The two layers are part of the thermosphere are exosphere and ionosphere.
What is the exosphere layer?The Exosphere is the topmost layer of the Earth's atmosphere.
and its gradually disappear into the vacuum of space.
It consist two parts that are:
exosphere and ionosphere.
Thus, option "A" is correct, the rest of the option is not a part of thermosphere.
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Answer:
hi hope your doing great the answer is A
Explanation:
its on Edge 2020
hope i helped :)
which law states that the pressure and absolue tempeture of a fixed quantity of gas are directly proportional under constant volume conditions?
Answer:
Gay lussacs law
Explanation:
Which response includes all the following processes that are accompanied by an increase in entropy? 1) 2SO 2(g) + O 2(g) → SO 3(g) 2) H 2O(l) → H 2O(s) 3) Br 2(l) → Br 2(g) 4) H 2O 2(l) → H 2O(l) + 1/ 2O 2(g)
Answer: Reaction (1) , (3) and (4) are accompanied by an increase in entropy.
Explanation:
Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.
(1) [tex]2SO_2(g)+O_2(g)\rightarrow SO_3(g)[/tex]
3 moles of reactant are changing to 1 mole of product , thus the randomness is increasing. Thus the entropy also increases.
2) [tex]H_2O(l)\rightarrow H_2O(s)[/tex]
1 mole of Liquid reactant is changing to 1 mole of solid product , thus the randomness is decreasing. Thus the entropy also decreases.
3) [tex]Br_2(l)\rightarrow Br_2(g)[/tex]
1 mole of Liquid reactant is changing to 1 mole of gaseous product , thus the randomness is increasing. Thus the entropy also increases.
4) [tex]H_2O_2(l)\rightarrow H_2O(l)+\frac{1}{2}O_2(g)[/tex]
1 mole of Liquid reactant is changing to half mole of gaseous product and 1 mole of liquid product, thus the randomness is increasing. Thus the entropy also increases.
Draw the structure of beeswax.beeswax is made from the esterfication of a saturated 16-carbon fatty acid and a 30 carbon straight chain primary alcohol.
Answer:
Triacontyl palmitate
Explanation:
In this case, we have a reaction between an acid and an alcohol. When we put together these kind of compounds an ester is produced. This reaction is called "esterification".
In our case, the alcohol is a structure with 30 carbon in which the "OH" group is bonded on carbon 1. The name of this compound is "n-triacontanol". The acid is a structure in which we have 16 carbon in which the "COOH" group is placed on carbon 1. The name of this compound is "palmitic acid". The ester produced by the acid and the alcohol is "Triacontyl palmitate".
See figure 1.
I hope it helps!
PV = nRT. If P = 1 atm, V = 5.0 liter, R = 0.0821 L.atm/mol.K, and T = 293 K; what is the value of n?
Answer:
n = 0.207 mole
Explanation:
We have,
P = 1 atm
V = 5 liter
R = 0.0821 L.atm/mol.K
T = 293 K
We need to find the value of n. The relation is as follows :
PV = nRT
Solving for n,
[tex]n=\dfrac{PV}{RT}\\\\n=\dfrac{1\times 5}{0.0821 \times 293}\\\\n=0.207\ \text{mol}[/tex]
So, the value of n is 0.207 mol.
Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K sp (MgF 2) = 6.9 × 10 –9]
Answer:
No precipitate is formed.
Explanation:
Hello,
In this case, given the dissociation reaction of magnesium fluoride:
[tex]MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-[/tex]
And the undergoing chemical reaction:
[tex]MgCl_2+2NaF\rightarrow MgF_2+2NaCl[/tex]
We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:
[tex]n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2[/tex]
Next, the moles of magnesium chloride consumed by the sodium fluoride:
[tex]n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2[/tex]
Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:
[tex]n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2[/tex]
Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:
[tex][Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M[/tex]
[tex][F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M[/tex]
Thereby, the reaction quotient is:
[tex]Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}[/tex]
In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.
Regards.
What is the frequency of a photon having an energy of 4.91 × 10–17 ? (c = 3.00 × 108 m/s, h = 6.63 × 10–34 J · s)
Answer:
The frequency of the photon is 7.41*10¹⁶ Hz
Explanation:
Planck states that light is made up of photons, whose energy is directly proportional to the frequency of radiation, according to a constant of proportionality, h, which is called Planck's constant. This is expressed by:
E = h*v
where E is the energy, h the Planck constant (whose value is 6.63*10⁻³⁴ J.s) and v the frequency (Hz or s⁻¹).
So the frequency will be:
[tex]v=\frac{E}{h}[/tex]
Being E= 4.91*10⁻¹⁷ J and replacing:
[tex]v=\frac{4.91*10^{-17} J}{6.63*10^{-34} J.s}[/tex]
You can get:
v= 7.41*10¹⁶ [tex]\frac{1}{s}[/tex]= 7.41*10¹⁶ Hz
The frequency of the photon is 7.41*10¹⁶ Hz
In the experiment students will create solutions with different ratios of ethanol and water. What is the mole fraction of ethanol when 10.00 mL of pure ethanol is combined with 2.00 mL of water
Answer:
[tex]x_{et}=0.6068[/tex]
Explanation:
Hello,
In this case, since the mole fraction of a compound, in this case ethanol, in a binary mixture, in this constituted by both water and ethanol, is mathematically defined as follows:
[tex]x_{et}=\frac{n_{et}}{n_{et}+n_{w}}[/tex]
Whereas [tex]n[/tex] accounts for the moles in the solution for each species, we must first compute the moles of both ethanol (density: 0.789 g/mL and molar mass: 46.07 g/mol) and water (density: 1g/mL and molar mass: 18.02 g/mol)
[tex]n_{et}=10.00mL\ et*\frac{0.789g\ et}{mL\ et} *\frac{1mol\ et}{46.07g\ et}=0.1713mol\ et\\ \\n_w=2.00mL\ w*\frac{1g\ w}{mL\ w} *\frac{1mol\ w}{18.02g\ w}=0.1110mol\ w[/tex]
Therefore, the mole fraction turns out:
[tex]x_{et}=\frac{0.1713mol}{0.1713mol+0.1110mol}\\\\x_{et}=0.6068[/tex]
Best regards.
clacium hydroxide is slightly soluable in water about 1 gram will dissolve in 1 liter what are the spectator ions in the reaction ions in the reaction of such a dilute solution of calcium hydroxide with hydrochloric acid
Answer:
Ca²⁺ and Cl⁻
Explanation:
In a chemical reaction, spectator ions are ions that are not involved in the reaction, that means are the same before and after the reaction.
In water, calcium hydroxide, Ca(OH)₂ is dissociated in Ca²⁺ and OH⁻. Also, hydrochloric acid, HCl, dissociates in H⁺ and Cl⁻. The reaction is:
Ca²⁺ + 2OH⁻ + 2H⁺ + 2Cl⁻ → 2H₂O + Ca²⁺ + 2Cl⁻
The ions that react are H⁺ and OH⁻ (Acid and base producing water)
And the ions that are not reacting, spectator ions, are:
Ca²⁺ and Cl⁻A sample of a hydrocarbon is found to contain 7.99g carbon and 2.01g hydrogen. What is the empirical formula for this compound
Answer:
The empirical formulae for the compound is CH3.
1. For the following reaction, 4.86 g of magnesium nitride are mixed with excess water. The reaction yields 7.18 g of magnesium hydroxide.
magnesium nitride(s) + water(1) –> magnesium hydroxide (aq) + ammonia (aq)
What is the ideal yield of magnesium hydroxide?
What is the percent yield for this reaction?
2. For the following reaction, 6.41 g of hydrogen gas are mixed with excess nitrogen gas. The reaction yields 26.2 g of ammonia.
nitrogen(g) + hydrogen(g) –> ammonia(g)
What is the ideal yield of ammonia?
What is the percent yield for this reaction?
3. For the following reaction, 3.79 g of water are mixed with excess chlorine gas. The reaction yields 8.70 g of hydrochloric acid.
chlorine(g) + water(1) –> hydrochloric acid(aq) + chloric acid (HCIO3)(aq)
What is the ideal yield of hydrochloric acid?
What is the percent yield for this reaction?
Answer:
See explanation
Explanation:
1)
Mg3N2(s) + 6H2O(l) ------------> 3Mg(OH)2 + 2NH3(g)
Number of moles of magnesium nitride= mass/molar mass= 4.86g/100.9494 g/mol = 0.048 moles
1 mole of magnesium nitride yields 3 moles of magnesium hydroxide
0.048 moles of magnesium nitride yields 0.048 moles × 3= 0.144 moles of magnesium hydroxide
Theoretical yield of magnesium hydroxide = 0.144 moles × 58.3197 g/mol = 8.398 g
Percent yield= actual yield/ theoretical yield × 100
Percent yield= 7.18/8.398 × 100/1 = 85.5%
2)
N2(g) + 3H2(g) -------> 2NH3(g)
Number of moles of hydrogen gas = mass/ molar mass = 6.41g/ 2gmol-1 = 3.205 moles of hydrogen gas.
From the balanced reaction equation;
3 moles of hydrogen gas yields 2 moles of ammonia
3.205 moles of hydrogen gas yields 3.205 × 2/3 = 2.1367 moles of ammonia
Theoretical yield of ammonia = 2.1367 moles × 17 gmol-1 = 36.3 g
Percent yield = actual yield/ theoretical yield ×100
Percent yield = 26.2/36.3 ×100
Percent yield = 72.2%
3)
3Cl2(g) + 3H2O(l) ------> HOCl3(aq) + 5HCl(aq)
Number of moles of water= mass/ molar mass = 3.79g/18 gmol-1 = 0.21 moles
Since
3 mole of water yields 5 mole of HCl
0.21 moles of water yields 0.21 × 5/3 = 0.35 moles of HCl
Theoretical yield of HCl = 0.35 moles × 36.5 gmol-1 = 12.775 g
Percent yield = actual yield/ theoretical yield × 100/1
Percent yield = 8.70/12.775 ×100
Percent yield = 68.1%
The combination of a carbonyl group and a hydroxyl group on the same carbon atom is called a ________ group.
a. carbamate group
b. carbonate
c. carboxlate
d. carboxyl
Answer:
d. carboxyl
Explanation:
The presence of carbonyl group (>C=O)) and a hydroxyl group ( (−OH) on the same carbon atom is called a "carboxyl" group. A carboxyl group is represented as COOH and acts as the functional group part of carboxylic acids.
For example:
Formic acid or Methanoic acid (H-COOH) Butanoic acid (C3H7-COOH)Hence, the correct option is "d. carboxyl ".
write the balanced nuclear equation for the radioactive decay of radium-226 to give radon-222, and determine the type of decay
Answer:
226Ra88→222Rn86+4He2
Explanation:
An α-particle usually consists of a helium nucleus which indicates the type of decay that was undergone in this radioactive process.
During α-decay(alpha decay), an atomic nucleus emits an alpha particle.
According to the following reaction, how many moles of ammonia will be formed upon the complete reaction of 31.2 grams of nitrogen gas with excess hydrogen gas ? nitrogen(g) + hydrogen(g) ammonia(g)
Answer:
4.46 mol of NH3
Explanation:
The equation of he reaction is given as;
2N + 3H2 --> 2NH3
From the stochiometry of the reaction, 1 mol of Nitrogen produces 2 mol of Ammonia.
Mass of Nitrogen = 31.2g
Molar mass of Nitrogen = 14g/mol
Number of moles = Mass / Molar mass = 31.2 / 14 = 2.23 mol
Since 1 mol of N = 2 mol of NH3;
2.23 mol of N2 would produce x
x = 2.23 * 2 = 4.46 mol of NH3
What is the oxidizing agent in the redox reaction represented by the following cell notation? Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)
Answer:
Silver.
Explanation:
Hello,
In this case, for the redox reaction:
[tex]Ni^0(s)+Ag^+(aq)\rightarrow Ni^{2+}+Ag^0(s)[/tex]
We can see the nickel is being oxidized as its oxidation state increases from 0 to 2+ whereas the oxidation state of silver decreases from +1 to 0, it means that the oxidizing agent is silver and the reducing agent is nickel.
Best regards.
The oxidizing agent in the redox reaction represented by the following cell notation is Silver.
Calculation of the oxidizing agent:The redox reaction is
Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)
here the nickel is being oxidized since its oxidation state rises from 0 to 2+ while on the other hand, the oxidation state of silver is reduced from +1 to 0, it means that the oxidizing agent is silver and the reducing agent is nickel.
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How many liters of CH₃OH gas are formed when 3.20 L of H₂ gas are completely reacted at STP according to the following chemical reaction? Remember 1 mol of an ideal gas has a volume of 22.4 L at STP.CO(g)+ H2(g) → CH3OH
Answer:
The correct answer is 1.60 Liters.
Explanation:
The given reaction:
CO (g) + H₂(g) ⇔ CH₃OH (g)
Based on the given reaction, two moles of H₂ reacts with one mole of CO and produce one mole of CH₃OH.
It is mentioned that 3.20 L of H₂ is reacted, therefore, there is a need to convert it into moles.
As 22.4 L at standard temperature and pressure is equivalent to 1 mole.
Therefore, 1 L at STP will be, 1/22.4 mole
Now 3.20 L at STP will be,
= 1/22.4 × 3.20
= 0.1428 mole
And as mentioned in the reaction that 2 moles of H₂ gives 1 mole of CH₃OH, therefore, 1 mole of H₂ will give 1/2 mole of CH₃OH
Now, 0.1428 mole of H₂ will give,
= 0.1428/2 = 0.071 mole of CH₃OH
= 0.071 × 22.4 = 1.60 L
The volume, in liters, of CH₃OH gas formed is 1.60 L
From the question,
We are to determine the volume of CH₃OH formed
The given chemical equation for the reaction is
CO(g)+ H₂(g) → CH₃OH
The balanced chemical equation for the reaction is
CO(g)+ 2H₂(g) → CH₃OH
This means
1 mole of CO reacts with 2 moles of H₂ to produce 1 mole of CH₃OH
Now, we will determine the number of moles of H₂ present in the 3.20 L H₂ at STP
1 mol of an ideal gas has a volume of 22.4 L at STP
Then,
x mole of the H₂ gas will have a volume of 3.20 L at STP
x = [tex]\frac{3.20 \times 1}{22.4}[/tex]
x = 0.142857 mole
∴ The number of mole of H₂ present is 0.142857 mole
Since
2 moles of H₂ reacts to produce 1 mole of CH₃OH
Then,
0.142857 mole of H₂ will react to produce [tex]\frac{0.142857}{2}[/tex] mole of CH₃OH
[tex]\frac{0.142857}{2} = 0.0714285[/tex]
∴ The number of moles of CH₃OH produced = 0.0714285 mole
Now, for the volume of CH₃OH formed
Since
1 mol of an ideal gas has a volume of 22.4 L at STP
Then,
0.0714285 mol of CH₃OH will have a volume of 22.4 × 0.0714285 at STP
22.4 × 0.0714285 = 1.5999984 L ≅ 1.60 L
Hence, the volume of CH₃OH gas formed is 1.60 L
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What class of organic product results when 1-heptyne is treated with a mixture of mercuric acetate in aqueous sulfuric acid, and then HOCH2CH2OH with catalytic sulfuric acid
Answer:
2-methyl-2-pentyl-1,3-dioxolane
Explanation:
In this case, we have two reactions:
First reaction:
1-heptyne + mercuric acetate -------> Compound A
Second reaction:
Compound A + HOCH2CH2OH -------> Compound C
First reaction
In the first reaction, we have as a main functional group a triple bond. We have to remember that mercuric acetate in sulfuric acid will produce a ketone. The carbonyl group (C=O) would be placed in the most substituted carbon of the triplet bond (in this case, carbon 2). With this in mind, we will have as a product: heptan-2-one. (See figure 1).
Second reaction
In this reaction, we have as reagents:
-) Heptan-2-one
-) Ethylene-glycol [tex]HOCH_2CH_2OH[/tex]
-) Sulfuric acid [tex]H_2SO_4[/tex]
When we put ethylene-glycol with a ketone or an aldehyde we will form a cyclic acetal. In this case, this structure would be formed on carbon 2 forming 2-methyl-2-pentyl-1,3-dioxolane. (See figure 2).
I hope it helps!
What would happen to the rate of a reaction with rate law rate = k [NO]2[Hz] if
the concentration of NO were doubled?
Answer:
The rate would have doubled
Explanation:
Determine whether each phrase describes carboxylic acids or esters.a. Do not form hydrogen bonds amongst themselves and have higher vapor pressureb. Form hydrogen bonds amongst themselves and have lower vapor pressurec. Notable for their pleasant fragrancesd. Their reactions with base are kn. own as saponificationse. Usually have a sour odorf. Their reactions with base are known as neutralizations
Explanation:
a. Esters
b. Carboxylic acids
c. Esters (ethyl hexanoate smells like pineapple)
d. Carboxylic acids (produces a carboxylic salt)
For the given phrases the following description is better.
a. Esters
b. Carboxylic acids
c. Esters (ethyl hexanoate smells like pineapple)
d. Carboxylic acids (produces a carboxylic salt)
Esters and carboxylic acids:An ester is a synthetic compound got from a corrosive in which somewhere around one - OH hydroxyl bunch is supplanted by an - O-alkyl (alkoxy) bunch, as in the replacement response of a carboxylic acid and a liquor.
Carboxylic acid is any of a class of natural mixtures in which a carbon (C) particle is clung to an oxygen (O) molecule by a twofold bond and to a hydroxyl bunch (―OH) by a solitary bond.
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g When considering the effects of temperature on spontaneity, if both ΔH and ΔS are positive, _______. Select the correct answer below: the process is spontaneous at all temperatures
Explanation:
The spontaneity of a system is deduced by the sign of the gibbs free energy value. If it is negative, it means the process / reaction is spontaneous however a positive value indicates the such process is not spontaneous.
Gibbs free energy, enthalpy and entropy are related by the following equation;
ΔG = ΔH - TΔS
A positive value of enthalpy, H and entropy, S means that G would always be a negative value at all temperatures.
A baseball has a mass of 0.145 kilograms. If acceration due to gravity is 9.8m/s,what is the weight of the baseball in newtons?
Answer:
I hope it works
Explanation:
As we know that
w=m*g
given m=0.145 , g=9.8
hence we get
w= (9.8)*(0.145)
w=1.421 m/sec 2
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Identify the term that matches each electrochemistry definition. The electrode where oxidation occurs Cathode The electrode where reduction occurs Choose... An electrochemical cell powered by a spontaneous redox reaction Choose... An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction Choose... A chemical equation showing either oxidation or reduction Choose...
Answer:
An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction
If we represent the equilibrium as:...N2O4(g) 2 NO2(g) We can conclude that: 1. This reaction is: A. Exothermic B. Endothermic C. Neutral D. More information is needed to answer this question. 2. When the temperature is increased the equilibrium constant, K: A. Increases B. Decreases C. Remains the same D. More information is needed to answer this question. 3. When the temperature is increased the equilibrium concentration of NO2: A. Increases B. Decreases C. Remains the same D. More information is needed to answer this question.
Answer:
1. This reaction is: B. Endothermic.
2. When the temperature is increased the equilibrium constant, K: A. Increases.
3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.
Explanation:
Hello,
In this case, considering the images, we can state that the red color at high temperature is due to the presence of nitrogen dioxide (product) and the lower coloring is due to the presence of dinitrogen tetroxide (reactant) at low temperature.
With the aforementioned, we can conclude that the chemical reaction:
[tex]N_2O_4(g) \rightleftharpoons 2 NO_2(g)[/tex]
Is endothermic since high temperatures favor the formation of the product and the low temperatures favor the consumption of the the reactant. thereby:
1. This reaction is: B. Endothermic.
2. When the temperature is increased the equilibrium constant, K: A. Increases. In this particular case, since the dinitrogen tetroxide has 1 molecule and nitrogen dioxide two molecules in the chemical reaction, the entropy change should be positive, therefore, by increasing the T, the Gibbs free energy of reaction becomes more negative:
[tex]G=H-TS[/tex]
As Gibbs free energy becomes more negative, the equilibrium constant becomes bigger given their relationship:
[tex]K=exp(-\frac{\Delta G}{RT} )[/tex]
3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.
Regards.
How many equivalent resonance structures can be drawn for the molecule of SO3 without having to violate the octet rule on the sulfur atom
Answer:
3
Explanation:
Resonance is a valence bond concept put forward by Linus Pauling to explain the fact that the observed properties of a molecule may be as a result of the fact that its actual structure lie somewhere between a given number of structural extremes called canonical structures or resonance structures.
There are three resonance structures for SO3 that obey the octet rule. All the S-O bonds in SO3 are equivalent in these resonance structures.
Seven equivalent resonance structures for the molecular of SO3 can be drawn without breaking the octet rule.
We can arrive at this answer because:
The octet rule is a rule that states that an atom must reach stability when it has eight electrons in the valence layer.This means that in bonds that cause the donation or sharing of electrons between atoms, each atom has eight electrons in the valence layer.In chemistry, resonance is a term that refers to structures created to represent the donation or sharing of electrons between the atoms of a molecule.These structures can be arranged in different ways, as long as they respect the octet rule.In an SO3 molecule, electrons are shared between atoms. This sharing can be done with seven resonance structures.
These structures are shown in the figure below.
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4. Given that the enthalpy of reaction for a system at 298 K is -292 kJ/mol and the entropy for that system is 224 J/mol K, what's the free energy for the system?
A.-87,793 kJ
B.-358 kJ
C.-225 kJ
D. -66,751 kJ
Answer:
[tex]\Delta G=-359\frac{kJ}{mol}[/tex]
Explanation:
Hello,
In this case, we must remember that the Gibbs free energy is defined in terms of the enthalpy, temperature and entropy as shown below:
[tex]\Delta G=\Delta H -T\Delta S\\[/tex]
In such a way, for the given data, we obtain it, considering the conversion from J to kJ for the entropy in order to conserve the proper units:
[tex]\Delta G=-292\frac{kJ}{mol} -(298)(224\frac{J}{mol}*\frac{1kJ}{1000J} )\\\\\Delta G=-359\frac{kJ}{mol}[/tex]
Best regards.
Answer:
B- 358 kj
Explanation: I took the test
The half-life of Zn-71 is 2.4 minutes. If one had 100.0 g at the beginning, how many grams would be left after 7.2 minutes has elapsed? Report your answer to 1 decimal place.
Answer:
12.5g
Explanation:
Half life = 2.4 Minutes.
The half life of a compound is the time it takes to decay to half of it's original concentration or mass.
Time lapsed= 7.2 minutes. This is equivalent to 3 half lives ( 3 * 2.4)
Initial mass = 100g
First half life;
100g --> 50g
Second half life;
50g --> 25g
Third half life;
25g --> 12.5g
The amount of Zn-71 that remains after 7.2 mins has elapsed is 12.5 g
We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
Half-life (t½) = 2.4 mins
Time (t) = 7.2 mins
Number of half-lives (n) =?[tex]n = \frac{t}{t_{1/2}} \\\\n = \frac{7.2}{2.4} \\\\[/tex]
n = 3Thus, 3 half-lives has elapsed.
Finally, we shall the amount remaining. This can be obtained as follow:Original amount (N₀) = 100 g
Number of half-lives (n) = 3
Amount remaining (N) =?[tex]N = \frac{N_{0}}{2^{n}} \\\\N = \frac{100}{2^{3 }}\\\\N = \frac{100}{8}\\\\[/tex]
N = 12.5 gThus, the amount of Zn-71 that remains after 7.2 mins is 12.5 g
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Which of the following pairs of chemical reactions are inverses of each other? Answer options: a. Hydrogenation and alkylation b.Halogenation and hydrolysis c. Ammoniation and alkylation d. Oxidation and reduction
Answer:
d. Oxidation and reduction
Explanation:
For this question we have to remember the definition of each type of reaction:
-) Hydrogenation
In this reaction, we have the addition of hydrogen to a molecule. Usually, an alkene or alkyne. In the example, molecular hydrogen is added to a double bond to produce an alkane.
-) Alkylation
In this reaction, we have the addition of a chain of carbon to another molecule. In the example, an ethyl group is added to a benzene ring.
-) Hydrolysis
In this reaction, we have the breaking of a bond by the action of water. In the example, a water molecule can break the C-O bond in the ester molecule.
-) Halogenation
In this reaction, we have the addition of a halogen (atoms on the VIIIA group). In the example, "Cl" is added to the butene.
-) Ammoniation
In this reaction, we have the addition of the ammonium ion ([tex]NH_4^+[/tex]). In the example, the ammonium ion is added to an acid.
-) Oxidation and reduction
In this reaction, we have opposite reactions. The oxidation is the loss of electrons and the reduction is the gain of electrons. For example:
[tex]Ag^+~+~e^-~->~Ag[/tex] Reduction
[tex]Al~->~Al^+^3~+~3e^-[/tex] Oxidation
Plz help!!!! Solve this by using factor labeling
Answer:
the answer is 2,000 nickels.
Explanation:
we multiplied 100 by 100, because there are 100 cents in a dollar, and we divided 10,000 by 5, because there are 5 cents in a nickel.
Calculate the pH of a solution that is 0.210 M in nitrous acid (HNO2) and 0.290 M in potassium nitrite (KNO2). The acid dissociation constant of nitrous acid is 4.50 × 10-4.
Answer:
pH = 3.49
Explanation:
We have a buffer system formed by a weak acid (HNO₂) and its conjugate base (NO₂⁻ coming from KNO₂). We can calculate the pH of a buffer ssytem using the Henderson-Hasselbach equation.
pH = pKa + log [base] / [acid]
pH = -log Ka + log [NO₂⁻] / [HNO₂]
pH = -log 4.50 × 10⁻⁴ + log 0.290 M / 0.210 M
pH = 3.49
The pH of the solution containing 0.210 M nitrous acid (HNO₂) and 0.290 M potassium nitrite (KNO₂) is 3.49
We'll begin by calculating the the pKa of acid. This can be obtained as follow:
Acid dissociation constant (Ka) = 4.50×10¯⁴
pKa =?pKa = –Log Ka
pKa = –Log 4.50×10¯⁴
pKa = 3.35Finally, we shall determine the pH of the solution.pKa = 3.35
Concentration of HNO₂, [HNO₂] = 0.210 M
Concentration of KNO₂, [KNO₂] = 0.290 M
pH =?The pH of the solution can obtain by using the Henderson-Hasselbach equation as illustrated below:
pH = pKa + log [base] / [acid]pH = pKa+ log [NO₂⁻] / [HNO₂]
pH = 3.35 + log (0.290 / 0.210)
pH = 3.49Thus, the pH of the solution is 3.49
Learn more: https://brainly.com/question/15911738
If the Ksp for Li3PO4 is 5.9×10−17, and the lithium ion concentration in solution is 0.0020 M, what does the phosphate concentration need to be for a precipitate to occur?
Answer:
7.4 × 10⁻⁹ M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for Li₃PO₄: 5.9 × 10⁻¹⁷
Concentration of lithium ion: 0.0020 M
Step 2: Write the reaction for the solution of Li₃PO₄
Li₃PO₄(s) ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)
Step 3: Calculate the phosphate concentration required for a precipitate to occur
The solubility product constant is:
Ksp = 5.9 × 10⁻¹⁷ = [Li⁺]³ × [PO₄³⁻]
[PO₄³⁻] = 5.9 × 10⁻¹⁷ / [Li⁺]³
[PO₄³⁻] = 5.9 × 10⁻¹⁷ / 0.0020³
[PO₄³⁻] = 7.4 × 10⁻⁹ M
Which option draws the correct conclusion from the following case study?
A patient with sickle-cell anemia and a fever goes to the emergency room and is given Tylenol to reduce
the fever. The patient has seizures and dies after taking the Tylenol. The physician writes up this case as
an interesting outcome for a patient with sickle-cell anemia.
The case study's validity is obvious because it describes a real-life situation.
The case study was influenced by bias, and led to incorrect conclusions being drawn
The case study was not intended to produce a generalized conclusion about treatment
Upon reading this case study, physicians should stop treating sickle cell patients with fevers using Tylenol
Answer:
I believe the answer The case study was influenced by bias, and led to incorrect conclusions being drawn. plz correct me if I am wrong
Explanation:
Answer: options B
Explanation: