Answer:
140 T or thymine base
Explanation:
Adenine pairs with Thymine in DNA thus the number of adenine will always equal number of thymine (unless some sort of mutation), therefore in this problem you have 140 A so you have 140 T as well. Remember: Adenine (A) and Thymine(T) is equal, & Cytosine (C) and Guanine (G) is equal
An attempt at synthesizing a certain optically active compound resulted in a mixture of its enantiomers. The mixture had an observed specific rotation of 14.1°. If it is known that the specific rotation of the R enantiomer is –28.4°, determine the percentage of each isomer in the mixture. g
Answer:
The percentage of the R-enantiomer is 26.18% while the percentage of the S-enantiomer is 73.82%
Explanation:
If the specific rotation of R enantiomer = -28.4, then the specific rotation of S = +28.4
Now, let us have x = % R, thus
% S = 100-x =y
Hence;
{- 28.4x + 28.4( 100 -x)}/100= 14.1
Thus;
-28.4x + 2840 -28.4x = 1410
-56.8x + 2840 = 1410
-56.8x = 1410-2840
-56.8x = -1430
x = 1430/56.8
x = 26.18%
y = 100-26.18% = 73.82%
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]
Identify the particle represented by each symbol as an alpha particle, a beta particle, a gamma ray, a positron, a neutron, or a proton.
a. 11P
b. 42He
c. +10e
Answer:
[tex]_1^{1} {P}[/tex] is symbol for proton emission in the nucleus.
[tex]_2^{4} {He}[/tex] symbolises alpha emission, equivalent to helium atom emission of a radioactive particle
[tex]+_1^{0} {e}[/tex] is the radiation symbol for positiron particle. which occurs when beta + radioactive decay occurs
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 compounds below are amines?
1. H4C-NH-CH3
2. H3C-NH-C-CH3
H2C-CH3
1 +
H3C-CH2-N-CH3
CH3
N
3. H
4.
Answer:
1. H4C-NH-CH3
2. H3C-NH-C-CH3
H2C-CH3
1 +
H3C-CH2-N-CH3
CH3
N
3. H
4.
.
.
.
.
.
.
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.
The calculated yield for the production of carbon dioxide was 73.4g. When the
experiment was performed in the lab, a yield of 72.3g was produced. What is the
percent yield of carbon dioxide?
Answer:10 grams of CO2
Explanation:
Yeild= exp. yeild÷ thoretical yeild × 100
Yeild= 73.3÷73.4 × 100
Yeild= 0.1 ×100
Yeild= 10
what would happen if you place two positive charges next to each other and let go. would they attract, stay still, or they would repel
Answer:
they would repel
Explanation:
unlike charges attract while like ones repel.
Quantum number of boron
Answer:
The answer is 5.
How many molecules of water are in 5 moles
Answer:
3.011x10^24 molecules
Explanation:
1 mole=6.022x10^23 molecules
5 moles*6.022x10^23 molecules/mole=3.011x10^24 molecules
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.
In this reaction: Mg (s) + I₂ (s) → MgI₂ (s)
If 2.34 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what is the percent yield?
Answer:
98.9%
Explanation:
2 moles of I₂ are required for each mole of Mg, so the reaction is limited by the available I₂. The 3.56 moles of I₂ should react with 1.78 moles of Mg to produce 1.78 moles of MgI₂. Instead, we get 1.76 moles of MgI₂.
The yield is 1.76/1.78 × 100% ≈ 98.876%
The yield is 98.9% of the quantity expected based on available reactants.
which best describes a mixture.
A it has a single composition and it has a set of characteristics
B it can have different compositions but it has a set of charactaristics that does not change
C it has a single composition but it has a set of characteristics that does change
D it can have different compositions and it has a set of characteristics that does change
Answer:
B) It can have different compositions, but it has a set of characteristics that does not change.
Explanation:
On e d g e n u i t y
I believe the answer is d lmk if im wrong or right
A 33.0−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a heat capacity of 9.20 J/K. If the final temperature of the system is 31.10°C, what is the specific heat capacity of the alloy? J g·°C
Answer:
THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K
Explanation:
Mass of alloy = 33 g
Initial temperature of alloy = 93°C
Mass of water = 50 g
Initail temp. of water = 22 °C
Heat capacity of calorimeter = 9.20 J/K
Final temp. = 31.10 °C
specific heat of alloy = unknown
specific heat capacity of water = 4.2 J/g K
Heat = mass * specific heat * change in temperature = m c ΔT
Heat = heat capcity * chage in temperature = Δ H * ΔT
In calorimetry;
Heat lost by the alloy = Heat gained by water + Heat of the calorimeter
mc ΔT = mcΔT + Heat capacity * ΔT
33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)
33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1
2042.7 C = 1911 + 83,72
C = 1911 + 83.72 / 2042.7
C = 1994.72 /2042.7
C =0.9765 J/g K
The specific heat of the alloy is 0.9765 J/ g K
How to treat stream water for drinking
Answer:
Explanation:
Boiling.
Use water filter
Use Ultraviolet Light.
Use chlorine drops
I would recomade boiling as the main
because its the easiest and cheapest Or water filter if you have one
Select the correct answer.
What effect does an increase in products have on the reaction rate of a mixture at equilibrium?
A.
The forward reaction rate increases.
B.
Both the forward and the reverse reaction rates decrease.
Both the forward and the reverse reaction rates increase.
D.
The reverse reaction rate increases.
Reset
Next
Answer:
At equilibrium the rate of the forward reaction is equal to the rate of the backward reaction.
When the product of a reaction at equilibrium is increased the equilibrium will shift left or to the reactant side. As a result the excess product will get converted to reactant. This is in accordance to Le Chatelier's principle.
Le Chatelier's principle states that when a system is subjected to stress the equilibrium will shift in a direction to minimize effect of the stress.
Thus the products added to the system at equilibrium will make the equilibrium shift to the reactant side, the rate of the reverse or backward reaction will increase.
Explanation:
Hope This Helps Amigo!
A new non-electrolyte molecule is discovered. When 241 mg of the molecule is dissolved in 250.0 mL of water, it has an osmotic pressure of 0.072 atm at 25 oC.What is the molar mass of the molecule
Answer:
327.89g/mol
Explanation:
Step 1:
The following data were obtained from the question:
Van 't Hoff factor (i) = 1 (since the molecule is non-electrolyte)
Temperature (T) = 25°C = 25°C + 273 = 298K
Gas constant (R) = 0.0821 atm.L/Kmol
Mass of molecule = 241mg
Volume of water = 250mL
Molarity (M) =?
Osmotic pressure (Π) = 0.072 atm
Step 2:
Determination of the molarity of the molecule.
This can be obtained as follow:
Π = iMRT
0.072 = 1 x M x 0.0821 x 298
Divide both side by 0.0821 x 298
M = 0.072 / (0.0821 x 298)
M = 2.94×10¯³ mol/L
Step 3:
Determination of the number of mole of the molecule. This can be obtained as follow:
Molarity = 2.94×10¯³ mol/L
Volume of water = 250mL = 250/1000 = 0.25L
Mole of molecule =..?
Molarity = mole /Volume
2.94×10¯³ = mole / 0.25
Cross multiply
Mole of molecule = 2.94×10¯³ x 0.25
Mole of molecule = 7.35×10¯⁴ mole.
Step 4:
Determination of the molar mass of the molecule.
Mole of molecule = 7.35×10¯⁴ mole.
Mass of molecule = 241mg = 241×10¯³g
Molar mass of molecule =..?
Mole = Mass /Molar Mass
7.35×10¯⁴ = 241×10¯³/ Molar Mass
Cross multiply
7.35×10¯⁴ x molar mass = 241×10¯³
Divide both side by 7.35×10¯⁴
Molar Mass = 241×10¯³/7.35×10¯⁴
Molar Mass = 327.89g/mol
Therefore, the molar mass of the molecule is 327.89g/mol
In redox half-reactions, a more positive standard reduction potential means I. the oxidized form has a higher affinity for electrons. II. the oxidized form has a lower affinity for electrons. III. the reduced form has a higher affinity for electrons. IV. the greater the tendency for the oxidized form to accept electrons.
Answer:
The 1st and 4th options are correct
I.the oxidized form has a higher affinity for electrons
IV. the greater the tendency for the oxidized form to accept electrons
Explanation:
Half reaction can be described as the oxidation or reduction reaction in a redox reaction.it is In the redox rection there is a change in the oxidation states of Chemical species involved. the oxidized form in the redox has a higher affinity for electrons and the greater the tendency for the oxidized form to accept electrons.
Standard reduction potential which is also referred to as standard cell potential can be described as the potential difference that exist between cathode and anode of the cell. In the standard reduction potential most times the species will be reduced which is usually analysed in a reduction half reaction.
(Standard Hydrogen Electrode) is utilized when determining the Standard reduction or potentials of a chemical specie. this is because of Hydrogen having zero reduction and oxidation potentials, as a result of this a measured potential of any species is compared with that of Hydrogen, the difference helps to know the potential reduction of that particular specie.
ions always have the same electronic structure as elements in which group of the periodic table?
Answer:
In 0 group of the periodic table
Explanation:
So they will not react with other atoms because they have a full outer shell of electrons and an overall charge of 0.
Hope it helps.
A growing concern in agricultural and food chemistry is the presence of residues in food. We use many forms of organic chemicals in agriculture and food chemistry and there is growing concern as to how safe these materials are. Choose an organic chemical used in agricultural of food chemistry and report on the functional groups contained in your compound, the uses of the compound, and the safety of that compound.
Answer:
Monosodium Glutamate (MSG) is a chemical which is used in agricultural of food chemistry.
Explanation:
Monosodium Glutamate (MSG) is a chemical that is used in types of different food as food additives. The functional group that is present in Glutamate are carboxylic acid and amine. This chemical is used in different types of foods which is responsible for enhancing the taste of the food. Monosodium Glutamate is safe if it is used in moderate dose but adversely affected when it is used in large amount.
what is the correct ionic equation, including all coefficients, charges, and phases for the following sets of reactants? Assume that the contribution of protons from H2SO4 is near 100%.
Ba(OH)2(aq)+H2SO4(aq) —>
help, I have no clue
Answer:
Ba(OH)2(aq)+H2SO4(aq) gives us 2BaH+H2O
Explanation:
The emission line used for zinc determinations in atomic emission spectroscopy is 214 nm. If there are 6.00×1010 atoms of zinc emitting light in the instrument flame at any given instant, what energy (in joules) must the flame continuously supply to achieve this level of emission?
Wine goes bad soon after opening because the ethanol dissolved in it reacts with oxygen gas to form water and aqueous acetic acid , the main ingredient in vinegar. Calculate the moles of water produced by the reaction of of oxygen. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.
Answer:
1.7 moles of ethanol would be needed.
Explanation:
* Calculate the moles of ethanol needed to produce 1.70mol of water. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
First off, we have to state the equation for the reaction.
So we know that;
ethanol + oxygen → acetic acid + water
This leads us to;
C2H5OH + O2 → CH3COOH + H2O
1 1 1 1
To obtain the moles of ethanol needed to produce 1.70mol of water, we look at the stoichiometry of the reaction above.
1 mol of ethanol produces 1 mole of water
x mol of thanol would produce 1.7 mol of water
Thus we have;
1 = 1
x = 1.7
x = 1.7 moles of ethanol would be needed.
Select the correct answer. What is heat of vaporization?
A. It is the heat required to change a substances temperature by 1C
B. It is the heat required to change a gram of substance from a solid to a liquid
C. It is the heat required to change a substance from a solid directly to a gas
D. It is the heat required to change a gram of substance from a liquid to a gas
E. It is the heat required to separate one substance into two substances
Answer:
D) it is the heat required to change a gram of substance from a liquid to a gas
Explanation:
idk i think its correct but if its wrong just let me know
Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. The concentration of fluoride ions after the addition of 5.00 mL of 0.0100 M HCl to 25.0 mL of this solution is __________ M.
a. 0.0980
b. 0.0817
c. 0.0167
d. 0.0850
e. 0.00253
Answer:
The answer is "Option b"
Explanation:
In this question first we calculates the moles in F-, HF, and in HCL, which can be defined as follows:
Formula:
[tex]\ Number \ of \ moles\ = \ Molarity \times \ Volume \ in \ litter[/tex]
[tex]\ moles \ in\ F- = 0.100 \ M \times 0.0250 L\\\\[/tex]
[tex]=\ 0.0025 \ moles[/tex]
[tex]\ moles \ in \ HF \ = 0.126M \times 0.0250 L[/tex]
[tex]= 0.00315 \ moles[/tex]
[tex]\ moles \ in \ HCl = 0.0100M \times 0.00500 L[/tex]
[tex]= 0.00005 \ moles[/tex]
[tex]\ Reaction: \\\\F - + H+ \rightarrow HF[/tex]
[tex]\Rightarrow \ moles \ in \ F- = 0.0025 \\\\\Rightarrow \ moles \ in \ H+ = 0.00005 \\\\ \Rightarrow \ moles \ in \ HF = 0.00315\\\\ \ total \ moles = 0.00250 -0.0000500 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00315 + 0.00005\\\\\ total \ moles =0.00245 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00245[/tex]
[tex]\ total \ volume \ in \ the \ solution = \ V = \ 0.0300 L\\\\ after \ addition \ of \ HCl \ the \ concentration \ of \ F- \ = 0.00245\ moles \div V[/tex]
[tex]=\frac{ 0.00245 \ moles }{0.0300L}\\\\= \frac{245 \times 10^4}{300 \times 10^5} \\\\= \frac{245}{3000} \\\\ = 0.0817 M[/tex]
Calculate the osmotic pressure of a solution prepared by dissolving 65.0 g of Na2SO4 in enough water to make 500 mL of solution at 20°C. (Assume no ion pairing – in other words, assume that the electrolyte completely dissociates into its constituent ions.)
Answer:
66.0 atm
Explanation:
We can calculate the osmotic pressure (π) using the following expression.
[tex]\pi = i \times M \times R \times T[/tex]
where,
i: van 't Hoff indexM: molarityR: ideal gas constantT: absolute temperatureStep 1: Calculate i
Sodium sulfate completely dissociates according to the following equation.
Na₂SO₄ ⇒ 2 Na⁺ + SO₄²⁻
Since it produces 3 ions, i = 3.
Step 2: Calculate M
We can calculate the molarity of Na₂SO₄ using the following expression.
[tex]M = \frac{mass\ of\ solute }{molar\ mass\ of\ solute\ \times liters\ of\ solution} = \frac{65.0g}{142.04g/mol \times 0.500L} =0.915M[/tex]
Step 3: Calculate T
We will use the following expression.
K = °C + 273.15
K = 20°C + 273.15 = 293 K
Step 4: Calculate π
[tex]\pi = 3 \times 0.915M \times \frac{0.08206atm.L}{mol.K} \times 293K =66.0 atm[/tex]
Fractionation of Crude Oil Select the correct ranking of the following alkanes according to the height reached in a fractionating column, highest first: butane, heptadecane, dodecane, ethane, decane Select the correct ranking of the following alkanes according to the height reached in a fractionating column, highest first:
butane, heptadecane, dodecane, ethane, decane
A. ethane > butane > decane > dodecane > heptadecane
B. heptadecane > > dodecane > decane butane > ethane
C. ethane > butane > decane> heptadecane >
D. dodecane butane > ethane > decane > dodecane > heptadecane
Answer:
A. ethane > butane > decane > dodecane > heptadecane
Explanation:
In fractionating column, crude oil is separated by means of fractional distillation due to the wide range of boiling point of the crude products such as ethane, propane, butane pentane etc.
The product with the least weight rises to top height while the product with highest weight will move down.
For the given hydrocarbon products, the ranking according to their molecular weight, starting with the lighter product to heavier product is
ethane (C2), butane (C4), decane(C10), dodecane (C12), heptadecane(C17).
Thus, the correct ranking, starting with the product that will rise highest is ethane > butane > decane > dodecane > heptadecane
A stock solution of HNO3 is prepared and found to contain 14.9 M of HNO3. If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, what is the concentration of the diluted solution
Answer:
[tex]0.745~M[/tex]
Explanation:
In this case, we have a dilution problem. So, we have to use the dilution equation:
[tex]C_1*V_1=C_2*V_2[/tex]
Now, we have to identify the variables:
[tex]C_1~=~14.9~M[/tex]
[tex]V_1~=~25~mL[/tex]
[tex]C_2~=~?[/tex]
[tex]V_2~=~0.5~L[/tex]
Now, we have different units for the volume, so we have to do the conversion:
[tex]0.5~L\frac{1000~mL}{1~L}=~500~mL[/tex]
Now we can plug the values into the equation:
[tex]C_2=\frac{14.9~M*25~mL}{500~mL}=0.745~M[/tex]
I hope it helps!
The core of the pyruvate dehydrogenase complex is made up of eight catalytic________that make up the_______component.
a. monomers; E1b. dimers; E2c. dimers; E3d. trimers; E2
Answer:
(D.)
The core of the pyruvate dehydrogenase complex is made up of eight catalytic trimers that make up the E2 component.
Explanation:
Eight trimers assemble as a hollow truncated cube, which forms the core of the multi-enzyme complex, known as the E2 complex in human pyruvate dehydrogenase complex.
Hi I have a lab for Chemistry I am struggling with. I have to do calculations given the following information
1. Mass of evaporating dish plus sample 26.57 g
2. Mass of evaporating dish 24.29 g
3. Mass of evaporating plus NaCl 68.66 g
4. Mass of evaporating dish 67.84 g
5. Mass of filter paper plus sand 37.69 g
6. Mass of filter paper 36.34 g
CALCULATIONS AND CONCLUSIONS
1. Calculate the mass of unknown mixture g
2. Calculate the mass of NaCl recovered g
3. Calculate the mass of sand recovered g
4. Calculate the percentage of NaCl in your unknown mixture %
5. Calculate the percent sand in your unknown mixture %
6. Calculate the total mass of sand and salt recovered g
7. Calculate the percent recovery of the components %
Answer:
1. 2.28 g
2. 0.82 g
3. 1.35 g
4. 36 %
5. 59 %
6. 2.17 g
7. 95 %
Explanation:
Hello,
1. In this case, the mass of the unknown mixture is obtained by subtracting the mass of the dish plus sample and the mass of the dish:
m = 26.57 g- 24.29 g = 2.28 g
2. In this case, the mass of the NaCl recovered is obtained by subtracting the mass of the dish plus NaCl and the mass of the dish:
m = 68.66 g- 67.84 g = 0.82 g
3. In this case, the mass of the sand recovered is obtained by subtracting the mass of the filter paper plus sand and the mass of the filter paper:
m = 37.69 g- 36.34 g = 1.35 g
4. The percentage of NaCl is computed by considering its mass and the mass of the unknown mixture:
% NaCl = 0.82 g / 2.28 g * 100 % = 36 %
5. The percentage of sand is computed by considering its mass and the mass of the unknown mixture:
% sand = 1.35 g / 2.28 g * 100 % = 59 %
6. Here,we have to add the mass of NaCl and sand:
m = 0.82 g + 1.35 g = 2.17 g
7. Finally, the percent recovery is obtained by diving the total recovered mass by the total obtained mass of the mixture:
% recovery = 2.17 g / 2.28 g * 100 % = 95 %
Best regards.
