Answer:
False
Explanation:
Electrophilic addition reactions may be under kinetic or thermodynamic control. Whether the reaction is under kinetic or thermodynamic control is easily deducible from the reaction time.
Shorter reaction time often reflect kinetic control while longer reaction reaction times favour thermodynamic control.
Hence, electrophilic addition reaction of conjugated dienes that occur at high temperature and/or long reaction times (reversible conditions) are said to be under thermodynamic and not kinetic control.
You are inside a room with a temperature of 11°C. You step outside and the temperature is 100°C. What is the AT?
A) 0.76°C
B) 157°C
C) 89°C
D) 6052°C
E) 1.31°C
F) 21°C
Answer:
89°c
Explanation:
i think this is the answer cause the temperature changed from 11 to 100 and so the atmospheric temperature would be the change in temperature
100-11=89°c
I hope this helps and sorry if it's wrong
#19.
An unknown sample weighs 45.2 g and takes 58.2 kJ to vaporize. What is
its heat of vaporization?
Manganese-55 has _____neutrons.
55 Mn
25
A. 55
B. 30
C. 25
QUESTION:- Manganese-55 has _____neutrons.
OPTIONS :-
A. 55
B. 30
C. 25
ANSWER:- NUMBER OF NEUTRONS IS EQUAL TO THE DIFFERENCE BETWEEN THE MASS IF THE ATOM AND ATOMIC NUMBER
SO DIFFERENCE IS EQUAL TO :- 55-25 = 30 NEUTRONS.
SO THERE IS 30 NEUTRONS IN SINGLE ATOM OF THE MANGANESE-55 ATOM.
Answer:
the mass of an atom is the sum of proton and neutron which are both concentrated in nocleus of an atom. from the question the mass is given as 55 and the proton is 25.
lution: What is the molarity of 245 g of H, SO4 dissolved in 1.00 L of solution?
[tex]\\ \large\sf\longmapsto H_2SO_4[/tex]
[tex]\\ \large\sf\longmapsto 2(1u)+32u+4(16u)[/tex]
[tex]\\ \large\sf\longmapsto 2u+32u+64u[/tex]
[tex]\\ \large\sf\longmapsto 98u[/tex]
[tex]\\ \large\sf\longmapsto 98g/mol[/tex]
Given mass=245g[tex]\boxed{\sf No\:of\:moles=\dfrac{Given\:mass}{Molar\:Mass}}[/tex]
[tex]\\ \large\sf\longmapsto No\:of\:moles=\dfrac{245}{98}[/tex]
[tex]\\ \large\sf\longmapsto No\:of\:moles=2.5mol[/tex]
Now
[tex]\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Volume\:of\:solution\;in\;L}}[/tex]
[tex]\\ \large\sf\longmapsto Molarity=\dfrac{2.5}{1}[/tex]
[tex]\\ \large\sf\longmapsto Molarity=2.5M[/tex]
Determine the number of atoms of O in 60.1 moles of Fe₂(SO₃)₃.
Answer:
3.310308*10^26
Explanation:
nO=9nFe2(SO3)3=9*60.1=540.9 moles
number of atoms: 540.9*6.02*10^23
Give the IUPAC and common name
Answer:
IUPAC Name:
N-ethyl-N-methylaniline
Common Name:
Benzenamine
When a 1:1 mixture of ethyl propanoate and ethyl butanoate is treated with sodium ethoxide, four Claisen condensation products are possible. Draw the structure(s) of the product(s) that have an ethyl group on the chiral center
Answer:
attached below
Explanation:
The Four Claisen condensation are grouped into :
Self Claisen condensation reaction Cross Claisen condensation reactionSelf Claisen condensation is when R = R'
Cross Claisen condensation is when R ≠ R'
attached below are the four Claisen condensation
Un sistema formado por una única sustancia, ¿será siempre homogéneo? ¿Porqué? Piensa a partir de las definiciones y trata de corroborar o negar usando ejemplos concretos.
Una sustancia homogénea es una sustancia que se compone de una sola fase.
Recordemos que definimos una fase en química como "cantidad química y físicamente uniforme u homogénea de materia que se puede separar mecánicamente de una mezcla no homogénea y que puede consistir en una sola sustancia o una mezcla de sustancias" según Ecyclopedia Britiannica.
El hecho de que un sistema esté compuesto por una sola sustancia no lo hace es autóctono. A veces, un sistema puede estar compuesto por partículas sólidas de una sustancia en equilibrio con su líquido. El sistema contiene solo una sustancia pero en diferentes fases, por lo tanto, el sistema contiene una sustancia pero no es homogéneo.
Por tanto, el hecho de que un sistema contenga una sola sustancia no significa necesariamente que sea homogéneo.
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Arrange these elements according to electron affinity.
a. He
b. C
c. F
Answer:
[tex]F>C> H_e[/tex]
Explanation:
From the question we are given
a. He
b. C
c. F
Generally
The electron affinity of an atom or molecule is defined as the amount of energy released when an electron is attached to a neutral atom or molecule in the gaseous state to form a negative ion.
Therefore
The order of electron Affinity is give as
[tex]F>C> H_e[/tex]
Elements arranged according to increasing electron affinity is He, C, F.
Electron affinity is a measure of the ability of an atom or molecule to attract and accept an electron. It quantifies the energy change that occurs when an electron is added to a neutral atom or ion in the gaseous state.
When an atom or ion gains an electron, it undergoes a process called electron attachment or electron capture. The electron affinity is the energy released or absorbed in this process. A higher electron affinity value indicates a stronger attraction for electrons.
The electron affinity is typically expressed in units of energy, such as electron volts (eV) or kilojoules per mole (kJ/mol). It can be either positive or negative. A positive electron affinity indicates that energy is released when an electron is added, while a negative electron affinity indicates that energy is required to add an electron.
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20. An oxide of osmium (symbol Os) is a pale yellow solid. If 2.89 g of the compound contains 2.16 g of osmium, what is its empirical formula?
The empirical formula is OsO₄ :
Explanation:
Osmium oxide contains osmium and oxygen only.
Thus, we shall determine the mass of oxygen in osmium oxide. This can be obtained as follow:
Mass of compound = 2.89 g
Mass of Os = 2.16 g
Mass of O =?Mass of O = (Mass of compound) – (Mass of Os)
Mass of O = 2.89 – 2.16
Mass of O = 0.73 g
Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:
Mass of Os = 2.16 g
Mass of O = 0.73 g
Empirical formula =..?Os = 2.16 g
O = 0.73 g
Divide by their molar mass of
Os = 2.16 / 190 = 0.011
O = 0.73 / 16 = 0.046
Divide by the smallest
Os = 0.011 / 0.011 = 1
O = 0.046 / 0.011 = 4
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The half-life of radon-222 is 3.8 days. How many grams of radon-222 remain
after 15.2 days if the original amount was 6.00 g?
A. 0.750 g
B. 0.375 g
C. 1.20 g
D. 3.00 g
The mass of radon-222 that will remain after 15.2 days given that it was originally 6 g is 0.375 g (Option B)
What is half life?This is the time taken for half a substance to decay.
How to determine the number of half-lives that has elapsedWe'll begin our calculation by calculating the number of half-lives that has elapsed after 15.2 days. This is illustrated below:
Half-life (t½) = 3.8 daysTime (t) = 15.2 day Number of half-lives (n) =?n = t / t½
n = 15.2 / 3.8
n = 4
Thus, 4 half-lives has elapsed.
How to determine the amount remainingOriginal amount (N₀) = 6 gNumber of half-lives (n) = 4Amount remaining (N) = ?The amount of radon-222 remaining can be obtained as illustrated below:
N = N₀ / 2ⁿ
N = 6 / 2⁴
N = 6 / 16
N = 0.375 g
Thus, the amount of radon-222 remaining after 15.2 days is 0.375 g
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Kamal was told by his mother to pour water through a thin cloth into another container to further purify the water.
a) What do you think will happen to the mud and sand when pouring the water? b)Do you think the water filtered by Kamal is safe to drink?
a].When sand is added to water it either hangs in the water or forms a layer at the bottom of the container. Sand therefore does not dissolve in water and is insoluble. It is easy to separate sand and water by filtering the mixture.
b]. The water filtered by kamal is not safe to drink .
If you run out of water, or cannot carry enough water with you for your entire trip, you may need to source drinking water from natural water sources.
An antacid tablet weighing 1.30g was fully neutralized at 42.00 mL(an excess amount) of 0.250MHCl. 10.00 mL of 0.100 M NaOH was then used to back titrate the excess HCl. How many moles of acid did the antacid neutralize
Answer:
0.0095 moles of acid were neutralized by the antiacid
Explanation:
The antiacid is a base that neutralize the acid in stomach. To find the moles of acid neutralized we need to find the moles of acid added initially. This acid is added in excess, then, the moles of NaOH added reacts to neutralize the moles of acid in excess. The difference between initial moles of HCl and moles of NaOH needed to titrate the excess = Moles of HCl that were neturalized by the antiacid as follows:
Moles HCl added:
42.00mL = 0.04200L * (0.250mol/L) = 0.0105 moles HCl
Moles NaOH to titrate the excess:
10.00mL = 0.01000L * (0.10mol/L) = 0.0010 moles NaOH = Moles HCl in excess.
Moles of acid that were neutralized:
0.0105 moles - 0.0010 moles =
0.0095 moles of acid were neutralized by the antiacidAssume you have 4 solids (A, B, C and D) of similar mass. Which of these requires the greatest energy input to melt?
polar covalent
covalent network
ionic compound
nonpolar covalent
The solid that require the greatest energy input to melt by mass is the option;
Covalent network
Reason for the above answer is as follows;
The elementary particles of a solid are held together by bonds that require
an input of energy to unlock, and once broken, the particles are then able
to change location within their containing vessels with less restrictions
Types of bonds
Polar covalent molecular solids have the following characteristics;a) Soluble in water b) Low melting point, b) Conduct electricity
Solids that are made up of a covalent network have the following characteristicsa) High melting point temperature b) Non conductive of electricity c) Not soluble in water
Solids of ionic compounds have the following characteristics;a) High melting point temperature b) The liquid state and solution
conducts electricity c) Soluble in water
Solids that have nonpolar covalent bonds have;a) Low melting point b) Normally in the gaseous or liquid state b) Not water soluble
Therefore, the covalent network, and the solids ionic compounds require the most energy to melt, however, the strength of the ionic bond in an ionic compound is a factor the charges present and the sizes of the atom, while
the covalent network solid, are combined to form essentially as a single
molecule and therefore require the greatest heat energy input break the bonds of the molecule down in order to melt
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g When 2.50 g of methane (CH4) burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion (in kJ) per mole of methane under these conditions
Answer:
-800 kJ/mol
Explanation:
To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).
First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:
Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol
moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄
Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:
ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol
Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).
When determining the amount of oxidant present by titration, you can use iodine/starch as an indicator. First, the oxidant, like hypochlorite, oxidizes Choose... When starch and iodine are both present, the solution is Choose... During the titration, a titrant like thiosulfate reduces the
The question is incomplete, the complete question is;
When determining the amount of an oxidant present by titration, you can use iodine and starch as an indicator.
First, the oxidant, like hypochlorite, oxidizes
Choose...
neutral iodine into iodide ion
iodide ion into neutral iodine
iodate polyatomic ion into iodide ion
When starch and iodine are both present, the solution is
Choose...
blue-black
brownish yellow
clear
During the titration, the titrant, like thiosulfate, reduces the
Choose...
iodide ion into iodate polyatomic ion
neutral iodine into iodide ion
iodide ion into neutral iodine
When the iodine has completely reacted at the endpoint of the titration, the solution should become
Choose...
clear
blue-black
brownish yellow
Answer:
1. iodide ion into neutral iodine
2. blue-black
3. neutral iodine into iodide ion
4. clear
Explanation:
Hypochlorite oxidizes the iodide ion to iodine molecule according to the reaction equation;
ClO-(aq) + 2H+(aq) + 2I-(aq) ---------> 6 I2(l) + Cl- (aq)+ H2O(l)
When iodine is added, the colour of the starch solution immediately changes to blue-black.
A reduction reaction occurs when the titrant, thiosulfate is added as follows;
I2 + 2S2O32- → 2I- + S4O62-
The solution at end point is found to become clear again.
Hydrogen is manufactured on an industrial scale by this sequence of reactions:
CH4(g) + H2O(g) ⇆ CO(g) + 3H2(g)
CO(g)+ H2O(g) ⇆ CO(g) + H2(g)
The net reaction is:
CH4(g) + H2O(g) ⇆ CO(g) + 4H2(g)
Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K1 and K2.
Answer:
[tex]K=K_1*K_2\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}[/tex]
Explanation:
Hello there!
In this case, for the given chemical reaction, it turns out firstly necessary to write the equilibrium expression for both reactions 1 and 2:
[tex]K_1=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \\\\K_2=\frac{[CO_2][H_2]}{[CO][H_2O]}[/tex]
Now, when we combine them to get the overall expression, we infer these two are multiplied to get:
[tex]K=K_1*K_2\\\\K=\frac{[CO][H_2]^3}{[CH_4][H_2O]} *\frac{[CO_2][H_2]}{[CO][H_2O]}\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}[/tex]
Regards!
……….is strong due to the ……………..between positive ions and negative delocalized electrons
Answer:
atom &bond
Explanation:
atom is strong due to the bond
Preparation the buffer solution: initial pH of buffer solution: ____ Titration of a weak acid with a strong base: initial pH of weak acid: ____ final pH of weak acid: ____ Amount of NaOH added: ____ Titration Curve for Weak Acid with a Strong Base (Paste curve here.)
Answer:
pH of buffer solution is 7.0
Initial pH of Weak acid is 3.27
Final pH of weak acid is 3.07
Amount of NaOH added is 1ml
Explanation:
Titration is a process in which acid and base are introduced together until a neutral solution is achieved whose pH value is near to buffer solution which is 7.0, the pH value for acid is below 7 while pH value for base is above 7.
Select the choice that best completes the following sentence: When cooled slowly, transformations near the melting temperature tend to yield ______ grains due to the formation of ______ nucleation sites followed by ______ grain growth.
Question Completion with Options:
O coarse...few...rapid
O fine...few...slow
O fine...multiple...rapid
O coarse...few...slow
O fine...multiple...slow
Answer:
The choice that best completes the sentence is:
O coarse...few...slow
Explanation:
Transformations near the melting temperature develop coarse grains because few nucleation sites are formed and the rate of the grain growth is usually slow. This is because of the process that starts with recrystallization, recovery, and nucleation before growth can occur. While recrystallization enables the grain to increase in size at high temperature, nucleation gives the grain the energy to irreversibly grow into larger-sized nucleus.
A chemist is preparing to carry out a reaction that requires 5.75 moles of hydrogen gas. The chemist pumps the hydrogen into a 10.5 L rigid steel container at 20.0 °C. To what pressure, in kPa, must the hydrogen be compressed? (Show all work for full credit and circle your final answer) *
Answer:
The hydrogen must be compressed to 1333.13302 kPa.
Explanation:
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P * V = n * R * T
In this case:
P= ?V= 10.5 Ln= 5.75 molesR= 0.082 [tex]\frac{atm*L}{mol*K}[/tex]T= 20 C= 293 K (being 0 C= 273 K)Replacing:
P* 10.5 L= 5.75 moles* 0.082 [tex]\frac{atm*L}{mol*K}[/tex] * 293 K
Solving:
[tex]P=\frac{5.75 moles* 0.082 \frac{atm*L}{mol*K} * 293 K}{10.5 L}[/tex]
P= 13.157 atm
If 1 atm is equal to 101.325 kPa, then 13.157 atm is equal to 1333.13302 kPa.
The hydrogen must be compressed to 1333.13302 kPa.
Give four examples illustrating each of the following terms. a. homogeneous mixture b. heterogeneous mixture c. compound e. physical change d. element f. chemical change
1. homogenous: sugar solution
2. heterogeneous: sand solution
3. compound: water
4. physical change: ice melting
5. element: hydrogen
6. chemical change: burning fire
3. (07.05 LC)
When zinc reacts with hydrochloric acid, it produces hydrogen gas. As the reaction proceeds, why does the rate of production of hydrogen gas decrease? (3 points)
the rate would decrease because the reactants are being depleted.
The electronic arrangement of an atom shows how electrons are distributed across the different energy levels. Which of the following elements is represented by the electron arrangement 2, 8, 18, 6?
А Сa
B. Mg
C. S
D Se
E. Ga
Answer:
The answer is D which is Selenium
What volume of water is produced when 38.5 g of ethanol reacts with oxygen at 500°C at 1.75 atm?
CH3CH2OH(g) + 3 O2(g)→ 2 CO2(g) + 3 H2O(g)
Answer:
90.99 or 91.0
Explanation:
Using the balanced equation, you convert 38.5g of ethanol to moles of water. From there, you plug the values into the Ideal Gas Equation: PV=nRT.
Answer: The volume of oxygen gas is 91.4 L.
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Given mass of ethanol = 38.5 g
Molar mass of ethanol = 46 g/mol
Plugging values in equation 1:
[tex]\text{Moles of ethanol}=\frac{38.5g}{46g/mol}=0.840 mol[/tex]
The given chemical equation follows:
[tex]CH_3CH_2OH(g)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(g)[/tex]
By stoichiometry of the reaction:
If 1 mole of ethanol produces 3 moles of water
So, 0.840 moles of ethanol will produce = [tex]\frac{3}{1}\times 0.840=2.52mol[/tex] of water
The ideal gas equation is given as:
[tex]PV=nRT[/tex] .......(2)
where
P = pressure = 1.75 atm
V = volume of oxygen gas = ?
n = number of moles= 2.52 moles
R = Gas constant = 0.0821 L.atm/mol.K
T = temperature of the tank = [tex]500^oC=[500+273]K=773K[/tex]
Putting values in equation 2, we get:
[tex]1.75 atm\times V=2.52mol\times 0.0821L.atm/mol.K\times 773K\\\\V=\frac{2.52\times 0.0821\times 773}{1.75}=91.4L[/tex]
Hence, the volume of oxygen gas is 91.4 L.
what is the mass of 4 moles of fluorine atoms?
Answer:
Vien, sometimes we make problems like this harder than they need to be. Suppose I asked you, “How many dozen wheels are on four dozen automobiles?” You would have no trouble answering 4 x 4 = 16 dozen.
A mole is just a quantity like a dozen. And in this case, instead of 4 wheels, each C2F6 molecule bears 6 fluorine atoms, right?
So 4 x 6 = 24 moles of fluorine atoms. You’ve got this, Vien!
(URGENT FOR BRAINLIEST!!)
The diagram below shows the movement of Earth plates.
The picture shows two Earth plates one beside the other. The plate on the left is shown moving towards the right and the plate
Which of these features is most likely formed as a result of the movement of Earth plates shown in the diagram? (2 points)
a canyon
a fault
a mountain
a ridge
Answer:
mountain
Explanation:
when plates move towards each other they create mountains
What is Bose Einstein state of matter and their examples
Answer:
A BEC ( Bose - Einstein condensate ) is a state of matter of a dilute gas of bosons cooled to temperatures very close to absolute zero is called BEC.
Examples - Superconductors and superfluids are the two examples of BEC.
Explanation:
công thức của định lý pytago
The sum of the squares of two sides of a right angle is equal to the square of the hypotenuse
What volume of each solution contains 0.14 mol of KCl? Express your answer using two significant figures.
1.8 M KCl
Answer:
Solution given:
1 mole of KCl[tex]\rightarrow [/tex]22.4l
1 mole of KCl[tex]\rightarrow [/tex]74.55g
we have
0.14 mole of KCl[tex]\rightarrow [/tex]74.55*0.14=10.347g
74.55g of KCl[tex]\rightarrow [/tex]22.4l
10.347 g of KCl[tex]\rightarrow [/tex]22.4/74.55*10.347=3.11litre
volume of each solution contains 0.14 mol of KCl contain 3.11litre.
1 mol of any gas contains 22.4L of volume at STP
Moles of KCl=0.14Volume of KCl:-
0.14(22.4)3.14L