Answer:
Explanation:
The specific rotation of the sample is -2 degrees/0.2 g/mL of mixture
This equals -10 degrees/g/mL of sample.
let the proportion of the R (+) enantiomer be x. The proportion of the S (-) enantiomer in the mixture will be given by (1-x).
specific rotation of the mixture = proportion of R enantiomer* its specific rotation + proportion of S enantiome * its specific rotation
i.e.
-10 = x *(+20) + (1-x)*(-20)
-10 = 20x-20 + 20x
-10+20 = 40x
+10 = 40 x
x=10/40 = 25%
Since the proportion of the other enantiomer is 1-x, it is 0.75 or 75%
So the mixture contains 25% R, 75% S, giving you an excess of 50%.
Answer:
10%
Explanation:
Enantiomeric excess is a way of describing how optically pure a mixture is by calculating the purity of the major enantiomer. It can range from 0%-100%. Enantiomeric excess ( ee ) can also be defined as the absolute difference between the mole fractions of two enantiomers.
Enantiomeric excess is also called optical purity. This is because chiral molecules cause the rotation of plane-polarized light and are said to be optically active. An enantiomerically pure sample has an enantiomeric excess of 100 percent
Enantiomeric excess = observed specific rotation/specific rotation of the pure enantiomer x 100
From the data given in the question;
observed specific rotation= -2°
specific rotation of the pure enantiomer = +20°
Therefore;
ee= 2/20 ×100
ee= 10%
a gas obeys the equation of state p(v-b)=RT.for the gas b=0.0391L/mol.calculate the fugacity coefficient for the gas at 1000°c and 1000atm
Answer:
The fugacity coefficient is [tex][\frac{f}{p} ] = 1.45[/tex]
Explanation:
From the question we are told that
The gas obeys the equation [tex]p(v-b) = RT[/tex]
The value of b is [tex]b = b = 0.0391 \ L /mol[/tex]
The pressure is [tex]p = 1000 \ atm[/tex]
The temperature is [tex]T= 1000^oC = 1273 K[/tex]
generally
[tex]RT ln[\frac{f}{p} ] = \int\limits^{p}_{o} [ {v_{r} -v_{i}} ]\, dp[/tex]
Where [tex]\frac{f}{p}[/tex] is the fugacity coefficient
[tex]v_r[/tex] is the real volume which is mathematically evaluated from above equation as
[tex]v_r = \frac{RT}{p} + b[/tex]
[tex]v_r = \frac{RT}{p} + 0.0391[/tex]
and [tex]v_{i}[/tex] is the ideal volume which is evaluated from the ideal gas equation (pv = nRT , at n= 1) as
[tex]v_{i} = \frac{RT}{p}[/tex]
So
[tex]RT ln[\frac{f}{p} ] = \int\limits^{1000}_{o} [[ \frac{RT}{p} + 0.0391] - [\frac{RT}{p} ]} ]\, dp[/tex]
=> [tex]RT ln[\frac{f}{p} ] = \int\limits^{1000}_{o} [0.391 ]\, dp[/tex]
=> [tex]RT ln[\frac{f}{p} ] = [0.391p]\left | 1000} \atop {0}} \right.[/tex]
=> [tex]RT ln[\frac{f}{p} ] = 38.1[/tex]
So
[tex]ln[\frac{f}{p} ] = \frac{39.1}{RT}[/tex]
Where R is the gas constant with value [tex]R = 0.082057\ L \cdot atm \cdot mol^{-1}\cdot K^{-1}[/tex]
[tex][\frac{f}{p} ] = \frac{39.1}{ 2.303 *0.082057 * 1273}[/tex]
[tex][\frac{f}{p} ] = 1.45[/tex]
Barium is a very reactive metal in the presence of oxygen and water, thus its density cannot be measured by water displacement. Instead, mesitylene (C9H12, density = 0.86370 g/mL (at 20 o C)) is used. 77.240 g of Ba is placed into a flask, and mesitylene is added so that together the total volume is 100.00 mL. The mass of the mesitylene and Ba together is 148.792 g. What is the density (in g/mL) of the Ba at 20 o C?
Answer:
The correct answer is 4.502 g per ml.
Explanation:
Based on the given question, the sum of the mass of mesitylene and barium together is 148.792 grams. The mass of barium given is 77.240 grams. Therefore, the mass of mesitylene will be,
Mass of mesitylene = Total mass - Mass of barium
= 148.792 - 77.240
= 71.552 grams
The density of mesitylene is 0.86370 g per ml. To calculate the volume of mesitylene, the formula to be used is,
Volume = mass / density. Now, putting the values we get,
Volume = 71.552 / 0.86370 = 82.8436 ml.
As the total volume is 100 ml, therefore, the volume of Ba will be,
Volume of Ba = 100-82.8436 = 17.1564 ml
The density of Ba at 20 degree C can be calculated by using the formula,
Density = mass / volume. Now putting the values we get,
Density = 77.240 g / 17.1564 ml
= 4.502 g per ml
Monel metal is a corrosion-resistant copper-nickel alloy used in the electronics industry. A particular alloy with a density of 8.80 g/cm3 and containing 0.090 % Si by mass is used to make a rectangular plate that is 15.0 cm long, 12.5 cm wide, and 3.50 mm thick and has a 2.50-cm-diameter hole drilled through its center such that the height of the hole is 3.50 mm .
The silicon in the plate is a mixture of naturally occurring isotopes. One of the those isotopes is silicon-30, which has an atomic mass of 29.97376 amu. The percent natural abundance, which refers to the atoms of a specific isotope, of silicon-30 is 3.10%.
Part A What is the volume of the plate?Express the volume numerically in cubic centimeters.
Part B How many silicon-30 atoms are found in this plate?
Express your answer numerically using two significant figures.
Answer:
Based on the given question, the dimensions of the plate is 15 cm in length, 12.5 cm in width, and 3.50 mm in thickness (0.350 cm). Now the volume of the plate will be,
V = 15 cm × 12.5 cm × 0.350 cm = 65.62 cm³
A hole of diameter 2.50 cm is drilled through the center of the plate, at the height of 3.50 mm or 0.350 cm. Now the volume of the hole is π(r)²h,
= 22/7 × (1.25 cm)² × 0.350 cm = 1.72 cm³
Thus, the volume of the plate will be determined by subtracting the volume of plate with the volume of hole, which will be,
65.62 cm³ - 1.72 cm³ = 63.9 cm³
The density of the alloy is 8.80 g/cm³, therefore, the mass of the alloy can be determined by using the formula, mass = density * volume
mass = 8.80 g/cm³ × 63.9 cm³ = 562.32 grams
Of the total alloy, 0.090 percent is Si, that is,
(0.090/100) × 562.32 g = 0.506 grams of Si
The natural abundance of the element is not determined by mass but by the number of atoms it possess. For this Avogadro's number and atomic mass of Si is used. Now the number of atoms of Si present is,
(0.506 g) (1 mol/28.0855 g) (6.023 × 10²³ atoms /mol) = 1.08 × 10²² Si atoms
Of these Si atoms, 3.10 percent are Si-30 so,
= (3.10 / 100) × (1.08 × 10²² atoms) / 1000 = 3.34 × 10²⁰ atoms of Si-30. or 3.4 × 10²⁰ atoms
Which of the following formulas represents an ionic compound?
1.HI 2.HCI 3.LiCI 4.SO2
Answer:
Numbers 4,3
Explanation:
Ionic bond is between nonmental and metals
An excess of sodium carbonate, Na2CO3, in solution is added to a solution containing 15.71 g CaCl2. After performing the experiment, 13.19 g of calcium carbonate, CaCO3, is produced. Calculate the percent yield of this reaction
Answer:
93.15 %
Explanation:
We have to start with the chemical reaction:
[tex]CaCl_2~+~Na_2CO_3~->~CaCO_3~+~NaCl[/tex]
Now, we can balance the reaction:
[tex]CaCl_2~+~Na_2CO_3~->~CaCO_3~+~2NaCl[/tex]
Our initial data are the 15.71 g of [tex]CaCl_2[/tex], so we have to do the following steps:
1) Convert from grams to moles of [tex]CaCl_2[/tex] using the molar mass (110.98 g/mol).
2) Convert from moles of [tex]CaCl_2[/tex] to moles of [tex]CaCO_3[/tex] using the molar ratio. ( 1 mol [tex]CaCl_2[/tex]= 1 mol of [tex]CaCO_3[/tex]).
3) Convert from moles of [tex]CaCO_3[/tex] to grams of [tex]CaCO_3[/tex] using the molar mass. (100 g/mol).
[tex]15.71~g~CaCl_2\frac{1~mol~CaCl_2}{110.98~g~CaCl_2}\frac{1~mol~CaCO_3}{1~mol~CaCl_2}\frac{100~g~CaCO_3}{1~mol~CaCO_3}=14.16~g~CaCO_3[/tex]
Finally, we can calculate the yield percent:
[tex]%~=~\frac{13.19~g~CaCO_3}{14.16~g~CaCO_3}*100=93.15~%[/tex]
I hope it helps!
The percentage yield obtained when excess sodium carbonate, Na₂CO₃, is added to a solution containing 15.71 g CaCl₂ is 93.2%
We'll begin by writing the balanced equation for the reaction. This is given below: [tex]Na_{2}CO_{3} + CaCl_{2} - > CaCO_{3} + 2NaCl[/tex]Molar mass of CaCl₂ = 40 + (35.5×2) = 111 g/mol
Mass of CaCl₂ from the balanced equation = 1 × 111 = 111 g
Molar mass of CaCO₃ = 40 + 12 + (16×3) = 100 g/mol
Mass of CaCO₃ from the balanced equation = 1 × 100 = 100 g
SUMMARY
From the balanced equation above,
111 g of CaCl₂ reacted to produce 100 g of CaCO₃
Next, we shall determine the theoretical yield of of CaCO₃. This can be obtained as follow:From the balanced equation above,
111 g of CaCl₂ reacted to produce 100 g of CaCO₃.
Therefore,
15.71 g of CaCl₂ will react to produce = [tex]\frac{15.71 * 100}{111} \\\\[/tex] = 14.15 g of CaCO₃.
Thus, the theoretical yield of of CaCO₃ is 14.15 g
Finally, we shall determine the percentage yield. This can be obtained as follow:Actual yield of CaCO₃ = 13.19 g
Theoretical yield of CaCO₃ = 14.15 g
Percentage yield =?[tex]Percentage yield = \frac{Actual}{Theoretical} * 100\\\\= \frac{13.19}{14.15} * 100\\\\[/tex]
= 93.2%Therefore, the percentage yield of the reaction is 93.2%
Learn more: https://brainly.com/question/13930222
Classify an element having the following ground state electron configuration as an alkali metal, alkaline earth metal, nonmetal, halogen, transition metal, or noble gas.
a. [Ne]3s1
b. [Ne]3s23p3
c. [Ar]4s23d104p5
d. [Kr]5s24d1
e. [Kr]5s24d105p6
Explanation:
Alkali metal refers to group1 elements.
Alkali earth metal refers to group 2 elements.
Non metals refers to elements in grouos 4 to group 7.
Halogen refers to group 17 elements
Transition Metal refers to group 3 to group 12 elements
Noble gases refer to elements in group 18.
To obtain the group number from the electronic configuration, we calculate the total number of electrons in the principal quantum number (coefficient of the letters).
a. [Ne]3s1
Principal quantum number = 3
Number of electrons present = 1
This element belongs to group 1. It is an Alkali Metal.
b. [Ne]3s23p3
Principal quantum number = 3
Number of electrons present = 2 + 3 = 5
This element belongs to group 15 (5A). It is a Non metal
c. [Ar]4s23d104p5
Principal quantum number = 4
Number of electrons present = 2 + 5 = 7
This element belongs to group 17 (7A). It is an Halogen.
d. [Kr]5s24d1
This configuration belongs to the element yttrium and has an incomplete d shell. Hence it is a transition metal.
e. [Kr]5s24d105p6
Principal quantum number = 5
Number of electrons present = 2 + 6 = 8
This element belongs to group 18 (8A). It is a Noble gas.
Two samples of the same rainwater are tested using two indicators at an environrnental lab. The first indicator, Methyl Orange, reveals a distinct yellow color when added to the sample. The second indicator, Litmus, turns red when placed in contact with the water sample.
Required:
a. Identify a possible pH value for the rainwater.
b. Explain, in terms of hydronium ions and hydroxide ion concentrations, the pH value of the rainwater.
Answer:
A. The pH value of rainwater is acidic about 4.4
B. The molar concentrations of the Hydronium ions are more than that of the hydroxide ions. That is why the rainwater is acidic with a pH of less than 7
Explanation:
A. Methyl orange is an acid indicator that is used to detect acidic solutions which have pH values that fall within the range of about 4.4 to 7. The distinct yellow colour change that was shown by the methyl orange as it was added to the water shows that the pH value is acidic, with a value above 4.4. (it has to be like this before methyl orange changes to yellow colour)
B. The Hydronium ( H30+) ion concentrations and the hydroxide (OH-) ion concentrations are used to measure the pH values of substances.
We can tell that the Hydronium ( H30+) ion concentrations are more than the hydroxide (OH-) ion concentrations in the sample of rainwater tested. This can be detected from the colour change that both the methyl orange and the litmus paper gave. The indicators showed that the rainwater solution was indeed acidic. Hence, the pH value will be less than 7, but greater than 4.4.
An organic chemistry student was studying the solubility of Methyl-N-acetyl-α-D-glucosaminide (1-O-methyl-GlcNAc), a derivative of glucosamine, in water but inadvertently added 1 equiv. of periodic acid instead. Based on your understanding of the reactions of monosaccharides with periodates, draw the organic product that the student obtained.
Complete Question
The diagram for this question is shown on the second uploaded image
Answer:
The organic product obtained is shown on the first uploaded image
Explanation:
The process that lead to this product formation is known as oxidative cleavage which is a reaction that involves the cleavage of a carbon to carbon bond at the same time this carbon which formed the carbon bond are oxidized i.e oxygen is been added to them
what type of bonds do compounds formed from non metal consist of?
Compounds formed from non-metals consist of molecules. The atoms in a molecule are joined together by covalent bonds. These bonds form when atoms share pairs of electrons.
A student has an unknown sample of solution X. This solution is placed in a 1.00 cm wide cuvet and inserted into the spectrometer, producing an absorbance reading of 0.275 at a wavelength of 525.0 nm. What is the concentration of solution X in the unknown sample
Answer:
The concentration of the sample is 3.564x10⁻³M.
Explanation:
Using Lambert-Beer law, absorbance of a sample is directely proportional to its concentration.
The general graph of the absorbance of the standards with different concentrations is:
Y = 75.9X + 0.0045
R² = 0.9946
Where Y is the absorbance of the sample and X its concentration in mole/L.
If a solution has an absorbance of 0.275:
0.275 = 75.9X + 0.0045
0.2705 = 75.9X
3.564x10⁻³M = X → The concentration of the sample.
A geochemist in the field takes a 46.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X. He notes the temperature of the pool, 21°C, and caps the sample carefully. Back in the lab, the geochemist filters the sample and then evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 0.87 g.
Required:
Using only the information above, can you calculate the solubility of X in water at 21°C? If yes, calculate it.
Answer: The solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.
Explanation:
The given data is as follows.
Volume of sample water = 46 ml
Temperature = [tex]21^oC[/tex]
After vaporization, washes and then drying the weight of mineral X = 0.87 g
This means that 46.0 ml of water contains 0.87 g of X. Therefore, grams present in 1 ml of water will be calculated as follows.
1 ml of water = [tex]\frac{0.87 g}{46.0 ml}[/tex]
= [tex]1.891 \times 10^{-2}[/tex] g/ml
Therefore, we can conclude that solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.
7.Which one of the following statements is not true?
1 point
O The molecules in a solid vibrate about a fixed position
O The molecules in a liquid are arranged in a regular pattern
The molecules in a gas exert negligibly small forces on each other, except during
collisions
The molecules of a gas occupy all the space available
Answer:
B. the molecules in liquid are loosely packed and scattered thus, they cannot be arranged
Given the density of iron (Fe) is 7.87 g/cm3, determine the mass of iron (in grams) in a rectangle block with the dimensions of 12.5 in long, 3.50 in wide, and 2.50 in high. (1in = 2.54 cm).
Answer:
[tex]m=14,105.71 g Fe[/tex]
Explanation:
Hello,
In this case, the first step is to compute the volume of the block considering the length, height and width:
[tex]V=L \times W \times H =12.5 in\times 3.50 in \times 2.50 in =109.375 in^3[/tex]
Then, we compute the volume in cubic centimetres:
[tex]V=109.375in^3\times \frac{16.3871 cm^3}{1in^3} =1792.34cm^3[/tex]
Finally, as the density is given by:
[tex]\rho =\frac{m}{V}[/tex]
We solve for the mass:
[tex]m=\rho \times V= 7.87\frac{g}{cm^3} \times 1792.34 cm^3\\\\m=14,105.71 g Fe[/tex]
Best regards.
Which of the following are not created by an arrangement of electric charges
or a current (the flow of electric charges)?
A. An electric field
B. A magnetic field
C. A quantum field
D. A gravitational field
Answer:
gravitational and quantum ARE NOT, but electric and magnetic ARE. there is a similar question to this but it's the exact opposite, so don't get confused
An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.35×10-2 M CH2Cl2, 0.173 M CH4 and 0.173 M CCl4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.155 mol of CH4(g) is added to the flask?
Answer:
[CH₂Cl₂] = 7.07x10⁻² M
[CH₄] = 0.319 M
[CCl₄] = 0.164 M
Explanation:
The equilibrium reaction is the following:
2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)
The equilibrium constant of the above reaction is:
[tex] K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{0.173 M*0.173 M}{(5.35 \cdot 10^{-2} M)^{2}} = 10.5 [/tex]
When 0.155 mol of CH₄(g) is added to the flask we have the following concentration of CH₄:
[tex] C = \frac{\eta}{V} = \frac{0.155 mol}{1.00 L} = 0.155 M [/tex]
[tex]C_{CH_{4}} = 0.328 M[/tex]
Now, the concentrations at the equilibrium are:
2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)
5.35x10⁻² - 2x 0.328 + x 0.173 + x
[tex]K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{(0.328 + x)(0.173 + x)}{(5.35 \cdot 10^{-2} - 2x)^{2}}[/tex]
[tex]10.5*(5.35 \cdot 10^{-2} - 2x)^{2} - (0.328 + x)*(0.173 + x) = 0[/tex]
Solving the above equation for x:
x₁ = 0.076 and x₂ = -0.0086
Hence, the concentration of the three gases once equilibrium has been reestablished is:
[CH₂Cl₂] = 5.35x10⁻² - 2(-0.0086) = 7.07x10⁻² M
[CH₄] = 0.328 + (-0.0086) = 0.319 M
[CCl₄] = 0.173 + (-0.0086) = 0.164 M
We took x₂ value because the x₁ value gives a negative CH₂Cl₂ concentration.
I hope it helps you!
A sample of chloroform, CHCl 3 , , was determined to have a molecular mass of 112.3g / (mol) . Its molecular mass is known to be 119.5g / (mol) . Calculate the absolute error and the percent error
Answer:
Explanation:
in your case ,
Meaured value = 112.3
actual value = 119.5
Absolute error= measured value - actual value
Percent error = [measured value - actual value / actual value ] x 100
Hope this help you to find the answer
Show work plzzz
Unknown Metal Bar #8
Mass of Unknown Metal bar 11.3g
Length of bar 13.90cm
Width of bar 2.9cm
Thickness of bar 0.081cm
1. Calculate the volume of the bar:
2. Calculate the (experimental) density of the bar:
3. Based on the provided list of (true) densities, what is the possible identity of the Unknown metal?
4. What is the percent difference between the true density of your metal and the calculated density?
= | − | ∗ 100%
Answer:
1= Volume
= Length x breath x height
= 13.90 x 2.9 x 0.081
=3.26511
2= Density = Mass ÷ volume
= 11.3 ÷ 3.26511
= 3.461 (3d.p)
idk the rest because you haven't shown a picture of the rest
Answer:
1. 3.3 cm³; 2. 3.5 g/cm³; 3. barium; 4. 4%
Explanation:
Experimental data:
Mass = 11.3 g
Length = 13.90 cm
Width = 2.9 cm
Thickness = 0.081 cm
Calculations:
1. Volume of bar
V = lwh = 13.90 cm × 2.9 cm × 0.081 cm = 3.3 cm³
2. Experimental density
[tex]\text{Density} = \dfrac{m}{V} = \dfrac{\text{11.3 g}}{\text{3.27 cm}^{3}} = \textbf{3.5 cm}^{\mathbf{3}}[/tex]
3. Identity of metal
The three most likely metals are scandium (3.00 g/cm³), barium (3.59 g/cm³), and yttrium (4.47 g/cm³)
The metal is probably barium.
4. Percent difference
[tex]\begin{array}{rcl}\text{Percent difference}&= &\dfrac{\lvert \text{ True - Calculated}\lvert}{ \text{True}} \times 100 \,\%\\\\& = & \dfrac{\lvert 3.59 - 3.5\lvert}{3.59} \times 100 \, \% \\\\& = & \dfrac{\lvert 0.1\lvert}{3.59} \times 100 \, \%\\ \\& = & 0.04 \times 100 \, \%\\& = & \mathbf{4 \, \%}\\\end{array}\\\text{The percent difference is $\large \boxed{\mathbf{4 \, \%} }$}[/tex]
C3H7-C(=O)-NH2 IUPAC NAME ?
Answer:
Amide
Explanation:
O=NH2 is the Amide group versus NH2, which is the amine group.
Answer:
Butamide
Explanation:
C3H7-C(=O)-NH2 IUPAC NAME
C4H9NO
H H H
H - C - C - C - C = O
H H H N - H
H
But amide
Amide because R-CO-NH2 ie C(=O)-NH2
But because 4 Cabon
Determine the amount of heat (in kJ) associated with the production of 5.71 × 104 g of ammonia according to the following equation. N2(g) + 3H2(g) 2NH3ΔH°rxn = −92.6 kJ Assume that the reaction takes place under standard-state conditions at 25°C.
Answer:
[tex]Q=-3.11x10^5kJ[/tex]
Explanation:
Hello,
In this case, for the given reaction, we are given the standard enthalpy of reaction per mole of ammonia that is -92.6 kJ, it means, that forming one mole of ammonia will release 92.6 kJ of energy. In such a way, for the formation of 5.71x10⁴ g of ammonia, the following amount of heat will be released:
[tex]Q=5.71x10^4gNH_3*\frac{1molNH_3}{17gNH_3}*-92.6\frac{kJ}{molNH_3}\\ \\Q=-3.11x10^5kJ[/tex]
Best regards.
The amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ
We'll begin by calculating the number of mole in 5.71×10⁴ g of NH₃
Mass of NH₃ = 5.71×10⁴ g
Molar mass of NH₃ = 14 + (3×1) = 17 g/mol
Mole of NH₃ =?Mole = mass / molar mass
Mole of NH₃ = 5.71×10⁴ / 17
Mole of NH₃ = 3358.82 moles Finally, we shall determine the heat required to produce 3358.82 moles (i.e 5.71×10⁴ g) of NH₃. This can be obtained as follow:N₂(g) + 3H₂(g) —> 2NH₃(g) ΔH°rxn = −92.6 kJ
Since reaction took place at standard conditions, it means:
1 moles of NH₃ required −92.6 kJ
Therefore,
3358.82 moles of NH₃ will require = 3358.82 × –92.6 = –311026.732 KJ
Thus, the amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ
Learn more: https://brainly.com/question/17332795
Now construct a different electrochemical cell. You put a zinc metal anode in contact with a 0.052 M solution of zinc nitrate and a silver cathode in contact with a 0.0042 M solution of silver(I) nitrate. What is the value of the electric potential at the moment the reaction begins
Answer:
[tex]1.66~V[/tex]
Explanation:
We have to start with the half-reactions for both ions:
[tex]Zn^+^2~+2e^-~->Zn[/tex] V= -0.76
[tex]Ag^+~e^-~->~Ag[/tex] V= +0.80
If we want a spontaneous reaction (galvanic cell) we have to flip the first reaction, so:
[tex]Zn~->~Zn^+^2~+2e^-~[/tex] V= +0.76
[tex]Ag^+~+~e^-~->~Ag[/tex] V= +0.80
If we want to calculate ºE we have to add the two values, so:
ºE=0.76+0.80 = 1.56 V
Now, we have different concentrations. So, if we want to calculate E we have to use the nerts equation:
[tex]E=ºE~+~\frac{0.059}{n}LogQ[/tex]
On this case, Q is equal to:
[tex]Q=\frac{[Zn^+^2]}{[Ag^+]^2}[/tex]
Because the total reaction is:
[tex]Zn~+~2Ag^+~->~Zn^+^2~+~2Ag[/tex]
So, the value of "Q" is:
[tex]Q=\frac{[0.052 M]}{[0.0042]^2}=2947.84[/tex]
Now, we can plug all the values in the equation (n=2, because the amount of electrons transferred is 2). So:
[tex]E=1.56~V~+~\frac{0.059}{2}Log(2947.84)=1.66~V[/tex]
I hope it helps!
Calculate the percentage of the void space out of the total volume occupied by 1 mole of water molecules. The density of water is assumed to be 1.0 g/mL that is 1.0 g/cm3. The molar mass of water is 18.0 g/mol. The atomic radius of hydrogen is 37 pm and of oxygen is 73 pm. The formula for the volume of a sphere is 4/3(r3
Answer:
The percentage of the void space out of the total volume occupied is 93.11%
Explanation:
A mole of water contains 2 atoms of hydrogen and 1 atom of oxygen.
To calculate the volume of a mole of water, we calculate 2 times the volume of the hydrogen atom and 1 times the volume of the oxygen atom
Let's calculate this one after the other.
For the hydrogen, formula for the volume will be
[tex]V_{hydrogen[/tex] = 2 × 4/3 × π × [tex]r_{H}^{3}[/tex]
where [tex]r_{H}^{3}[/tex] = 37 pm which is read as 37 picometer (1 picometer = 10^-12 m) = 37 × [tex]10^{-12}[/tex] meters
Volume of the hydrogen = 8/3 × (37 × 10^-12)^3 = 4.05 * 10^-31
we multiply this by the avogadro's number = 6.02 * 10^23
= 4.05 * 10^-31 * 6.02 * 10^23 = 2.6 * 10^-8 m^3
We do same for thr oxygen, but this time we do not multiply the volume of the oxygen by 2 as we have only one atom of oxygen
Volume of oxygen = 4/3 * π * (73 * 10^-12) ^3 * avogadro's number = 9.81 * 10^-7 m^3
adding both volumes together, we have 1.24 * 10^-6 m^3 or simply 1.24 ml ( 0.01 m = 1 ml)
Dividing the molar mass of one mole of water by its density, we can get the volume of 1 mole of water
= (18g/mol)/(1 g/ml) = 18 ml/mol
Now we proceed to calculate the volume of void = Total volume - volume of molecule = 18 - 1.24 = 16.76 ml
Now, the percentage of void = volume of void/total volume * 100%
= 16.76/18 * 100% = 93.11%
Pick the odd one out?
Ethanol
Hexane
Oil
Carbon tetrachloride
Answer: Ethanol is the odd one out.
Explanation:
A polar compound is defined as the compound which is formed when there is a difference of electronegativities between the atoms. It is also defined as the bond which is formed due to the unequal sharing of electrons between the atoms.
Non-polar compound is defined as the compound which is formed when there is no difference of electronegativities between the atoms or the polarities cancel out.
Hexane [tex](C_6H_{14}), Oil (mixture of hydrocarbons) and carbon tetrachloride [tex](CCl_4)[/tex] all are non polar whereas ethanol is polar due to electronegative difference between hydrogen and oxygen.
A mixture of compounds containing diethylamine, phenol, ammonia, and acetic acid is separated using liquid-liquid extraction as follows: Step 1: Concentrated HCl is added followed by draining the aqueous layer. Step 2: Dilute NaOH is added to the organic layer followed by draining the aqueous layer. Step 3: Concentrated NaOH is added to the organic layer followed by draining the aqueous layer. Which compound would you expect to be extracted into the aqueous layer after the addition of dilute HCl, step 1? Group of answer choices
Complete Question
The complete question is shown on the first uploaded image
Answer:
The correct option is ammonia
Explanation:
The mixture contains two base compound which are
ammonia,
and diethylamine
Now the addition of HCl which is a strong acid in step 1 will cause the protonation of the two base compound , which makes the soluble hence resulting in them being extracted to the aqueous layer as represented in below
[tex]NH_3 + HCl\to NH_4 ^{+} + Cl^-[/tex]
and
[tex](CH 3CH 2) 2NH + HCl \to (CH 3CH 2) 2NH_2^{+} + Cl[/tex]
Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH(aq). Calculate the amount of Ga(s) that can be deposited from a Ga(III) solution using a current of 0.850 A that flows for 60.0 min.
Answer:
Mass of Ga = 0.73694 gram
Explanation:
Given:
Current = 0.850 A
Time = 60 minutes
Find:
Amount of gas deposit.
Computation:
Total charge = Current × Time in second
Total charge = 0.850 × 60 × 60
Total charge = 3,060 C
Mole of electron = Total charge / Faraday constant [Faraday constant = 96,485.3329]
Mole of electron = 3,060 / 96,485.3329
Mole of electron = 0.0317146
Moles of Ga = 1/3 [Mole of electron]
Moles of Ga = 1/3 [0.0317146]
Moles of Ga = 0.01057
Mass of Ga = molar mass × Moles of Ga
Mass of Ga = 69.72 × 0.01057
Mass of Ga = 0.73694 gram
How many kg of gas fill a 11.6 gal gas tank
Answer:
43.964
Explanation:
i think i used a calculator so let me know if its wrong
Answer:
39.49 kg
Explanation:
:)
How is excitation in spectroscopy brought about
Answer: the exciation of molecules is brount by absorption of energy in spectroscpy
Explanation:
The standard free energy change, ΔG°', for this reaction is +6.7 kJ/mol. However, the observed free energy change (ΔG) for this reaction in pig heart mitochondria is +0.8 kJ/mol. What is the ratio of [isocitrate]/[citrate] in these mitochondria at 25.0 °C?
Best example of potential energy?
Answer:
water stored in a dam
Explanation:
when the water is in dam it is ready to move bit is not moving
Which statement BEST describes how a golf club does "work" on a golf ball?
(A) When the club hits the ball the club transfers all of its kinetic energy to the ball.
(B) All of the kinetic energy from the club is transferred to the ball as they both move through the air.
(C)
Some of the kinetic energy from the golf club is transferred to the ball and some transforms into sound
and heat, but the total energy remains the same.
(D) The golf club loses kinetic energy when it hits the ball and the ball gains kinetic energy from the air as it
travels
Answer:
C
Explanation:
It looks pretty reasonable to me
g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete the mass balance expressions for the sodium and arsenate species in the final solution.
Answer:
The concentration of the sodium and arsenate ions at the end of the reaction in the final solution
[Na⁺] = 0.05512 M
[HAsO₄²⁻] = 0.00185 M
[AsO₄³⁻] = 0.01714 M
Explanation:
Complete Question
A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.
Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O
From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)
Concentration in mol/L = (Number of moles) ÷ (Volume in L)
Number of moles = (Concentration in mol/L) × (Number of moles)
For Na₂HAsO₄
Concentration in mol/L = 0.03798 M
Volume in L = (500/1000) = 0.50 L
Number of moles = 0.03798 × 0.5 = 0.01899 mole
For NaOH
Concentration in mol/L = 0.03428 M
Volume in L = (500/1000) = 0.50 L
Number of moles = 0.03428 × 0.5 = 0.01714 mole
Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.
Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O
0.01899 0.01714 0 0 (At time t=0)
(0.01899 - 0.1714) | 0 → 0.01714 0.01714 (end)
0.00185 | 0 → 0.01714 0.01714 (end)
Hence, at the end of the reaction, the following compounds have the following number of moles
Na₂HAsO₄ = 0.00185 mole
This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction
NaOH = 0 mole
Na₃AsO₄ = 0.01714 moles
This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction
H₂O = 0.01714 moles
So, at the end of the reaction
Na⁺ has 0.0037 + 0.05142 = 0.05512 mole
(HAsO₄)²⁻ has 0.00185 mole
(AsO₄)³⁻ has 0.01714 mole.
And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L
Hence, the concentration of the sodium and arsenate ions at the end of the reaction is
[Na⁺] = 0.05512 M
[HAsO₄²⁻] = 0.00185 M
[AsO₄³⁻] = 0.01714 M
Hope this Helps!!!
Answer:
[tex]\rm [Na^{+}]= \text{0.055 12 mol/L}[/tex]
[tex]\rm [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}] = \text{0.018 99 mol/L}[/tex]
Explanation:
The overall equation for the reaction is
Na₂HAsO₄ + NaOH ⟶ Na₃AsO₄ + H₂O
1. Mass balance for Na
All the Na⁺ comes from the Na₂HAsO₄ and the NaOH.
The mass balance equation for Na is
[tex]\rm c_{Na^{+}} = 2[Na^{+}]_{Na_{2}HAsO_{4}} + [Na^{+}]_{NaOH}[/tex]
At the moment of mixing and before the reaction started, the total volume had doubled, so the concentrations of each component were halved.
[Na₂HAsO₄] = ½ × 0.037 98 =0.018 99 mol·L⁻¹
[NaOH] = ½ × 0.034 28 = 0.017 14 mol·L⁻¹
[tex]\rm c_{Na^{+}} = 2\times 0.01899 + 0.01714 = 0.03798 + 0.01714\\c_{Na^{+}}= \textbf{0.055 12 mol/L}[/tex]
2. Mass balance for arsenate species
All the arsenate species come from the Na₂HAsO₄.
The reactions involved are
HAsO₄²⁻+ OH⁻ ⇌ AsO₄³⁻ + H₂O
HAsO₄²⁻ + H₂O ⇌ H₂AsO₄⁻ + OH⁻
H₂AsO₄⁻ + H₂O ⇌ H₃AsO₄ + OH⁻
The mass balance equation for arsenate species is
[tex]\rm c_{\text{arsenate}} = [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}][/tex]
At the moment of mixing, the concentration of Na₂HAsO₄ had halved.
[Na₂HAsO₄] = ½ × 0.039 78 = 0.018 99 mol·L⁻¹
[tex]\rm [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}] = \textbf{0.018 99 mol/L}[/tex]
