Answer:
A 2.8 liters
Explanation:
Step 1: Write the balanced equation
N₂ + 3 H₂ ⇄ 2 NH₃
Step 2: Establish the appropriate volume ratio
At the same temperature and pressure, the volume ratio of H₂ to NH₃ is 3:2.
Step 3: Calculate the volume of ammonia produced from 4.2 L of hydrogen
4.2 L H₂ × 2 L NH₃/3 L H₂ = 2.8 L
Explain how the existence of isotopes relates to the number of neutrons within the nuclei of an element.
Answer:
because of it less attraction and its neutral position
Answer:
The existence of isotopes relates to the number of neutrons within the nuclei of an element because isotopes contain the same amount of protons (based on what element they are) but different number of neutrons in their nuclei. Because they have a different amount of neutrons, than the original element
,they also have a different atomic mass.
Explanation:
hope it helps!
Determine the molecular formula of a compound if it is composed of 40.92% carbon, 4.58% hydrogen, and 54.50% oxygen. The molar mass is 176.12 g/mol.
A) C3H8O3.
B) CH2O.
C) C2H3O2.
D) C3H4O3.
Answer:
No correct answer listed. See explanation for defense.
Explanation:
Given
C: 40.92% => 40.92g/100wt => (40.92/12)moles C = 3.41 moles O
H: 4.58% => 4.58g/100wt => (4.58/1)moles H = 4.58 moles H
O: 54.50% => 54.5g/100wt => (54.5/16)moles O = 3.41 moles O
Empirical ratio => C : H: O => (3.41/3.41) : (4.58/3.41) : (3.41/3.41) => 1 : 1.34 : 1
=> C : H : O => 3(1 : 1.34 : 1) => 3 : 4 : 3 => Empirical Formula C₃H₄O₃
Molecular Weight = Empirical Formula Wt x N
176.12 = 88 x N
N = whole number multiple of empirical formula = 176.12/88 = 2
∴ Molecular Formula => (C₃H₄O₃)₂ => C₆H₈O₆
Note => Only ionic compounds (salts) have subscripts reduced to lowest whole number ratios. Molecular compounds as C₆H₈O₆ are not reduced to lowest whole number ratios. Therefore, there is no correct answer in the answer choice list for the 'Molecular Formula'. Doc :-)
the -OH group cannot exhibit inductive effect? true/false, and reason for ur choice
Answer:
false
Explanation:
The inductive effects are know as the ability of the atom or a group to create polarization and electronic density long the covalent bond and it needs a higher density. The -OH group cannot exhibit the indictive effects as it becomes --O.Consider the reaction: CaCO3(s)CaO(s) CO2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.58 moles of CaCO3(s) react at standard conditions.
Answer:
the entropy change for the system when 1.58 moles of CaCO3(s) react at standard conditions is 253.748 J/K
Explanation:
Given the data in the question;
CaCO₃(s) → CaO(s) + CO₂(g)
1.58 moles 1.58 moles 1.58 moles
Since 1 mole of CaCO₃ gives 1 mole of CaO and 1 mole of CO₂
Thus, 1.58 mole of CaCO₃ gives 1.58 moles of CaO and 1.58 moles of CO₂.
Now,
At 298 K, standard entropy values are;
ΔS° ( CaCO₃ ) = 92.9 J/mol.K
ΔS° ( CaO ) = 39.8 J/mol.K
ΔS° ( CO₂ ) = 213.7 J/mol.K
So,
ΔS°[tex]_{system[/tex] = ∑ΔS°( product ) - ∑ΔS°( reactant )
ΔS°[tex]_{system[/tex] = [ ΔS°(CaO) + ΔS°( CO₂ ) ] - ΔS°( CaCO₃ )
we substitute
ΔS°[tex]_{system[/tex] = [ 39.8 J/mol.K + 213.7 J/mol.K ] - 92.9 J/mol.K
ΔS°[tex]_{system[/tex] = 160.6 J/mol.K
i.e, for 1 mol CaCO₃, ΔS°[tex]_{system[/tex] = 160.6 J/mol.K
Now, for 1.58 mol CaCO₃,
ΔS°[tex]_{system[/tex] = 1.58 mol × 160.6 J/mol.K
ΔS°[tex]_{system[/tex] = 253.748 J/K
Therefore, the entropy change for the system when 1.58 moles of CaCO3(s) react at standard conditions is 253.748 J/K
Flag A solution of the weak acid, HF, and a solution of the strong acid, HCl, have the same pH. Which solution will require the most sodium hydroxide, NaOH, to neutralize
Answer:
C) Both will require the same amount because the concentrations are equal.
Explanation:
The pH of a solution is defined as:
pH = -log [H+]
This H+ is the ion that reacts with OH- (From NaOH) as follows:
H+ + OH- → H2O
When all H+ reacts, we can say the solution was neutralized.
Now, as both, the solution with the weak acid and the solution with strong acid have the same pH, we can say that their [H+] is the same. Assuming the volume of both solutions is the same:
Both will require the same amount because the concentrations are equal.
Categorize the following reaction as an acid-base neutralization, precipitation, combination, decomposition, combustion, displacement, or disproportionation reaction.
Ba(C2H3O2)2(aq) + Na2CO3(aq) → BaCO3(s) + 2 NaC2H3O2(aq)
Answer:
Precipitation
Explanation:
Let's consider the balanced chemical equation between barium acetate and sodium carbonate to form barium carbonate and sodium acetate.
Ba(C₂H₃O₂)₂(aq) + Na₂CO₃(aq) → BaCO₃(s) + 2 NaC₂H₃O₂(aq)
Both products and reactants are salts. But, among the products, barium carbonate is solid. This allows us to classify it as a precipitation reaction.
For the reaction of oxygen and nitrogen to form nitric oxide, consider the following thermodynamic data :
ΔH∘rxn 180.5kJ/mol
ΔS∘rxn 24.80J/(mol⋅K)
Required:
a. Calculate the temperature in kelvins above which this reaction is spontaneous
b. Calculate the equilibrium constant for the following reaction at room temperature, 25°C
Answer:
a. 7278 K
b. Kc = 4.542 × 10⁻³¹
Explanation:
a.
The reaction is spontaneous when ΔG° < 0. We can calculate ΔG° using the following expression.
ΔG° = ΔH° - T × ΔS°
Then, the reaction will be spontaneous when,
ΔH° - T × ΔS° < 0
T > ΔH°/ΔS
T > (180.5 × 10³ J/mol)/(24.80J/mol⋅K)
T > 7278 K
b.
First, we will calculate ΔG° at 25 °C (298 K)
ΔG° = ΔH° - T × ΔS°
ΔG° = (180.5 × 10³ J/mol) - 298 K × (24.80J/mol⋅K) = 1.731 × 10⁵ J/mol
Then, we will calculate the equilibrium constant (Kc) using the following expression.
ΔG° = - R × T × ln Kc
-ΔG°/R × T = ln Kc
-(1.731 × 10⁵ J/mol)/(8.314 J/mol.K) × 298 K = ln Kc
Kc = 4.542 × 10⁻³¹
Enzyme catalyzing breakdown of atp to adp
Answer:
ATP hydrolase
Explanation:
Enzymes are biological catalysts which perform diverse functions in the body. Enzymes are specific in their mode of action because an enzyme fits into its substrate as a key fits into a lock.
The particular enzyme that catalyzes the breakdown of ATP to ADP is ATP hydrolase. The phosphate released by the action of this enzyme is used in the phosphorylation of other compounds thereby making them more reactive.
1. Why is it necessary to equalize the pressure(i.e, have the water level the same in each tube) before taking a volume reading?
2. Why is it important to use water that has been pre-saturated with CO2 in the gas burettes?
3.If your antacid sample had been contaminated by moisture, what effect(if any )would you expect this to have on your result
4.Explain why an'antacid is called as such,what is the role of the NAHCO3 or CACO3 in reactions?
Answer:
If you contact water with a gas at a certain temperature and (partial) pressure, the concentration of the gas in the water will reach an equilibrium ('saturation') according to Henry's law.
Explanation:
This means: if you increase the pressure (e.g. by keeping the vial closed), the CO2 concentration will increase. So it simply depends what concentration you need for your assay: 'CO2-saturated' water at low pressure or 'CO2-saturated' water at high pressure.
Predict the Normality of H2SO4 if 75 ml of 96.6 % pure H2SO4 added to 425 ml water. The density of H2SO4 is 1.83 g/cm3?
Explanation:
Normality is one of the concentration terms.
It is expressed as:
[tex]N=\frac{mass of the substance}{equivalent mass}* \frac{1}{volume of solution in L.}[/tex]
The volume of the solution is 425 mL.
Mass of sulfuric acid given is:
[tex]mass=volume * purity* density\\ = 75 mL * 0.966 * 1.83 g/mL\\\\=132.5 grams\\[/tex]
The equivalent mass of sulfuric acid is 49.0g/equivalents
Hence, the normality of the given solution is:
[tex]N=\frac{132.5g}{49.0g/equi.} *\frac{1000}{425mL} \\Normality=6.36N[/tex]
Answer is: 6.36N.
Write the separation scheme for the isolation of triphenylmethanol from the reaction mixture once the reaction is complete. The separation begins after the addition of HCl and water to the reaction and includes the column chromatography procedure to further purify crude triphenylmethanol isolated in the day 1 procedure.
````````````````````
What is the oxidation state of nitrogen in N ?
Answer:
+5
Explanation:
The diagram above shows the hydrides of groups 14, 15, 16, and 17 elements. Why does H20, HF, and NH3 have much higher boiling points than the rest of the molecules in their groups?
Answer:
Hydrogen Bonding
Explanation:
Hydrogen Bonding occurs when a hydrogen atom is bonded to N, O, and F atoms.
The molecules H₂O, HF, and NH₃ all experience hydrogen bonding, which is a relatively strong IMF, causing the molecules to have stronger attraction to each other. Having a stronger attraction between molecules results in more energy required to separate them, thus these molecules will have a higher boiling point than the rest of the molecules in their group.
Name the following molecule
Answer:
It is a Biological Molecule
how does iron I differ from iron II
Answer:
Metals tend to form positive oxidation states. Here, Iron (I) has an oxidation state of +1 while Iron (II) has an oxidation state of +2. Similarly, Lead (I) has an oxidation state of +1 while Lead(II) has an oxidation state of +2. A change in oxidation state can rather cause significant changes in the compound.
A researcher is attempting to produce ethanol using an enzyme catalyzed batch reactor. The ethanol is produced from corn starch by first-order kinetics with a rate constant of 0.05 hr-1. Assuming the concentration of ethanol initially is 1 mg/L, what will be the concentration of ethanol (in mg/L) after 24 hours
Answer:
The correct solution is "3.32 gm/L".
Explanation:
Given:
Rate constant,
[tex]K = 0.05 \ hr^{-1}[/tex]
Time,
[tex]t = 24 \ hours[/tex]
Concentration of ethanol,
[tex]C_o= 1 \ mg/L[/tex]
Now,
The concentration of ethanol after 24 hours will be:
⇒ [tex]C_o=C\times e^{-K\times t}[/tex]
By putting the values, we get
[tex]1=C\times e^{-0.05\times 24}[/tex]
[tex]1=C\times 0.30119[/tex]
[tex]C= 3.32 \ gm/L[/tex]
A rigid, sealed container that can hold 26 L of gas is filled to a pressure of
5.97 atm at 374 °C. The pressure suddenly decreases to 3.64 atm. What is
the new temperature inside the container, in units of °C?
Answer:
121 °C
Explanation:
From the question given above, the following data were obtained:
Initial pressure (P₁) = 5.97 atm
Initial temperature (T₁) = 374 °C
Final pressure (P₂) = 3.64 atm
Final temperature (T₂) =?
NOTE: Volume = constant
Next, we shall convert 374 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Initial temperature (T₁) = 374 °C
Initial temperature (T₁) = 374 °C + 273
Initial temperature (T₁) = 647 K
Next, we shall determine the final temperature. This can be obtained as follow:
Initial pressure (P₁) = 5.97 atm
Initial temperature (T₁) = 647 K
Final pressure (P₂) = 3.64 atm
Final temperature (T₂) =?
P₁ / T₁ = P₂ / T₂
5.97 / 647 = 3.64 / T₂
Cross multiply
5.97 × T₂ = 647 × 3.64
5.97 × T₂ = 2355.08
Divide both side by 5.97
T₂ = 2355.08 / 5.97
T₂ = 394 K
Finally, we shall convert 394 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
Final temperature (T₂) = 394 K
Final temperature (T₂) = 394 – 273
Final temperature (T₂) = 121 °C
Thus, the new temperature is 121 °C
Suppose we have two rock samples, A and B. Rock A was subject to both physical and chemical weathering while rock B was subject to chemical weathering only. Which rock would experience more chemical weathering? Why? (2pts) (Hint: consider the effect of surface area on the rate of chemical weathering)
Answer:
Rock A will have far more chemical weathering than Rock B due to the rise in area effect
Explanation:
Rock A undergoes both Physical and Chemical weathering. So, thanks to physical weathering there'll appear cracks within the rock, which can, in turn, increase the area of rock on which weathering is occurring. So, Chemical weathering will happen much faster now as there's a rise in the area. within the case of Rock B, there's only chemical weathering therefore the increase in the area won't be that very much like compared to Rock A.
Based on the "Reactivity in Substitution Reactions" experiment, which molecule would be expected to react the fastest using AgNO3 in water-ethanol ?
Answer:
C) EtOH 1% AgNO3
how does lead resemble chromium?
Write the cell notation for an electrochemical cell consisting of an anode where Mn (s) is oxidized to Mn2 (aq) and a cathode where Co2 (aq) is reduced to Co (s) . Assume all aqueous solutions have a concentration of 1 mol/L.
Answer:
Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Explanation:
In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.
For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.
Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
10g of a non-volatile and non-dissociating solute is dissolved in 200g of benzene.
The resulting solution boils At temperature of 81.20oC. Find the molar mass of solute.
Given that the BP of pure benzene is 80.10oC and Its elevation boiling point constant = 2.53 oC/m.
Answer: The molar mass of solute is 115 g/mol.
Explanation:
Elevation in the boiling point is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent.
The expression for the calculation of elevation in boiling point is:
[tex]\text{Boiling point of solution}-\text{boiling point of pure solvent}=i\times K_b\times m[/tex]
OR
[tex]\text{Boiling point of solution}-\text{Boiling point of pure solvent}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)
where,
Boiling point of pure solvent (benzene) = [tex]80.10^oC[/tex]
Boiling point of solution = [tex]81.20^oC[/tex]
i = Vant Hoff factor = 1 (for non-electrolytes)
[tex]K_b[/tex] = Boiling point elevation constant = [tex]2.53^oC/m[/tex]
[tex]m_{solute}[/tex] = Given mass of solute = 10 g
[tex]M_{solute}[/tex] = Molar mass of solute = ? g/mol
[tex]w_{solvent}[/tex] = Mass of solvent = 200 g
Putting values in equation 1, we get:
[tex]81.20-80.10=1\times 2.53\times \frac{10\times 1000}{M_{solute}\times 200}\\\\M_{solute}=\frac{1\times 2.53\times 10\times 1000}{1.1\times 200}\\\\M_{solute}=115g/mol[/tex]
Hence, the molar mass of solute is 115 g/mol.
5. How many grams of tin metal can be produced from smelting (heating) of a 4.5 kilograms of tin (IV) oxide? (Note: Elemental tin and oxygen gas are the only products of this reaction).
Answer:
About 3500 grams of tin.
Explanation:
We want to determine amount of tin metal (in grams) that can be produced from smelting 4.5 kilograms of tin(IV) oxide.
First, write the chemical compound. Since our cation is tin(IV), it forms a 3+ charge. Oxygen has a 2- charge, so we will have two oxygen atoms. Hence, tin(IV) oxide is given by SnO₂.
By smelting it, we acquire elemental tin and oxygen gas. Hence:
[tex]\text{SnO$_2$}\rightarrow \text{Sn} + \text{O$_2$}[/tex]
(Note: oxygen is a diatomic element.)
The equation is balanced as well.
To convert from SnO₂ to only Sn, we can first convert from grams of SnO₂ to moles, use mole ratios to convert to moles of Sn, and then from there convert to grams.
Since Sn has a molar mass of 118.71 g/mol and oxygen has a molar mass of 15.999 g/mol, the molar mass of SnO₂ is:
[tex](118.71)+2(15.999) = 150.708\text{ g/mol}[/tex]
Therefore, given 4.5 kilograms of SnO₂, we can first convert this into grams using 1000 g / kg and then using the ratio:
[tex]\displaystyle \frac{1\text{ mol SnO$_2$}}{150.708\text{ g SnO$_2$}}[/tex]
We can convert this into moles.
Next, from the chemical equation, we can see that one mole of SnO₂ produces exactly one mole of Sn (and also one mole of O₂). So, our mole ratio is:
[tex]\displaystyle \frac{1\text{ mol Sn}}{1\text{ mol SnO$_2$}}[/tex]
With SnO₂ in the denominator to simplify units.
Finally, we can convert from moles Sn to grams Sn using its molar mass:
[tex]\displaystyle \frac{118.71\text{ g Sn}}{1\text{ mol Sn}}[/tex]
With the initial value and above ratios, we acquire:
[tex]\displaystyle 4.5\text{ kg SnO$_2$}\cdot \frac{1000 \text{ g SnO$_2$}}{1\text{ kg SnO$_2$}}\cdot \displaystyle \frac{1\text{ mol SnO$_2$}}{150.708\text{ g SnO$_2$}}\cdot \displaystyle \frac{1\text{ mol Sn}}{1 \text{ mol SnO$_2$}} \cdot\displaystyle \frac{118.71\text{ g Sn}}{1\text{ mol Sn}}[/tex]
Cancel like units:
[tex]=\displaystyle 4.5\cdot \frac{1000}{1}\cdot \displaystyle \frac{1}{150.708}\cdot \displaystyle \frac{1}{1} \cdot\displaystyle \frac{118.71\text{ g Sn}}{1}[/tex]
Multiply. Hence:
[tex]\displaystyle = 3544.5696...\text{ g Sn}[/tex]
Since we should have two significant figures:
[tex]=3500 \text{ g Sn}[/tex]
So, about 3500 grams of tin is produced from smelting 4.5 kg of tin(IV) oxide.
Answer:
3546g
Explanation:
start w/ tin (IV) oxide n elemental tin and oxygen gas are the only products of this reaction
SnO2 -> Sn + O2
Sn molecular wt: 119
O2 molecular wt: 32
SnO2 molecular wt: 119+32 = 151
so Sn / SnO2 wt ratio = 119 / 151
4.5 kilograms of tin (IV) oxide will produce:
= 4.5 * 119 / 151
= 3.546 kg
or 3546 grams of tin metal
no need to involve moles ;)
Choose all the answers that apply. Silicon (Si) has 14 protons and an atomic mass of 28. Silicon has _____. three electron shells 14 electrons 14 neutrons two electron shells 28 electrons
Answer:
three electron shells
14 electrons
14 neutrons
Explanation:
Silicon has three electron shells arranged as follows; 2, 8, 4. This corresponds to the fact that silicon is a member of group 14 of the periodic table.
Note that, the number of protons in an atom is the same as the number of electrons in the neutral atom. Since Silicon has 14 protons, it also has 14 electrons likewise.
The mass number of silicon is 28 but number of neutrons= mass number - number of protons. Since mass number = 28, then there are 14 neutrons in silicon.
If the temperature of a volume of dieal gas ncreases for 100 to 200, what happens to the average kinetic energy of the molecules?
Answer:
It increases but less than double
Explanation:
As the temperature of a gas increase, the average kinetic energy of the gas increases. The kinetic energy of a gas is the thermal energy that the gas contains.
We know, the kinetic energy of an ideal gas is given by :
[tex]$V_{avg} = \sqrt{\frac{8R}{\pi M}}$[/tex]
where, R = gas constant
T = absolute temperature
M = molecular mass of the gas
From the above law, we get
[tex]$V_{avg} \propto \sqrt{T}$[/tex]
Thus, if we increase the temperature then the average kinetic energy of the ideal gas increases.
In the context, if the temperature of the ideal gas increases from 100°C to 200°C, then
[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{\frac{T_2}{T_1}}$[/tex]
[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{\frac{473.15}{373.15}}$[/tex]
[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{1.26}$[/tex]
[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =1.12$[/tex]
[tex]$(V_{avg})_2 = 1.12\ (V_{avg})_1$[/tex]
Therefore, [tex]$(V_{avg})_2 > (V_{avg})_1$[/tex]
Thus the average kinetic energy of the molecule increases but it increases 1.12 times which is less than the double.
Thus, the answer is " It increases but less that double".
please help!
What is the definition of thermal chemistry?
a.The study of change that involves warm objects
b.The study of change that involves heat
c.The study of change that involves cool objects
d.The study of change that involves temperature
7. There are 7. 0 ml of 0.175 M H2C2O4 , 1 ml of water , 4 ml of 3.5M KMnO4 what is the molar concentration ofH2C2O4 ?
8. Using the data from question 7 what is the molar concentration of KMnO4 ?
10. From question number 7, what effect increasing the volume of water has on the reaction rate?
Answer:
7. 0.1021 M
8. 1.167 M
10. Increase in volume of water would lower the rate of reaction
Explanation:
7. What is the molar concentration of H₂C₂O₄ ?
Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.
Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L
So, the molar concentration of H₂C₂O₄, M = number of moles of H₂C₂O₄/volume = n/V
= 1.225 × 10⁻³ mol/12 × 10⁻³ L
= 0.1021 mol/L
= 0.1021 M
8. Using the data from question 7 what is the molar concentration of KMnO₄ ?
Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.
Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L
So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V
= 14 × 10⁻³ mol/12 × 10⁻³ L
= 1.167 mol/L
= 1.167 M
10. From question number 7, what effect increasing the volume of water has on the reaction rate?
Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.
Name of this product
Answer:
Explanation:
ethyl 3-methylbenzoate
Why does the dehydration of an alcohol more often use concentrated sulfuric acid, H 2 S O 4 HX2SOX4, as the acid catalyst rather than dilute hydrochloric acid, H C l HCl
KAnswer:
See explanation
Explanation:
It is more common to use H2SO4 for dehydration reaction rather than HCl because HCl contains a good nucleophile,the chloride ion.
Owing to the presence of the chloride ion, a substitution reaction involving the chloride ion may also proceed also thereby affecting the elimination reaction.
Also, concentrated H2SO4 is a very good drying agent thus, as long as it is used, the alcohol substrate is completely dehydrated to yield the alkene.
Note that HCl is not a dehydrating agent.
A sample of hellium has a volume of 500 mL at STP. What will be its new volume be in mL if the temperature is increased to 325 K and its pressure is increased to 125 kPa?
Answer:
[tex]V_2=482.5mL[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by using the combined gas law due to the fact that we are dealing with variable volume, temperature and pressure:
[tex]\frac{P_2V_2}{T_2}=\frac{P_1V_1}{T_1}[/tex]
In such a way, we solve for the final volume, V2, considering that the initial volume, V1, is 500 mL, the initial temperature, T1, is 273 K (STP), the initial pressure, P1, is 1 atm (STP) and the final temperature, T2, is 325 K and the final pressure, P2, is 125 kPa (1.23 atm):
[tex]V_2=\frac{P_1V_1T_2}{T_1P_2} \\\\V_2=\frac{(1atm)(500mL)(325K)}{(273K)(1.23atm)} \\\\V_2=482.5mL[/tex]
Regards!
