Answer:
A. 15.6 days
Explanation:
We have a general formula used to solve for questions dealing with half life.
N(t) = No (1/2)^t/t½
Where
N(t) = Quantity of sample left after t days
No = Initial amount of sample
t½ = Half life of sample
t = Duration or time required for sample to decay
In the above question, we are asked to determine how long(time) that the sample has been in storage.
The formula for time (t) has been derived as:
t =[ t½ × In(Nt/No)] ÷ - In 2
Where
No = 237 grams
N(t) = 29.625 grams
t½ = 5.2 days
t = ????? Unknown
t = [ t½ × In(Nt/No)] ÷ - In 2
t = [ 5.2 × In(29.625/237)] ÷ - In 2
t = [5.2 × -2.0794415417] ÷ - In 2
t = -10.813096017 ÷ - ln 2
t = 15.6 days
Therefore, the sample has been in storage for 15.6 days.
what is the volume in cubic centimeters of a tablet weighing 210 mg
Answer:
1mg=0.001
210
210×0.001÷1
=0.21
A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentration of I2 was 6.29×10^−4M. Calculate the Kc at 1000K for:
H2(g)+I2(g)⇌2HI(g)
Answer:
The answer is "29.081"
Explanation:
when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.
[tex]\text{calculating the initial HI}= \frac{mol}{V}[/tex]
[tex]=\frac{9.3 \times 10 ^ -3}{2}[/tex]
[tex]=0.00465 \ Mol[/tex]
[tex]\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }[/tex]
Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:
[tex]HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\[/tex]
It is defined that:
[tex]I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\[/tex]
[tex]HI \ eq= 0.00465 - 2x \\[/tex]
[tex]=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M[/tex]
Now, we calculate the position:
For the reaction [tex]H 2(g) + I 2(g)\rightleftharpoons 2HI(g)[/tex], you can calculate the value of Kc at 1000 K.
data expression for Kc
[tex]2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}[/tex]
[tex]= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386[/tex]
calculating the reverse reaction
[tex]H_2(g) + I_2(g)\rightleftharpoons 2HI(g)[/tex]
[tex]Kc = \frac{1}{Kc} \\\\[/tex]
[tex]= \frac{1}{0.034386}\\ \\= 29.081\\[/tex]
The Kc of the reaction is 40.
Molarity of the HI = 9.30×10^−3mol/ 2.00 L = 4.65 × 10^-3 M
Let the concentrations of I2 and H2 be x, but we are told in the question that 6.29×10^−4M was present at equilibrium.
The molarity of HI at equilibrium now becomes; 4.65 × 10^-3 M - 6.29×10^−4M
= 4 × 10^-3 M
But;
Kc = [HI]^2/[H2] [I2]
Kc = ( 4 × 10^-3)^2/(6.29×10^−4)^2
Kc = 40
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If 5.0 mL of a Sports Drink with an absorbance reading of 0.34 was diluted with water to 10.0 mL and was read by the colorimeter, what is the expected absorbance of the solution
Answer:
the expected absorbance of the solution = 0.17
Explanation:
From the information given:
Using Beer's Lambert Law, we have
A = ∈CL
where;
A = Absorbance
∈ = extinction coefficient
C = concentration
L = cell length
Since Absorbance is associated with concentration.
Assuming the measurement were carried out in the same solution; Then ∈ and L will be constant and A ∝ C ----- (1)
Let consider the concentration to be C (mol/L)
5.0 mL of a Sports Drink = 5.0 mL × C (mol)/1000 mL
= 5C/1000 mL
was diluted with water to 10.0 mL
So, when diluted with water to 10.0 mL; we have:
The new concentration to be : [tex]\dfrac{(5 C \times 1000) \ mol }{(1000 \times 10 \times 1000)\ mL}[/tex]
Since :1000mL = 1 L
The new concentration = [tex]\dfrac{C \ mol }{2 \ L}[/tex]
As stated that the initial absorbance reading [tex]A_1[/tex] = 0.34
The expected absorbance reading will be [tex]A_2[/tex] = ???
From (1)
A ∝ C
∴
[tex]\dfrac{A_2}{A_1}=\dfrac{C_2}{C}[/tex]
[tex]A_2 = \dfrac{A_1}{C}[/tex]
[tex]A_2 = \dfrac{0.34}{2}[/tex]
[tex]A_2 = 0.17[/tex]
Thus ; the expected absorbance of the solution = 0.17
What is the activation energy for a reaction which proceeds 50 times as fast at 400 K as it does at 300 K? Answer in units of J/mol rxn."
Answer:
Activation energy for the reaction is 39029J/mol
Explanation:
Arrhenius equation is an useful equation that relates rate of reaction at two different temperatures as follows:
[tex]ln\frac{K_2}{K_1} = \frac{-Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1} )[/tex]
Where K₁ and K₂ are rate of reaction, Ea is activation energy and R is gas constant (8.314J/molK
If the reaction at 400K is 50 times more faster than at 300K:
K₂/K₁ = 50 where T₂ = 400K and T₁ = 300K:
[tex]ln50 = \frac{-Ea}{8.314J/molK} (\frac{1}{400K} -\frac{1}{300K} )[/tex]
[tex]ln 50 = 1x10^{-4}Ea[/tex]
Ea = 39029 J/mol
Activation energy for the reaction is 39029J/mol
The activation energy for this chemical reaction is equal to 39,029.24 J/mol.
Given the following data:
Rate of reaction = 50Final temperature = 400 KInitial temperature = 300 KIdeal gas constant, R = 8.314 J/molK
To determine the activation energy for this chemical reaction, we would use the Arrhenius' equation:
Mathematically, Arrhenius' equation is given by the formula:
[tex]ln\frac{K_2}{K_1} = \frac{-E_a}{R} (\frac{1}{T_2} - \frac{1}{T_1})[/tex]
Where:
K is the rate of chemical reaction.[tex]E_a[/tex] is the activation energy.R is the ideal gas constant.T is the temperature.Substituting the given parameters into the formula, we have;
[tex]ln50 = \frac{-E_a}{8.314} (\frac{1}{400} - \frac{1}{300})\\\\3.9120 = \frac{-E_a}{8.314} (\frac{-1}{1200})\\\\3.9120 = \frac{E_a}{9976.8} \\\\E_a = 9976.8 \times 3.9120\\\\E_a = 39,029.24 \;J/mol[/tex]
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Which statement best describes the types and locations of particles that make up the atom? A. The neutral-charged neutrons and positive-charged protons are found within the nucleus, and the negative-charged electrons orbit outside the nucleus of the atom. B. The neutral-charged neutrons, positive-charged protons, and negative-charged electrons are all found within the nucleus of the atom. C. The negative-charged electrons and positive-charged protons are found within the nucleus, and the neutral-charged neutrons orbit outside the nucleus of the atom. D. The negative-charged neutrons and positive-charged protons are found within the nucleus, and the neutral-charged electrons orbit outside the nucleus of the atom.
Answer:
A. The neutral-charged neutrons and positive-charged protons are found within the nucleus, and the negative-charged electrons orbit outside the nucleus of the atom.
Explanation:
Before examining the options, it's important to know that the atom is made up of three sub atomic particles which are; Protons, Neutrons and Electrons.
The protons are positively charged and are found in the nucleus of an atom. The Neutron A=are neutral in terms of charge and are also found in the nucleus of an atom. The electrons on the other hand are negatively charged and are found outside the nucleus if an atom, more specifically on the orbitals.
The option that best describes this is; option A.
Which is an application of the trimethylsilyl (TMS) group in organic synthesis?
Answer:
Protecting group
Explanation:
TMS groups can be used as protecting and leaving groups for the synthesis of siloxane-based molecules. It is also used as protecting group for alcohols.
A protecting group is a temporary group added during organic synthesis to prevent a portion of molecule from reacting.
Which of the following statements is not true regarding acids? (2 points) Acids can be corrosive, causing damage to skin or clothing. Many fruits contain weak acids, giving them a sour taste. Acids react with bases in neutralization reactions to produce water. The hydroxide ion concentration is greater than the hydronium ion concentration.
Answer:
Acids do not react with bases in neutralization reactions to produce water.
Explanation:
Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. After a certain amount of time, it was found that 6.23 mL of O2 had passed through the membrane, but only 3.85 mL of of the unknown gas had passed through. What is the molar mass of the unknown gas?
Answer:
identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. After a certain amount of time, it was found that 6.23 mL of O2 had passed through the membrane, but only 3.85 mL of of the unknown gas had passed through. What is the molar mass of the unknown gas
identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. After a certain amount of time, it was found that 6.23 mL of O2 had passed through the membrane, but only 3.85 mL of of the unknown gas had passed through. What is the molar mass of the unknown gas
If 20.6 grams of ice at zero degrees Celsius completely change into liquid water at zero degrees Celsius, the enthalpy of phase change will be positive. TRUE FALSE
Answer:
TRUE.
Explanation:
Hello,
In this case, since the fusion enthalpy of ice is +333.9 J/g and the fusion entropy is defined as:
[tex]\Delta _{fus}S=\frac{m*\Delta _{fus}H}{T_{fus}}[/tex]
We can compute it considering the temperature (0 °C) in kelvins:
[tex]\Delta _{fus}S=\frac{20.6g*333.9J/g}{(0+273)K}\\\\\Delta _{fus}S=25.2J/K[/tex]
Therefore answer is TRUE.
Best regards.
The 3d energy level in hydrogen has how many distinct states with different values of the quantum number m
The question is incomplete, the complete question is;
The 3d energy level in hydrogen has how many distinct states with different values of the quantum number m?
A. 3
B. 5
C. 6
D. 2
E. 4
Answer:
B. 5
Explanation:
The magnetic quantum number is used in describing the actual orientation of orbitals in space. The name 'magnetic quantum number' was coined because it describes the effect of different orientations of orbitals which was initially observed in the presence of an external magnetic field.
The d orbital can exhibit five orientations corresponding to five values of the magnetic quantum number, these are; -2,-1,0,1,2 hence the answer above.
What is the molar mass of
CH4?
(C = 12.011 amu, H = 1.008 amu)
Answer:
16.0 g/mol (3 s.f.)
Explanation:
Molar mass is the mass of a mole of substance.
1 mole of CH₄ has 1 C atom and 4 H atoms.
Since the molar mass is numerically equal to the molecular mass of a compound, let's find the molecular mass of CH₄ first.
Molecular mass= sum of all the atomic mass in a molecule
Molecular mass of CH₄
= 12.011 amu +4(1.008 amu)
= 16.043 amu
Thus the molar mass of CH₄ is 16.043g/mol, or 16.0g/mol to 3 significant figures.
There are always attractive forces between a collection of atoms or molecules which may not be negligible as suggested by the kinetic molecular theory. The strength of these forces depend upon the nature of the atom or molecule. If a mole of gas is at STP and there are very strong attractive intermolecular forces between the gas particles, the volume will be _________.
Answer:
Negligible
Explanation:
According to the kinetic theory of gases, the degree of intermolecular interaction between gases is minimal and gas molecules tend to spread out and fill up the volume of the container.
If the attraction between gas molecules increases, then the volume of the gas decreases accordingly. This is because, gas molecules become highly attracted to each other.
This intermolecular attractive force may be so strong, such that the actual volume of the gas become negligible compared to the volume of the container.
Three 15.0-mL acid samples—0.10 M HA, 0.10 M HB, and 0.10 M H2C¬are all titrated with 0.100 M NaOH. If HA is a weak acid, HB is a strong acid, and H2C is a diprotic acid, which statement is true of all three titrations?
Answer:
All three titrations require the same volume of NaOH to reach the first equivalence point.
Explanation:
Statements are:
All three titrations have the same final pH.
All three titrations require the same volume of NaOH to reach the first equivalence point.
All three titrations have the same pH at the first equivalence point.
All three titrations have the same initial pH.
The pH of the titration depends of the nature of the acid: If the acid is a strong acid, pH at the equivalence of the titration is 7. For a weak acid equivalence point depends of the nature of the conjugate base and initial pH of the weak acid. For a diprotic acid also depends of the nature of the acid.
Thus:
All three titrations have the same initial pH, All three titrations have the same pH at the first equivalence point. and All three titrations have the same final pH. are the three FALSE.
As the concentrations of the acids is 0.10M and are titrated with 0.100M NaOH, The volume to reach the first equivalence point is the same for all the three acids.
Thus:
All three titrations require the same volume of NaOH to reach the first equivalence point.In this reaction: Mg (s) + I₂ (s) → MgI₂ (s), if 10.0 g of Mg reacts with 60.0 g of I₂, and 57.84 g of MgI₂ form, what is the percent yield?
Answer:
88.1% (to 3 s.f.)
Explanation:
Please see attached picture for full solution.
f 24.7 g of NO and 13.8 g of O₂ are used to form NO₂, how many moles of excess reactant will be left over? 2 NO (g) + O₂ (g) → 2 NO₂ (g)
Answer:
0.02 moles of O₂ will be leftover.
Explanation:
The reaction is:
2NO(g) + O₂(g) → 2NO₂(g) (1)
We have the mass of NO and O₂, so we need to find the number of moles:
[tex] n_{NO} = \frac{m}{M} = \frac{24.7 g}{30.01 g/mol} = 0.82 moles [/tex]
[tex] n_{O_{2}} = \frac{m}{M} = \frac{13.8 g}{31.99 g/mol} = 0.43 moles [/tex]
From equation (1) we have that 2 moles of NO reacts with 1 mol of O₂ to produce 2 moles of NO₂, so the excess reactant is:
[tex] n_{NO} = \frac{2}{1}*0.43 moles = 0.86 moles [/tex]
[tex]n_{O_{2}} = \frac{1}{2}*0.82 moles = 0.41 moles[/tex]
Hence, from above we can see that the excess reactant is O₂ since 0.41 moles react with 0.86 moles of NO and we have 0.43 moles in total for O₂.
The number of moles of excess reactant is:
[tex]n_{T} = 0.43 moles - 0.41 moles = 0.02 moles[/tex]
Therefore, 0.02 moles of O₂ will be leftover.
I hope it helps you!
The number of moles of excess reactant that would be left over is 0.0197 mole
From the question,
We are to determine the number of moles of excess reactant that would be left over.
The given balanced chemical equation for the reaction is
2NO(g) + O₂(g) → 2NO₂ (g)
This means,
2 moles of NO is needed to completely react with 1 mole of O₂
Now, we will determine the number of moles of each reactant present
For NOMass = 24.7 g
Molar mass = 30.01 g/mol
From the formula
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]
∴ Number of moles of NO present = [tex]\frac{24.7}{30.01}[/tex]
Number of moles of NO present = 0.823059 mole
For O₂
Mass = 13.8 g
Molar mass = 32.0 g/mol
∴ Number of moles of O₂ present = [tex]\frac{13.8}{32.0}[/tex]
Number of moles of O₂ present = 0.43125 mole
Since,
2 moles of NO is needed to completely react with 1 mole of O₂
Then,
0.823059 mole of NO is will react with completely react with [tex]\frac{0.823059 }{2}[/tex] mole of O₂
[tex]\frac{0.823059 }{2} = 0.4115295[/tex]
∴ Number of moles of O₂ that reacted is 0.4115295 mole
This means O₂ is the excess reactant and NO is the limiting reactant
Now, for the number of moles of excess reactant left over
Number of moles of excess reactant left over = Number of moles of O₂ present - Number of moles of O₂ that reacted
∴ Number of moles of excess reactant left over = 0.43125 mole - 0.4115295 mole
Number of moles of excess reactant left over = 0.0197205 mole
Number of moles of excess reactant left over ≅ 0.0197 mole
Hence, the number of moles of excess reactant that would be left over is 0.0197 mole
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Calculate the solubility of Co(OH)2, in g/L, in solutions that have been buffered to the following pHs.
Ksp=1.6*10^-15.
a. 7.00
b. 10.00
c. 4.00
Answer:
a. 14.9g/L
b. 1.49x10⁻⁶
c. 1.49x10⁷
Explanation:
You can write the buffer Ksp of Co(OH)₂ as follows:
Co(OH)₂(s) ⇄ Co²⁺ + 2OH⁻
Ksp = 1.6x10⁻¹⁵ = [Co²⁺] [OH⁻]²
To have buffered the solutions means [OH⁻] is fixed. From the equilibrium of water we can relate [OH⁻] with pH as follows:
[OH⁻] = 10^[14-pH]
With [OH⁻] and Ksp we can solve for [Co²⁺]. Its concentration is equal to solubility (That is the amount of Co(OH)₂ that can be dissolved).
[Co²⁺] is in mol/L. With molar mass of Co(OH)₂ -92.948g/mol-, We can obtain, in the end, its solubility in g/L.
-Molar concentration of [Co²⁺] and solubility:
a. [OH⁻] = 10^[14-7.00] = 1x10⁻⁷
[Co²⁺] = 1.6x10⁻¹⁵ / [1x10⁻⁷]²
[Co²⁺] = 0.16mol / L = Solubility.
In g/L = 0.16mol / L ₓ(92.948g/mol) =
14.9g/L
b. [OH⁻] = 10^[14-10.00] = 1x10⁻⁴
[Co²⁺] = 1.6x10⁻¹⁵ / [1x10⁻⁴]²
[Co²⁺] = 1.6x10⁻⁸mol / L = Solubility.
In g/L = 1.6x10⁻⁸mol / L ₓ(92.948g/mol) =
1.49x10⁻⁶g/L
c.[OH⁻] = 10^[14-4.00] = 1x10⁻¹⁰
[Co²⁺] = 1.6x10⁻¹⁵ / [1x10⁻¹⁰]²
[Co²⁺] = 1.6x10⁵mol / L = Solubility.
In g/L = 1.6x10⁵mol / L ₓ(92.948g/mol) =
1.49x10⁷g/L
As you can see, and as general rule, all hydroxides are solubles in acids.
Write the equilibrium constant expression for this reaction: 2H+(aq)+CO−23(aq) → H2CO3(aq)
Answer:
Equilibrium constant expression for [tex]\rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq) \rightleftharpoons H_2CO_3\, (aq)[/tex]:
[tex]\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{[\mathrm{H_2CO_3}]}{\left[\mathrm{H^{+}\, (aq)}\right]^{2} \, \left[\mathrm{CO_3}^{2-}\right]}[/tex].
Where
[tex]a_{\mathrm{H_2CO_3}}[/tex], [tex]a_{\mathrm{H^{+}}}[/tex], and [tex]a_{\mathrm{CO_3}^{2-}}[/tex] denote the activities of the three species, and [tex][\mathrm{H_2CO_3}][/tex], [tex]\left[\mathrm{H^{+}}\right][/tex], and [tex]\left[\mathrm{CO_3}^{2-}\right][/tex] denote the concentrations of the three species.Explanation:
Equilibrium Constant ExpressionThe equilibrium constant expression of a (reversible) reaction takes the form a fraction.
Multiply the activity of each product of this reaction to get the numerator.[tex]\rm H_2CO_3\; (aq)[/tex] is the only product of this reaction. Besides, its coefficient in the balanced reaction is one. Therefore, the numerator would simply be [tex]\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)[/tex].
Similarly, multiply the activity of each reactant of this reaction to obtain the denominator. Note the coefficient "[tex]2[/tex]" on the product side of this reaction. [tex]\rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq)[/tex] is equivalent to [tex]\rm H^{+}\, (aq) + H^{+}\, (aq) + {CO_3}^{2-}\, (aq)[/tex]. The species [tex]\rm H^{+}\, (aq)[/tex] appeared twice among the reactants. Therefore, its activity should also appear twice in the denominator:
[tex]\left(a_{\mathrm{H^{+}}}\right)\cdot \left(a_{\mathrm{H^{+}}}\right)\cdot \, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right = \left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right[/tex].
That's where the exponent "[tex]2[/tex]" in this equilibrium constant expression came from.
Combine these two parts to obtain the equilibrium constant expression:
[tex]\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \quad\begin{matrix}\leftarrow \text{from products} \\[0.5em] \leftarrow \text{from reactants}\end{matrix}[/tex].
Equilibrium Constant of ConcentrationIn dilute solutions, the equilibrium constant expression can be approximated with the concentrations of the aqueous "[tex](\rm aq)[/tex]" species. Note that all the three species here are indeed aqueous. Hence, this equilibrium constant expression can be approximated as:
[tex]\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{\left[\mathrm{H_2CO_3\, (aq)}\right]}{\left[\mathrm{H^{+}\, (aq)}\right]^2\cdot \left[\mathrm{{CO_3}^{2-}\, (aq)}\right]}[/tex].
Discuss whether each of the following is a mixture or a pure substance. If it is a mixture, please state if it homogeneous or heterogeneous. Please give your reasoning for each case. If it is difficult to tell, explain why.
1. Seawater.
2. Chocolate.
3. Table sugar.
4. An apple.
5. A sheet of paper
6. The spice paprika.
7. Seven Up.
Answer:
Pure substance:
Seven upTable sugarSeawater (H20 + NaCl + impurities)Mixtures:
Chocolate (homogeneous)Paprika spice (homogeneous)Seawater (heterogeneous, because of impurities)A sheet of paper (homogeneous)An apple (heterogeneous)Explanation:
Pure substances exists as either elements or compounds.
A homogeneous mixture has even distribution and a single phase composition of its constituents; whereas heterogeneous mixtures are not uniform and constituents exist in different phases.
write the balanced nuclear equation for the radioactive decay of radium-226 to give radon-222, and determine the type of decay
Answer: DEAR THE ANSWER TO YOUR QUESTION IS,
Explanation: Consider the equation for the decay of radium-226 to radon-222, with the simultaneous loss of an alpha particle and energy in the form of a gamma ray. Radium-226 is the reactant; radon, an alpha particle, and a gamma ray are the products. The equation is:
shown in the attach figure
TYPE OF DECAY: as α-particle emmit in this reaction hence its the α-decay
It decays by emitting an alpha particle composed of two protons and two neutrons. The radium nucleus turns into radon-222 nucleus, itself radioactive, containing two protons and two neutrons less. The disintegration releases 4.6 million electronvolts of energy
PLS GIVE ME RATING IF YOU FIND IT HELPFULL SO THAT OTHER BE BENIFIT OF THAT ANSWEER
THANKS....
Which alkyl halide is least reactive with magnesium metal?
A. CH_3 CH_2 CH_2 Br
B. CH_3 CH_2 CH_2 F
C. CH_3 CH_2 CH_2 I
D. CH_3 CH_2 CH_2 CI
Answer:
B. CH_3 CH_2 CH_2 F
Explanation:
The alkyl group contains all univalent group derived from alkanes by the loss of a hydrogen atom. When an alkyl group combines or is bonded with an halogen family (such as Flourine, Chlorine , Bromine , Iodine) , we have an alkyl halide such as the options given in the question.
Alkyl halide bond with an sp³ Carbon. Depending on where the halide is bonded to , we can have the primary halide, secondary halide and the tertiary halide.
When alkyl halide react with magnesium metal, we have a reaction that is known as Grignard Reaction. The order of reactivity of alkyl halide with grignard reagents is :
F < Cl < Br < I .
This is because of the nature of the electronegativity and the bond strength.
As we move than the group from the top to the bottom, the electronegativity decreases and the bond strength increase, so as the size. Thus, this is exactly what the reactivity of alkyl halide with Grignard reaction is established for.
Thus since the fluorine have the highest electronegativity and the smallest bond size in the halogen family , then it will be the least reactive alkyl halide in reacting with magnesium metal.
What indicators of a chemical reaction occurred in the following equation?
C6H12O6 (s) + 602 (g) → 6 CO2 (g) + 6H2O (1) +
energy
Answer:
1) evolution of gas
2) evolution of heat
Explanation:
In this reaction, glucose is broken down into its constituents; carbon dioxide and water. The question is to decipher indicators of a chemical reaction from the equation.
If we look at the equation carefully, we will notice that a gas was evolved (CO2). The evolution of a gas indicates that a chemical reaction must have taken place. Secondly, energy is given off as heat. This is another indication that a chemical reaction has taken place.
What volume, in mL, of 4.50 M NaOH is needed to prepare 250. mL of 0.300 M NaOH?
Answer:
16.7 mL
Explanation:
Convert 250 mL to L.
250 mL = 0.250 L
Calculate the amount of moles of NaOH in 250 mL of 0.300 M NaOH.
0.250 L × 0.300 M = 0.075 mol
Using this amount of moles, you need to find out what volume of 4.50 M will give you that many moles. You can do this by dividing the amount of moles by the molarity.
(0.075 mol)/(4.50 M) = 0.0167 L
Convert from L to mL.
0.0167 L = 16.7 mL
A 50.0 mL sample containing Ni+2 was treated with 25.0 mL of .050 M EDTA to complex all of the Ni+2 and leave some excess EDTA in solution. the excess EDTA was then titrated with .050 M Zn+2, requiring 5.00 mL to reach equivalence point. What was the concentration of Ni+2 in the original solution? (EDTA forms complexes with metals in a 1:1 stoichiometry)
A solution contains 90 milliequivalents of HC1 in 450ml. What is its normality?
Answer:
Normality N = 0.2 N
Explanation:
Normality is the number of gram of equivalent of solute divided of volume of solution, where the number of gram of equivalent of solute is weight of the solute divided by the equivalent weight.
Normality is represented by N.
Mathematically, we have :
[tex]\mathbf{Normality \ N = \dfrac{Number \ of \ gram \of \ equivalent\ of\ solute }{volume \ of \ solution}}[/tex]
Given that:
number of gram of equivalent of solute = 90 milliequivalents 90 × 10⁻³ equivalent
volume of solution (HCl) = 450 mL 450 × 10⁻³ L
[tex]\mathbf{Normality \ N = \dfrac{90 \times 10^{-3}}{450 \times 10^{-3}}}[/tex]
Normality N = 0.2 N
What is the freezing point of an aqueous glucose solution that has 25.0 g of glucose?
Answer:
[tex]T_f=-2.58\°C[/tex]
Explanation:
Hello,
In this case, we can compute the the freezing point depression by using the following formula:
[tex]T_f-T_0=-i*m*Kf[/tex]
Whereas the freezing point of pure water is 0 °C van't Hoff factor for glucose is 1, the molality is computed as shown below and the freezing point constant of water is 1.86 °C/m:
[tex]m=\frac{25.0g\ glucose*\frac{1mol\ glucose}{180g\ glucose} }{100g*\frac{1kg}{1000g} }\\ \\m=1.39m[/tex]
Thus, the freezing point of the solution is:
[tex]T_f=T_0-i*m*Kf\\\\T_f=0\°C-1*1.39m*1.86\frac{\°C}{m}\\ \\T_f=-2.58\°C[/tex]
Regards.
Calculate the molarity of a solution containing 9.25 mol H2SO4 in 2.75 L of solution.
A) 9.25
B) 6.50
C) 3.36
D) 25.4
E) 33.6
Answer:
THE MOLARITY OF THE SOLUTION IS 3.36 MOLE/L
Explanation:
First we must understand what molarity is.
Molarity is the number of mole per unit volume of solution. In this question, 9.25 mole of H2SO4 was given in 2.75 L of solution.
Molarity is written in mole per dm3 or L.
So we can calculate the molarity:
9.25 ole of H2SO4 = 2.75 L of solution
The number of mole in 1 L of solution will be:
= 9.25 mole / 2.75 L
= 3.3636 mole/ L
In conclusion, the molarity of the solution is approximately 3.36 mole/L
under certain water conditions, the free chlorine (hypochlorous acid, hocl) in a swimming pool decomposes according to the law of uninhibited decay. after shocking a pool, the pool boy, geoff, tested the water and found the amount of free chlorine to be 2.6 parts per million (ppm). twenty-four hours later, geoff tested the water again and found the amount of free chlorine to be 2.1 ppm. what will be the reading after 2 days (that is, 48 hours)
Answer:
1.7 ppm
Explanation:
Original amount N' = 2.6 ppm
time to testing t = 24 hr
final amount N = 2.1 ppm
Using exponential inhibited decay, we have
N = N'e^(-kt)
Where
N is the new reading
N' is the original reading
t is the decay time
k is the decay constant
Substituting, we have
2.1 = 2.6 x e^(-k x 24)
2.1 = 2.6 x e^(-24k)
0.808 = e^(-24k)
We take the natural log of both sides of the equation
Ln 0.808 = Ln (e^(-24k))
-0.213 = - 24k
K = 0.213/24 = 0.00886
After 48 hrs, the reading of free chlorine will be
N = 2.6 x e^(-0.00886 x 48)
N = 2.6 x e^(-0.425)
N = 2.6 x 0.654
N = 1.7 ppm
Place the following atmospheric gases in order of abundance, from highest to lowest.
a. Carbon Dioxide
b. Other trace gases
c. Argon
d. Oxygen
e. Nitrogen
Explanation:
The percentage abundance of the following gases in the atmosphere is given as;
Carbon dioxide = 0.04 percent
Other trace gases = about a tenth of one percent of the atmosphere.
Argon = 0.93 percent
Oxygen = 21 percent
Nitrogen = 78 percent
Te order from highest to lowest is given as;
Nitrogen
Oxygen
Argon
Carbon dioxide
Other trace gases
Identify the Brønsted acid in the following equation:
H2SO4(aq) + 2NH3(aq)
(NH4)2SO4aq
Answer:
H₂SO₄
Explanation:
A Brønsted acid is a proton donor. It loses protons.
This equation may be easier to understand if we write it ionically.
[tex]\underbrace{\hbox{H$_{2}$SO$_{4}$}}_{\hbox{Br$\o{}$nsted acid}} + 2 \text{NH}_{3} \longrightarrow \, \underbrace{\hbox{SO$_{4}^{2-}$}}_{\hbox{Br$\o{}$nsted base}} + \text{2 NH}_{4}^{+}[/tex]
We see that the H₂SO₄ has lost two protons to become SO₄²⁻, so it is a Brønsted acid.
What is the molar mass of butane if 4.49 x 1016 molecules of butane weigh 8.29?
Answer: The molar mass of butane will be [tex]11.2\times 10^{7}g[/tex]
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP, contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles. and weighs equal to the molar mass.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given molecules}}{\text {avogadro's number}}=\frac{4.49\times 10^{16}}{6.023\times 10^{23}}=0.74\times 10^{-7}moles[/tex]
Now [tex]0.74\times 10^{-7}moles[/tex] weigh = 8.29 g
Thus 1 mole will weigh = [tex]\frac{8.29}{0.74\times 10^{-7}}\times 1=11.2\times 10^{7}g[/tex]
The molar mass of butane will be [tex]11.2\times 10^{7}g[/tex]