Catalysts are substances that increase the rate of reaction but can be recovered unchanged at the end of the reaction. Catalysts can be classified as either homogeneous (same state as reactants) or heterogeneous (different state than reactants).
Platinum is used to catalyze the hydrogenation of ethylene:
H2(g)+CH2CH2(g)−⟶Pt(s)CH3CH3(g)
Chlorofluorocarbons (CFCs) catalyze the conversion of ozone (O3) to oxygen gas (O2):
2O3(g)−⟶CFC(g)3O2(g)
Magnesium catalyzes the disproportionation of hydrogen peroxide to produce water and oxygen:
2H2O2(aq)−⟶Mg(s)2H2O(l)+O2(g)
What type of catalysts are platinum, CFCs, and magnesium under these conditions?

Answer:

THE HEAT REQUIRED TO CHANGE 2 KG OF ETHANOL FROM 26 °C TO THE BOILING POINT AND TO VAPOR AT THAT TEMPERATURE IS 1965.175 KJ.

Explanation:

Boiling point of ethanol = 78.29 °C = 78.29 + 273 K = 351.29 K

Mass = 2 kg = 2000 g

Final temp. = 26.0 °C = 26 + 273 K= 299 K

Change in temperature = (78.29 - 26) °C = 52.29 °C

1. Heat required to raise the temperature from 26 °C to the boiling point?

Heat = mass * specific heat * change in temperature

Heat = 2000 * 2.44 * 52.29

Heat = 255 175.2 J

2. Heat required to change the liquid to vapor at that temperature?

Heat = mass * enthalphy of vaporization

Heat = 2000 * 855

Heat =1 710000 J

The total heat required to raise the temperature of 2 kg of ethanol from 26 °C to the boiling point and then to change the liquid to vapor at that temperature will be:

Heat = mcT + m Lv

Heat = 255 175.2 J + 1710000 J

Heat = 1965175.2 J

Heat = 1965.175 kJ of heat.

Answer:

an atom with a half-filled subshell - hydrogen

an atom with a completely filled outer shell - argon

an atom with its outer electrons occupying a half-filled subshell and a filled subshell- copper

Explanation:

The outermost shell or the valence shell of the atom is the last shell in the atom. Chemical reactions occur at this outer most shell. The number of electrons on the outermost shell of an atom determines the group to which it belongs in the periodic table as well as its chemical properties.

Hydrogen has a half filled 1s sublevel. Only one electron is present in this sublevel.

Let us consider argon

1s2 2s2 2p6 3s2 3p6

The outermost ns and np levels are completely filled. Thus the outermost shell is completely filled.

In the last case; let us look at the electronic configuration of nitrogen;

1s2 2s2 2p3

The outermost 2p subshell is exactly half filled while the 2s sublevel is fully filled. The outermost shell of nitrogen is made up of 2s2 and 2p3 sublevels.

Answer:

[tex]2.11\ * 10^{-2}[/tex]  is the correct answer to the given question.

Explanation:

Given k=6.40 x 10-3 min-1.

According to the first order reaction .

The concentration of time can be written as

[tex][\ A\ ]\ = \ [\ A_{0}\ ] * e \ ^\ {-kt}[/tex]

Here [tex][\ A\ ]_{0}[/tex] = Initial concentration.

So  [tex][\ A\ ]_{0}= 0.0314 M[/tex]

Putting this value into the above equation.

[tex]0.0314 \ *\ e^{6.40 x 10^{-3} \ * \ 62.0 }[/tex]

=0.211 M

This can be written as

[tex]=\ 2.11 *\ 10^{-2}[/tex]

Answer:

  C

Explanation:

The molecule has 8 carbon atoms joined by 7 C-C bonds.

The first two diagrams show 6 carbon atoms, not 8.

The last two diagrams show line segments representing C-C bonds. Only choice C shows 7 such segments.

The appropriate choice is C.

Answer:

C.

Explanation:

Answer:

simple cations (monatomic cations of elements of only one possible charge)

Explanation:

Simple cations (monatomic cations of elements of only one possible charge)  are named using the unmodified element name and adding the word "ion"

For example, the Na+ is named the sodium ion.

An atom or molecule with a net electric charge as a result of the loss or gain of one or more electrons is known as an ion.

Answer:

(a) [tex]m=2.69m[/tex]

(b) [tex]x_{LiBr}=0.099[/tex]

(c) [tex]\% LiBr=18.9\%[/tex]

Explanation:

Hello,

In this case, given the molality in mol/L, we can compute the required units of concentration assuming a 1-L solution of acetonitrile and lithium bromide that has 1.80 moles of lithium bromide:

(a) For the molality, we first compute the grams of lithium bromide in 1.80 moles by using its molar mass:

[tex]m_{LiBr}=1.80mol*\frac{86.845 g}{1mol}=156.32g[/tex]

Next, we compute the mass of the solution:

[tex]m_{solution}=1L*0.826\frac{g}{mL}*\frac{1000mL}{1L}=826g[/tex]

Then, the mass of the solvent (acetonitrile) in kg:

[tex]m_{solvent}=(826g-156.32g)*\frac{1kg}{1000g}=0.670kg[/tex]

Finally, the molality:

[tex]m=\frac{1.80mol}{0.670kg} \\\\m=2.69m[/tex]

(b) For the mole fraction, we first compute the moles of solvent (acetonitrile):

[tex]n_{solvent}=669.68g*\frac{1mol}{41.05 g} =16.31mol[/tex]

Then, the mole fraction of lithium bromide:

[tex]x_{LiBr}=\frac{1.80mol}{1.80mol+16.31mol}\\ \\x_{LiBr}=0.099[/tex]

(c) Finally, the mass percentage with the previously computed masses:

[tex]\% LiBr=\frac{156.32g}{826g}*100\%\\ \\\% LiBr=18.9\%[/tex]

Regards.

Answer:

No additional particle was produced during the decay.

Explanation:

The equation of decay is given as;

¹⁰₆C  + ⁰₋₁ e → ¹⁰₅B + x

To identify x, we have to calculate its atomic and mass number.

In the reactants side;

Atomic Number = 6 + (-1) = 5

Mass number = 10 + 0 = 10

In the products side;

Atomic Number = 5 + x

Mass Number = 10 + x

Generally, reactant = product

Atomic Number;

5 = 5 + x

x = 5 - 5 = 0

Mass Number;

10 = 10 + x

x = 10 - 10 = 0

This means no additional particle was produced during the decay.

Answer:

AT STOICHIOMETRIC POINT, THE VOLUME OF ACID ADDED IS 0.01 L

AT HALF-WAY POINT, THE VOLUME OF ACID IS 0.0050 L

Explanation:

In solving titration problems, you must remember this formula;

MaVa = MbVb

Since M a= 0.005 M

Mb = 0.010 M

Vb = 5 mL = 5 /1000 = 0.005 L

Va = unknown.

Solving for Va, we have:

Va = MbVb / Ma

Va = 0.010 * 0.005 / 0.005

Va = 0.01 L

So therefore, the volume of acid added at:

1. the stoichiometric point is 0.01 L

2. half-way point of titration is 0.01 /2 = 0.0050 L

For the pH:

Since HCl is a strong acid, it dissociate into {H30}+ ion.

First calculate the number of moles of hydronium ion

number of mole = concentration of hydronium ion {H30}+ * Volume

n = 0.005 * 0.01  = 0.00005 moles

A. At initial point of the titration, the volume of base added is 0 L

{H30]+ = n(H+)/ V = 0.00005 / 0.01 = 0.005 M

pH  = - log {0.005}

pH = 2.3

B. At the final point, since the volumes and concentrations of acid and base are the same, the pH is equal to 7.

n(H+) = n(OH^-)

pH = 7