Ortho and para hydrogen are nuclei forms
How many atom in protons
Answer:
Its atomic number is 14 and its atomic mass is 28. The most common isotope of uranium has 92 protons and 146 neutrons. Its atomic number is 92 and its atomic mass is 238 (92 + 146).
Consider the reaction of 2-chloro-2-methylpentane with sodium iodide.
Assuming no other changes, how would it affect the rate if one simultaneously doubled the concentration of 2-chloro-2-methylpentane and sodium iodide?
A) No effect.
B) It would double the rate.
C) It would triple the rate.
D) It would quadruple the rate.
E) It would increase the rate five times.
Answer:
Explanation:
The reaction between 2 chloro- 2 methyl pentane and sodium iodide takes place through SN2 mechanism . iodide ion is the nucleophile which attacks the substrate . The rate of such reaction depends upon concentration of both the nucleophile and the substrate .
Hence rate of reaction will be increased by 2 x 2 = 4 times.
option D ) is correct.
Explanation:
The given reaction represents the reaction between a tertiary alkyl halide that is 2-chloro-2-methylpentane and a nucleophile that is NaI.
This reaction favors SN1 mechanism which has order one.
So, the given reaction follows first-order kinetics.
For a first-order reaction, the rate law is:
rate =k [A]
That means the rate of the reaction is dependent on the concentration of reactants.
So, when the concentration of the reactant is doubled then, the rate of the reaction is also doubled.
Among the given options the correct answer is option B) It would double the rate.
which of these statements is true about planets? Planets
A. revolve around the sun
B. are spherical in shape
C. rotate in its axis
D. all of the above
Answer:
D. all of the above
Explanation:
A and C are verified by Keplar's laws of planetary motion.
B is verified by the equatorial and polar aces of the Planet.
Answer:
c
Explanation:
they dont have to orbit the sun specifically and are commonly more ovoid than spherical
A second-order reaction has a half-life of 12 s when the initial concentration of reactant is 0.98 M. The rate constant for this reaction is ________ M-1s-1. A) 12
Answer: 0.085 (Ms)⁻¹
Explanation: Half life = 12 s
is the initial concentration = 0.98 M
Half life expression for second order kinetic is:
k = 0.085 (Ms)⁻¹
The rate constant for this reaction is 0.085 (Ms)⁻¹ .
Question 9 of 25
How many hydrogen atoms are in a molecule of table sugar (C12H,2011)?
O A. 12
B. 45
C. 11
D. 22
SUBMIT
D.22
is my answer than welcome
When solid Ni metal is put into an aqueous solution of Pb(NO3)2, solid Pb metal and a solution of Ni(NO3)2 result. Write the net ionic equation for the reaction.
Answer:
[tex]Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to write the complete molecular equation as shown below:
[tex]Pb(NO_3)_2(aq)+Ni(s)\rightarrow Ni(NO_3)_2(aq)+Pb(s)[/tex]
Now, we can separate the nitrates in ions as they are aqueous to obtain:
[tex]Pb^{2+}(aq)+2(NO_3)^-(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+2(NO_3)^-(aq)+Pb(s)[/tex]
And then, we cancel out the nitrate ions as the spectator ones, for us to obtain the net ionic equation:
[tex]Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)[/tex]
Best regards!
Identify the compound in the following group that is most soluble in water. Match the words in the left column to the appropriate blanks in the sentences on the right. ResetHelp Of the three compounds butanoic acid, butane, and butanone, the one that is most soluble in water is . This is because its functional group can form the intermolecular forces with polar water.
butanoic acid strongest hydrocarbon butanone butane alcohol carboxylic acid What carboxylic acid is found in each of the following substances? Drag the appropriate descriptions to their respective bins. Reset rancid butter stinging red ants Methanoic acid Butanoic acid Propanoic acid Ethanoic acid Review Co Draw the structure of methyl butanoate, Draw the molecule on the canvas by choosing buttons from the Tools (for bonds and charges), Atoms toolbars. H: 129 uxo com H o с + N 1 0 S a CH Br р
Answer:
Following are the responses to the given points:
Explanation:
For question 1:
Butanoic acid, butane, and butanone are also the three chemicals most dissolve in water. Its intermolecular force forces are produced by carboxylic acid functional groups with water.
For question 2:
Butanoic acid is a rancid buffer.
Methanoic acid is responsible for the stinging red ants
For question 3:
Methyl butanoate's chemical structure.
The following compound can be identified as
Answer:
3: Lactone
Explanation:
Lactones are defined as carboxylic esters that contain the structure (−C(=O)−O−) which is essentially showing that an ester has now become part of the chemical structure of the ring.
Thus, looking at the question, it has the structure as defined in Lactones.
Thus, we can say that the compound is a Lactone.
what is the hybridisation of the central carbon in CH3C triple bonded to N
Explanation:
the carbon would be sp3 hybridized, and it doesn't matter which carbon, since either of them have a full octet
Which does not result in deviations from linearity in a Beer's law plot of absorbance versus concentration?a. light losses at the cell interface b. all are sources of nonlinearity c. stray radiation d. equilibrium between different forms of the analyte e. a wide bandwidth relative to the width of the absorption band
Answer:
a
Explanation:
Beer-Lambert Law shows the relationship between the factors affecting the absorbance of a sample in relation to the concentration. These factors are:
the concentration c, path length (l), and the molar absorptivity (ε).
As a result, more radiation is assimilated as the concentration rises, and the absorbance rises as well. However, the longer the path length, the increase in the number of molecules and the higher the absorbance.
Thus, the straight-line equation for Beer-Lambert's law is:
A = εcl
From the above explanation, the option that doesn't relate to the deviations from linearity of Beer's law plot is in Option (a).
Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction.
half-reaction identification
Cu+(aq)--->Cu2+(aq) + e- _________
I2(s) + 2e--->2I-(aq) _________
Answer:
Cu+(aq)--->Cu2+(aq) + e- : oxidation
reason: there is loss of electrons.
I2(s) + 2e--->2I-(aq) : reduction
reason: There is reduction of electrons.
which of the following kb values represents the weakest base?
Answer:
the weakest base will have a higher Kb value since it will be closer to an acid than a base
7 kb values represents the weakest base.
What is kb value?Kb is the base dissociation constant which is a measure of how completely a base dissociates into its component ions in water. pKb is define as the negative base-10 logarithm of the base dissociation constant (Kb) of a solution.
Ka is define as the acid dissociation constant while pKa is the -log of this constant. Kb is define as the base dissociation constant, while pKb is the -log of the constant.
The acid and base dissociation constants are usually expressed in terms of moles per liter (mol/L).
Thus, 7 kb values represents the weakest base.
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URGENT- please do by 14th July if possible!!!
1. How do metals react with acids?
2. What are the similarities and differences in the way different metals react with water and acids?
3. Why are some metal is more reactive than others
4. Why is the reactivity of metals so important to us?
5. What the displacement reactions?
6. Why do you displacement reactions happen?
7. Why are they important to us?
8. How are displacement reactions explained as redox reactions?
Thank you!
Answer:
Acids react with most metals to form hydrogen gas and salt. ... When an acid reacts with metal, salt and hydrogen gas are produced
Write the functional isomers of C2H6O?
Answer:
See explanation
Explanation:
Isomers are molecules which have the same molecular formula but different structural formulas. Sometimes, isomers may even contain different functional groups.
The formula C2H6O may refer to an ether or an alcohol. The compound could be CH3CH2OH(ethanol) or CH3OCH3(methoxymethane).
Hence, the functional isomers of the formula C2H6O are ethanol and methoxymethane.
what are the products in a chemical equation located?
Answer:
they are the end results so they are after the yields symbol
Explanation:
why beta carbon hydrogen is easily replaceable but not alpha carbon hydrogen
Answer:
Four common types of reactions involving carbonyl reactions: 1) nucleophilic addition; 2) nucleophilic acyl substitution; 3) alpha substitution; 4) carbonyl condensations. The first two were previously discussed and the second two involve the properties of the carbon directly adjacent to the carbonyls, α carbons.
Alpha-substitution reactions results in the replacement of an H attached to the alpha carbon with an electrophile.
The nucleophile in these reactions are new and called enols and enolates.
Explanation:
The carbon in the carbonyl is the reference point and the alpha carbon is adjacent to the carbonyl carbon.
Hydrogen atoms attached the these carbons denoted with Greek letters will have the same designation, so an alpha hydrogen is attached to an alpha carbon.
Aldehyde hydrogens not given Greek leters.
α hydrogens display unusual acidity, due to the resonance stabilization of the carbanion conjugate base, called an enolate.
Tautomers are readily interconverted constitutional isomers, usually distinguished by a different location for an atom or a group, which is different than resonance.
The tautomerization in this chapter focuses on the carbonyl group with alpha hydrogen, which undergo keto-enol tautomerism.
Keto refers to the tautomer containing the carbonyl while enol implies a double bond and a hydroxyl group present in the tautomer.
The keto-enol tautomerization equilibrium is dependent on stabilization factors of both the keto tautomer and the enol tautomer, though the keto form is typically favored for simple carbonyl compounds.
The 1,3 arrangement of two carbonyl groups can work synergistically to stabilize the enol tautomer, increasing the amount present at equilibrium.
The positioning of the carbonyl groups in the 1,3 arrangement allows for the formation of a stabilizing intramolecular hydrogen bond between the hydroxyl group of the enol and the carbonyl oxygen as well as the alkene group of the enol tautomer is also conjugated with the carbonyl double bond which provides additional stabilization.
Aromaticity can also stabilize the enol tautomer over the keto tautomer.
Under neutral conditions, the tautomerization is slow, but both acid and base catalysts can be utilized to speed the reaction up.
Biological enol forming reactions use isomerase enzymes to catalyze the shifting of a carbonyl group in sugar molecules, often converting between a ketose and an aldose in a process called carbonyl isomerization.
You have 10 pounds of egg whites. You need 6oz to make one serving of cosomme. How many servings can you make?
Answer:
I think you can make 26, hope this helped.
Explanation:
g A high altitude balloon is filled with 1.41 x 104 L of hydrogen gas (H2) at a temperature of 21oC and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is -48oC and the pressure is 63.1 torr
Answer:
1.27 × 10⁵ L
Explanation:
Step 1: Given data
Initial pressure (P₁): 745 TorrInitial volume (V₁): 1.41 × 10⁴ LInital temperature (T₁): 21 °CFinal pressure (P₂): 63.1 TorrFinal volume (V₂): ?Final temperature (T₂): -48 °CStep 2: Convert the temperatures to the Kelvin scale
We will use the following expression.
K = °C + 273.15
K = 21 °C + 273.15 = 294 K
K = -48 °C + 273.15 = 225 K
Step 3: Calculate the final volume of the balloon
We will use the combined gas law.
P₁ × V₁ / T₁ = P₂ × V₂ / T₂
V₂ = P₁ × V₁ × T₂/ T₁ × P₂
V₂ = 745 Torr × 1.41 × 10⁴ L × 225 K/ 294 K × 63.1 Torr
V₂ = 1.27 × 10⁵ L
Next, the chemist measures the volume of the unknown liquid as 0.610 L and the mass of the unknown liquid as 972. g.
Calculate the density of the liquid. Round
your answer to 3 significant digits.
1593.4 g / cm
10
Given the data above, is it possible to identify yes
Answer:
Density of liquid = 1.59 g/cm³
Explanation:
We'll begin by converting 0.610 L to cm³. This can be obtained as follow:
1 L = 1000 cm³
Therefore,
0.610 L = 0.610 L × 1000 cm³ / 1 L
0.610 L = 610 cm³
Finally, we shall determine the density of the liquid. This can be obtained as follow:
Volume of liquid = 610 cm³
Mass of liquid = 972 g
Density of liquid =?
Density = mass / volume
Density of liquid = 972 / 610
Density of liquid = 1.59 g/cm³
The density of any given liquid is equivalent to the mass of the liquid divided by the volume of the liquid.
From the given information, we have:
The mass of the unknown liquid to be = 972 g
The volume of the unknown liquid to be = 0.610 L = 610 cm³
If the formula for calculating [tex]\mathbf{Density = \dfrac{Mass}{Volume}}[/tex]
Then;
[tex]\mathbf{Density \ of \ the \ unknown \ liquid = \dfrac{972 \\ g}{610 \ cm}}[/tex]
[tex]\mathbf{Density \ of \ the \ unknown \ liquid = 1.593442623 \ g/cm}[/tex]
The Density of the unknown liquid ≅ 1.593 g/cm³
In conclusion, the density of the unknown liquid is 1.593 g/cm³ to 3 significant figures.
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Suppose that a certain atom possesses only four distinct energy levels. Assuming that all transitions between levels are possible, how many spectral lines will this atom exhibit
Answer:
Following are the response to the given question:
Explanation:
The number of shells
n = 4
Calculating the spectral line:
[tex]= \frac{n(n-1)}{2}\\\\ = \frac{4(4-1)}{2} \\\\= \frac{4\times 3}{2}\\\\ = \frac{12}{2}\\\\ = 6[/tex]
Which type of element is almost always found as a single atom in nature?
O A. Alkaline earth metal
O B. Halogen
c. Noble gas
D. Oxygen family element
Noble gas elements are almost always found as a single atom in nature.
What is inert gas?An inert gas is a gas that does not undergo chemical reactions under a set of given conditions. The noble gases often do not react with many substances and were historically referred to as inert gases.
All noble gases have the maximum number of electrons in their outer shell; i.e. 2 electrons for helium and 8 for the other five.
Noble gases are monoatomic, which means they exist as single atoms. This is because of their electronic stability.
Thus, noble gas elements are almost always found as a single atom in nature. Hence, option C is correct.
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You will observe a weak acid-strong base titration in this experiment. Select all statements that are true about weak acid-strong base titrations.
A. Weak acid-strong base titrations always start at a higher pH than strong acid-strong base titrations, no matter the initial concentration.
B. The pH is less than 7 at the equivalence point.
C. The pH is greater than 7 at the equivalence point.
D. Half way to the equivalence point, a buffer region is observed.
Answer:
The pH is greater than 7 at the equivalence point.
Explanation:
Equivalence point is the point where the acid reacts with the base as stipulated in the equation of the reaction.
When a weak acid and a strong base are titrated, the pH of the solution at equivalence point is actually found to be around about pH ~ 9.
Hence, for a weak acid and strong base titration, The pH is greater than 7 at the equivalence point.
A titration between a weak acid and a strong base yields a solution whose pH is greater than 7 at the equivalence point.
What are weak acids?Weak acids are acids which only ionize partially in aqueous solutions.
When weak acids are dissolved in water, they produce only few hydrogen ions.
A strong base on the other hand ionizes completely to produce hydroxide ions in aqueous solutions.
The titration of a weak acid and a strong base gives a solution whose pH is greater than 7 at equivalence point.
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Which process takes place when recharging a rechargeable battery?
a- Oxidation occurs at the positive anode.
b- Oxidation occurs at the positive cathode.
c- Oxidation occurs at the negative anode.
d- Oxidation occurs at the positive cathode.
Answer:
the answer is d.) potassium
Explanation:
2. For each of the ionic compounds in the table below, name the compound and explain the rule that you
used in formulating your name for the compound.
Name:
Rule for naming compound:
-PbF4
-NH4NO3
-Li2S
Answer:
2
Explanation:
Lead(|V) fluoride
Ammonium Nitrate
Lithium sulfide
For the rules, I don't know what you were taught. I just do it intuitively since I have done so much chemistry.
The first one the roman numerals represents the charge of the lead which much match the 4- charge from the 4 fluorides.
The second one is just two polyatomic ions which you just have to remember.
The last one is the typical ionic compound naming technique i guess.
For the following reaction, 15.4 grams of chlorine gas are allowed to react with 49.6 grams of sodium iodide. chlorine (g) sodium iodide (s) sodium chloride (s) iodine (s) What is the maximum amount of sodium chloride that can be formed
Answer:
19.3 g of NaCl can be produced
Explanation:
We state the reaction:
Cl₂ (g) + 2NaI (s) → 2NaCl (s) + I₂ (s)
We need to determine limiting reagent:
15.4 g . 1mol /70.9g = 0.217 moles of chlorine
49.6 g . 1mol / 149.89g = 0.331 moles of NaI
Ratio is 1:2. 1 mol of chlorine reacts to 2 moles of NaI
0.217 moles may react to (0.217 . 2)/1 = 0.434 moles of NaI
It is ok to say the NaI is the limting reactant because we need 0.434 moles of it and we only have 0.331.
Ratio is 2:2.
0.331 moles of NaI can produce 0.331 moles of NaCl
We convert mass to moles: 0.331 mol . 58.45g /mol = 19.3 g
2. 27.8 mL of an unknown were added to a 50.0-mL flask that weighs 464.7 g. The total mass of the flask and the liquid is 552.4 g. Calculate the density of the liquid in Lbs/ in3.
Answer:
[tex]d=4.24x10^{-4}\frac{lb}{in^3}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:
[tex]d=\frac{m}{V}[/tex]
Thus, we can find the mass of the unknown by subtracting the total mass of the liquid to the mass of the flask and the liquid:
[tex]m=552.4g-464.7g=87.7g[/tex]
So that we are now able to calculate the density in g/mL first:
[tex]d=\frac{87.7g}{27.8mL}=3.15g/mL[/tex]
Now, we proceed to the conversion to lb/in³ by using the following setup:
[tex]d=3.15\frac{g}{mL}*\frac{1lb}{453.6g}*\frac{1in^3}{16.3871mL}\\\\d=4.24x10^{-4}\frac{lb}{in^3}[/tex]
Regards!
CAN HF USED TO CLEAVE ETHERS EXPLAIN
Answer:
no
Explanation:
Fluoride is not nucleophilic (having the tendency to donate electrons) enough to allow for the use of HF to cleave ethers in protic media(protic solvents are polar liquid compounds that have dissociable hydrogen atoms). The rate of reaction is comparably low, so that heating of the reaction mixture is required.
Oleic acid and elaidic acid are isomeric alkenes.
a. True
b. False
Answer:
False
Explanation:
Because Elaidic acid is an isomer of oleic acid. I really hope this helps you.
What will be the equilibrium temperature when a 245 g block of lead at 300oC is placed in 150-g aluminum container containing 820 g of water at 12.0oC?
Answer:
The correct approach is "12.25°C".
Explanation:
Given:
Mass of lead,
mc = 245 g
Initial temperature,
tc = 300°C
Mass of Aluminum,
ma = 150 g
Initial temperature,
ta = 12.0°C
Mass of water,
mw = 820 g
Initial temperature,
tw = 12.0°C
Now,
The heat received in equivalent to heat given by copper.
The quantity of heat = [tex]m\times s\times t \ J[/tex]
then,
⇒ [tex]245\times .013\times (300-T) = 150\times .9\times (T-12.0) + 820\times 4.2\times (T-12.0)[/tex]
⇒ [tex]3.185(300-T) = 135(T-12.0) + 3444(T-12.0)[/tex]
⇒ [tex]955.5-3.185T=135T-1620+3444T-41328[/tex]
⇒ [tex]43903.5 = 3582.185 T[/tex]
⇒ [tex]T = 12.25^{\circ} C[/tex]
Write the formulas of all species in solution for the following ionic compounds by writing their dissolving equations:
(Use the lowest possible coefficients.)
1. Rubidium hydroxide: __--__+___
2. Sodium carbonate: __--__+__
3. Ammonium selenite:__--__+__
Answer:
1. RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)
2. Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)
3. (NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)
Explanation:
Let's consider the dissolving equations for the following compounds.
1. Rubidium hydroxide
RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)
2. Sodium carbonate
Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)
3. Ammonium selenite
(NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)
