Answer:
The percentage yield is 50%
A Carbon-10 nucleus has 6 protons and 4 neutrons. Through radioactive beta decay, it turns into a Boron-10 nucleus, with 5 protons and 5 neutrons, plus another particle. What kind of additional particle, if any, is produced during this decay
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.
Which of the following is named using the unmodified element name and adding the word "ion"? Select the correct answer below:
a. simple cations (monatomic cations of elements of only one possible charge)
b. simple anions (monatomic anions of elements of only one possible charge)
c. simple protons
d. simple neutrons
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.
Give two examples (i.e. list 2 elements that are examples) of: a. an atom with a half-filled subshell b. an atom with a completely filled outer shell c. an atom with its outer electrons occupying a half-filled subshell and a filled subshell
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.
Ethanol, , boils at 78.29 °C. How much energy, in joules, is required to raise the temperature of 2.00 kg of ethanol from 26.0 °C to the boiling point and then to change the liquid to vapor at that temperature? (The specific heat capacity of liquid ethanol is 2.44 J/g ∙ K, and its enthalpy of vaporization is 855 J/g.)
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.
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:
- Platinum acts as a heterogeneous catalyst in the hydrogenation of ethylene.
- CFCs act as homogeneous catalysts in the conversion of ozone to oxygen gas.
- Magnesium acts as a heterogeneous catalyst in the disproportionantion of hydrogen peroxide.
Explanation:
Hello,
For the given reactions, considering the definition of homogeneous and heterogeneous catalyst, we can identify that is each catalyst behave as follows:
- Platinum acts as a heterogeneous catalyst in the hydrogenation of ethylene as all the reactants are gaseous but it remains solid.
- CFCs act as homogeneous catalysts in the conversion of ozone to oxygen gas as it remains gaseous as well as both ozone and oxygen.
- Magnesium acts as a heterogeneous catalyst in the disproportionantion of hydrogen peroxide as it is solid whereas the other species are aqueous, liquid and gaseous
Best regards.
At 25 oC, the rate constant for the first-order decomposition of a pesticide solution is 6.40 x 10-3 min-1. If the starting concentration of pesticide is 0.0314 M, what concentration will remain after 62.0 minutes at 25 oC? 3.12 x 10-2 M 47.4 M 2.11 x 10-2 M 4.67 x 10-2 M 8.72 M
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]
Acetonitrile, CH3CN, is a polar organic solvent that dissolves many solutes, including many salts. The density of a 1.80 M acetonitrile solution of LiBr is 0.826 g/mL. Calculate the concentration of the solution in units of (a) molality; (b) mole fraction of LiBr; (c) mass percentage of CH3CN.
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.
(a) Titration curve for the titration of 5.00 mL 0.010 M NaOH(aq) with 0.005 M HCl(aq), indicating the pH of the initial and final solutions and the pH at the stoichiometric point.
What volume of HCl has been added at
(b) the stoichiometric point
(c) the halfway point of the titration?
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
Which diagram represents this molecule?
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:
