of F = 18.998 g/mol

% moles of F = 18.64 / 18.998 = 0.9812

Taking the simplest ratio for C, Cl and F as:

0.9816 : 1.9623 : 0.9812

= 1 : 2 : 1

The empirical formula is = $CCl_2F$

Also, Given that:

Pressure = 21.3 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 21.3 / 760 = 0.02803 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25 + 273.15) K = 298.15 K

Volume = 458 mL = 0.458 L (1 mL = 0.001 L)

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.02803 atm × 0.458 L = n × 0.0821 L.atm/K.mol × 298.15 K

⇒n = 0.00052445 moles

Given that :

Amount = 0.107 g

Molar mass = ?

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.00052445= \frac{0.107\ g}{Molar\ mass}$

$Molar\ mass= 204.0233\ g/mol$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12.0107 + 2×35.453 + 1×18.998 = 101.9147 g/mol

Molar mass = 204.0233 g/mol

So,

Molecular mass = n × Empirical mass

204.0233 = n × 101.9147

⇒ n = 2

The molecular formula = $C_2Cl_{4}F_2$

6 0

4 years ago

Please help make sure its correct thanks

Wewaii [24]

The empirical formula of metal iodide : CoI₃(Cobalt(III) Iodide)

<h3>Further explanation</h3>

13.02 g sample of Cobalt , then mol Co(MW=58.933 g/mol) :

$\tt mol=\dfrac{mass}{Ar}=\\\\mol=\dfrac{13.02~g}{58.933}\\\\mol=0.221$

Mass of metal iodide formed : 97.12 g, so mass of Iodine :

$\tt =mass~metal~iodide-mass~Cobalt\\\\=97.12-13.02\\\\=84.1~g$

Then mol iodine (MW=126.9045 g/mol) :

$\tt \dfrac{84.1}{126.9045}=0.663$

mol ratio of Cobalt and Iodine in the compound :

$\tt 0.221\div 0.663=1\div 3$

5 0

3 years ago

A. Convert the following Fahrenheit temperatures to Celsius (you may use a calculator):

Bezzdna [24]

37C
20C
-6.67C
50F
86F
-40F

4 0

3 years ago

What does it mean to optimize a solution ? How are solutions optimize

Romashka [77]

Question

What does it mean to optimize a solution?

To find out best possible solution for a given problem within the given constraint is generally termed as optimization

How are solution optimize ?

To solve an optimization problem, begin by drawing a picture and introducing variables. Find an equation relating the variables. Find a function of one variable to describe the quantity that is to be minimized or maximized. Look for critical points to locate local extrema.

6 0

3 years ago