A molecule of composition is replicated in a solution containing unlabeled (not radioactive) GTP, CTP, and TTP plus adenine nucleoside triphosphate with all its phosphorus atoms in the form of the radioactive isotope 32P. Will both daughter molecules be radioactive
Answer:
Please find the complete question in the attached file.
Explanation:
It would only be radioactive if the DNA molecule that employed the poly-T rand as templates. Its other molecule of the daughter would not have been radioactive as it did not need dATP for its replication. While each strand of the second molecule includes t, simultaneous reproduction dATP from both daughter molecules is needed so that each of those is radioactive.
Two planets have similar masses but differ in their distances from the Sun.
Which planet experiences a greater gravitational force with the Sun?
A. The planet that is smaller in volume
B. The planet that is larger in volume
C. The planet that is farther from the Sun
D. The planet that is closer to the Sun
Answer:
D. The planet that is closer to the Sun
A 250ml sample of sir at 373.15k is warmed at 473.15 at constant pressure. What is the volume of the air sample at the new temperature
Answer:
317 mL.
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 250 mL
Initial temperature (T₁) = 373.15 K
Final temperature (T₂) = 473.15 K
Pressure = Constant
Final volume (V₂) =?
The final volume of the air sample can be obtained by using the Charles' law equation as illustrated below:
V₁/T₁ = V₂/T₂
250 / 373.15 = V₂ / 473.15
Cross multiply
373.15 × V₂ = 250 × 473.15
373.15 × V₂ = 118287.5
Divide both side by 373.15
V₂ = 118287.5 / 373.15
V₂ = 317 mL
Therefore, the final volume of the air sample is 317 mL
Question 13 of 32
The pH of a sample of seawater is 8.0. What is the hydrogen ion concentration of seawater?
A. 8.0 M
B. 1 x 108 M
C. 6.0 M
D. 1 x 10-8 M
Answer:
D
Explanation:
Given the balanced reaction: Zn + 2HCl → H2 + ZnCl2
If 5 grams of each reactant are available for the reaction and HCl is known
to be the limiting reactant, which of the following is correct?
O Both reactants will be completely used up.
O There will be excess of both reactants remaining.
O HCl will be completely used up while Zn will remain in excess.
O Zn will be completely used up while HCl will remain in excess.
Answer:
O HCl will be completely used up while Zn will remain in excess.
Explanation:
Zn + 2HCl → H₂ + ZnCl₂In reactions involving two reactants, if one of them is the limiting reactant then the other one has to be the reactant in excess.
Meaning that in this case, the reaction will proceed until HCl is completely used up, and a certain amount of Zn will remain (thus being the reactant in excess).
Identify the compound with the lowest dipole moment. Identify the compound with the lowest dipole moment. CH3CH2CH3 CH3OCH3 CH3CHO CH3OH CH3CN
Answer:
CH3CH2CH3
Explanation:
Dipole moment is the measure of the polarity of a chemical bond. It is the extent of charge separation in a molecule.
Dipole moment is the product of the magnitude of charge and the distance separating the charges from each other.
The molecule having the lowest dipole moment among the options is the molecule that has the least polarity. The least polar molecule among the options is CH3CH2CH3, it has no polar bonds in its structure.
We have that for the Question "Identify the compound with the lowest dipole moment. Identify the compound with the lowest dipole moment. CH3CH2CH3 CH3OCH3 CH3CHO CH3OH CH3CN "
it can be said that
[tex]CH_3CH_2CH_3 have the lowest dipole moment[/tex]From the question we are given
CH3CH2CH3
CH3OCH3
CH3CHO
CH3OH
CH3CN
Generally alkanes have the lowest dipole moment, they have C-H bond which are non polar.
Therefore,
[tex]CH_3CH_2CH_3[/tex] have the lowest dipole moment
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help me plsssssssss.... im timed
Answer:
a
Explanation:
a
Answer:
B. Isotopes of the element.
Explanation:
Isotopes are basically atoms of the same element that contain the same number of protons, but a different number of neutrons.
2KClO3 (s)⇄2KCl (s)+ 3O2 (aq) equilibrium constant
Answer: The equilibrium constant for the given chemical reaction is [tex][O_2]^3[/tex]
Explanation:
The equilibrium constant is defined as the ratio of the concentration of the products to the concentration of reactants each raised to the power of their stoichiometric coefficients.
The concentration of all the solids and liquids are considered to be 1 in the expression of equilibrium constant
For the given chemical equation:
[tex]2KClO_3(s)\rightleftharpoons 2KCl(s)+3O_2(aq)[/tex]
The expression of equilibrium constant follows:
[tex]K_{eq}=[O_2]^3[/tex]
Hence, the equilibrium constant for the given chemical reaction is [tex][O_2]^3[/tex]
As a result of the particles in a gas being in constant motion, gas has a _______.
variable volume
variable Pressure
variable Shape
variable mass
Answer:
i think it's variable pressure
if not soo advance sorry :)
An elementary step is defined as a chemical collision in a reaction mechanism. A collection of different types of collisions makes up the reaction mechanism, so elementary steps provide a molecular view of the overall reaction.
a. True
b. False
what climate zone is asia?
Answer:
The Tundra Climate
Explanation:
:) hope this helps
Phosphorus -32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The half-life of phosphorus-32 is 14.3 days. What mass of phosphorus-32 is left from an original sample of 175 mg of Na332PO4 after 35.0 days
Answer: The mass of P-32 left from the original sample is 32.07 mg
Explanation:
All radioactive decay processes follow first-order reactions.
Calculating rate constant for first order reaction using half life:
[tex]t_{1/2}=\frac{0.693}{k}[/tex] .....(1)
[tex]t_{1/2}[/tex] = half life period = 14.3 days
k = rate constant = ?
Putting values in equation 1:
[tex]k=\frac{0.693}{14.3days}\\\\k=0.0485days^{-1}[/tex]
The integrated rate law equation for first-order kinetics:
[tex]k=\frac{2.303}{t}\log \frac{a}{a-x}[/tex] ......(2)
Given values:
a = initial concentration of reactant = 175 mg
a - x = concentration of reactant left after time 't' = ? mg
t = time period = 35.0 days
Putting values in equation 2:
[tex]0.0485days^{-1}=\frac{2.303}{35.0 days}\log (\frac{175}{a-x})\\\\\log (\frac{175}{a-x})=\frac{0.0485\times 35.0}{2.303}\\\\\log (\frac{175}{a-x})=0.737\\\\\frac{175}{a-x}=10^{0.737}\\\\a-x=\frac{175}{5.457}=32.07mg[/tex]
Hence, the mass of P-32 left from the original sample is 32.07 mg
1. Complete the following chart for the following atoms:
Element
Mass
number
Number of
electrons
Number of
neutrons
Atomic
Number
(number of
protons)
Potassium
16
56
26
What is the atomic symbol (the nuclide) for the isotope with 15 protons and 16
neutrons? Your answer must be in the form shown:
AX
You must show the correct numbers for A and Z
and have the correct symbol (Z). (3 points)
Explanation:
If there are 15 protons, 15 nuclear particles of unit positive charge, then
Z
=
15
. Now
Z
≡
the atomic number
, and you look at your copy of the Periodic Table, and you find that for
Z
=
15
, the element phosphorus is specified.
But we are not finished. Along with the 15 defining protons, there are also 16 neutrally charged, massive nuclear particles, 16 neutrons, and the protons and neutrons together determine the atomic mass. The isotope is thus
31
P
, which is almost 100% abundant, and an important nucleus for
NMR spectroscopy.
The limiting reactant in a chemical reaction is the reactant __________ Select one: A. for which you have the lowest mass in grams. B. which has the lowest coefficient in the balanced equation. C. which has the lowest molar mass. D. which is left over after the reaction has gone to completion. E. None of the above.
Answer:
i think its A
is it a physical or chemical change when a candle is lit
Determine the electron geometry, molecular geometry, and idealized bond angles for each of the following molecules. CF4CF4 NF3NF3 OF2OF2 H2SH2S In which cases do you expect deviations from the idealized bond angle
Answer:
CF4
Molecular geometry- tetrahedral
Electron geometry- tetrahedral
NF3
-molecular geometry - trigonal pyramidal
Electron geometry - tetrahedral
OF2
Molecular geometry - bent
Molecular geometry - tetrahedral
H2S
Molecular geometry- bent
Electron geometry - tetrahedral
Explanation:
According to Valence Shell Electron Pair Repulsion Theory, the shape of a molecule depends on the number of electron pairs on the valence shell of the central atom in the molecule.
For all the compounds listed, the central atom has four points of electron density. This correspond to a tetrahedra electron pair geometry. The presence of lone pairs on the central atom of OF2,NF3 and H2S accounts for the departure of the observed molecular geometry from the geometry and idealized bond angle predicted on the basis of the VSEPR theory.
Which commercial technology commonly uses plasmas?
a radio
a race car
a television
a microwave oven
Answer:
A television is commercial technology commonly uses plasmas.The information code that an organism inherits can best be referred to as its -
O A genotype
B. territory
C. species
D. kingdom
Answer:
it will be no.A genotype
How many grams of Al2O3 is extracted from 250. g of FeO?
Answer:
[tex]m_{Al_2O_3}=118.27gAl_2O_3[/tex]
Explanation:
Hello there!
In this case, if we consider the following chemical reaction, whereby Al2O3 is produced from Al and FeO:
[tex]3FeO+2Al\rightarrow 3Fe+Al_2O_3[/tex]
Thus, since there is 3:1 mole ratio of FeO to Al2O3, it turns out feasible for us to use their molar masses, 71.844 g/mol and 101.96 g/mol respectively, to obtain the grams of the latter as follows:
[tex]m_{Al_2O_3}=250.gFeO*\frac{1molFeO}{71.844gFeO}*\frac{1molAl_2O_3}{3molFeO} *\frac{101.96gAl_2O_3}{1molAl_2O_3}\\\\m_{Al_2O_3}=118.27gAl_2O_3[/tex]
Regards!
A 15.0 mL urine from a dehydrated patient has a density of 1.019g/mL. What is the mass of the sample, reported in mg?
Answer:
Mass of sample in mg = 15,285 mg
Explanation:
Given:
Volume of urine sample = 15 ml
Density of sample = 1.019 g/ml
FInd:
Mass of sample in mg
Computation:
Mass = density x volume
Mass of sample in mg = Volume of urine sample x Density of sample
Mass of sample in mg = 1.019 x 15
Mass of sample in mg = 15.285 gram
Mass of sample in mg = 15.285 x 1,000
Mass of sample in mg = 15,285 mg
Why is iodine always Used in a solution containing excess I2
Answer:
If a standard iodine solution is used as a titrant for an oxidizable analyte, the technique is iodimetry. If an excess of iodide is used to reduce a chemical species while simultaneously forming iodine.
Iodine always used in a solution excess KI is given to aid in the solubilization of free iodine, which would be insoluble in clean water during normal circumstances.
What is Iodine?
Iodine is a kind of element which are mainly used in iodometry titration. It can be represented by I.
What is solution?A solution would be a homogenous mixture of two components, usually a solute as well as a solvent.
Iodimetry would be a technique that uses a standard iodine solution as a titrant for such an oxidizable analyte. When an excessive amount of iodide is used to decrease a chemical while somehow producing iodine.
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Please select the word from the list that best fits the definition
Include skin, the respiratory system, the circulatory system, and inflammation
A. antibody
B. antigen
C.Natural Defences
D.active immunity
Answer:
I choose D option because may be it's correct
I believe it is c
Explanation:
because if u think about it and also do some research you would see that the circulatory system is a strong part of your body which can help u through natural defences if this sounds weird it's all in research but if it ain't c dont blame me for ruining your life- lol but yeah I think its c
gress.
Newton's third law of motion
Is the following statement true or false?
When a large truck pushes a small car with a given force, the small car is applying an equal and opposite
force on the truck.
Choose 1 answer:
True
False
Answer:
true
Explanation:
the small car also has gravity making it heavy
If 0.5 L of O2(g) reacts with H, to produce 1 L of H2O(g), what is the volume of
H2O(g) obtained from 1 L of O2(g)?
-
0.5 L
2.5 L
2 L
1.5 L
Answer:
2 L
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2H₂ + O₂ —> 2H₂O
From the balanced equation above, we can see clearly that 1 L of O₂ reacted to 2 L of H₂O.
This implies that 2 L of H₂O can be obtained by the reaction of 1 L of O₂.
Thus, option 3 gives the correct answer to the question.
Hydrogen bonds within liquid water are attractions between protons and hydroxide ions. are dipole-dipole attractions. are ion-induced dipole attractions. are attractions between protons and oxygen nuclei. are attractions between two hydrogen atoms.
Answer:
true because the bonds cannot be broken down
Calculate the mass of hydrogen formed when 26.98 g of aluminum reacts with excess hydrochloric acid according to the following balanced chemical equation: 2 Al + 6 HCl → 2 AlCl3 + 3 H2
Answer: The mass of hydrogen formed when 26.98 g of aluminum reacts with excess hydrochloric acid according to the given balanced equation is 3.03 g.
Explanation:
The given balanced reaction equation is as follows.
[tex]2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}[/tex]
Here, the mole ration of Al and hydrogen produced is 2 : 3
As mass of aluminum is given as 26.98 g. So, moles of aluminum (molar mass = 26.98 g/mol) is as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{26.98 g}{26.98 g/mol}\\= 1 mol[/tex]
So, when 1 mole of Al reacted then 1.5 moles of hydrogen is produced as per the given mole ratio.
Therefore, mass of hydrogen formed is calculated as follows.
[tex]mass = moles \times molar mass\\= 1.5 mol \times 2.02 g/mol\\= 3.03 g[/tex]
Thus, we can conclude that the mass of hydrogen formed when 26.98 g of aluminum reacts with excess hydrochloric acid according to the given balanced equation is 3.03 g.
The sp of strontium carbonate, SrCO3, is 5.60×10−10 . Calculate the solubility of this compound in g/L.
Answer:
3.50 × 10⁻³ g/L
Explanation:
Step 1: Make an ICE chart for the solution of SrCO₃
"S" represents the molar solubility.
SrCO₃(s) ⇄ Sr²⁺(aq) + CO₃²⁻(aq)
I 0 0
C +S +S
E S S
The solubility product constant (Ksp) is:
Ksp = [Sr²⁺] [CO₃²⁻] = S²
S = √Ksp = √5.60 × 10⁻¹⁰ = 2.37 × 10⁻⁵ M
Step 2: Convert "S" to g/L
The molar mass of SrCO₃ is 147.63 g/mol.
2.37 × 10⁻⁵ mol/L × 147.63 g/mol = 3.50 × 10⁻³ g/L
Compound A has the formula C8H8. It reacts rapidly with acidic KMnO4 but reacts with only 1 equivalent of H2 over a palladium catalyst. On hydrogenation under conditions that reduce aromatic rings, A reacts with 4 equivalents of H2, and hydrocarbon B, C8H16, is produced. The reaction of A with KMnO4 gives CO2 and a carboxylic acid C, C7H6O2.
Required:
Draw the structure of compound B below.
Answer:
C8H16 (Ethylcyclohexane).
Explanation:
From the given information:
Compound A is an alkene because it interacts with 1 unit of hydrogen across a palladium catalyst.
Also, we are given another hint that:
Compound A needs 4 equivalence of H2 to hydrogenate under circumstances that decrease aromatic rings, indicating that it is a phenyl substituted alkene.
Compound A with formula C8H8 reacts instantly with KMnO4 to produce CO2, as well as carboxylic acid, points out that Compound acts as a terminal alkene.
Therefore, we can opine that compound A is a terminal phenyl substituted alkene whose formula = C8H8 (Styrene)
The diagrammatic expression of the compound can be seen below.
However, in the presence of the palladium catalyst, the reduction of Compound A with 4 units of hydrogen produces Compound B: C8H16 (Ethylcyclohexane).
the density of gold is 19.3 g/ml. if you had a cube made of pure gold that weighs 58.1 pounds what would be the lengh of a side of this cube
Answer:
11.1 cm
Explanation:
First we convert 58.1 pounds into grams:
58.1 lb * [tex]\frac{453.592g}{1lb}[/tex] = 26353.69 gThen we calculate the volume of the gold cube, using the given density:
26353.69 g ÷ 19.3 g/mL = 1365.48 mL
1365.48 mL is equal to 1365.48 cm³.With the volume of a cube we can calculate the side length:
Length = ∛(1365.48 cm³) = 11.1 cma solution that contains a large amount of solute would be described as what
Answer:
A concentrated solution is one that has a relatively large amount of dissolved solute. A dilute solution is one that has a relatively small amount of dissolved solute.
