Answer:
1 cubic mile = 1.101 * 10^12 US gallons
1 US bbl oil = 42 US gallons = 3.8143*10^ -11 cubic miles
Explanation:
The number of the exponent of the oil reserve is not very well shown in the question so, I provide you the conversion of bbl oil into cubic mile, the only thing you have to do is multiply the number of bbls of the reserve for the conversion in cubic miles and you'll have the answer.
A space shuttle moves through the atmosphere on its return from space. At point 1 on its wing, the air temperature and pressure are measured to be T1 = 300°C and p1 = 10,000 Pa while at point 2 they are T, = 700°C and p2 = 575,000 Pa. Calculate the air density in kg/m3 at both points 1 and 2.
Answer:
The air density in at points 1 is [tex]60.7 kg/m^3[/tex] and 2 is [tex]2060 kg/m^3[/tex].
Explanation:
Average molecular weight of an air ,M= 28.97 g/mol
[tex]PV=nRT[/tex]
or [tex] PM=dRT[/tex]
P = Pressure of the gas
n = moles of gas
T = Temperature of the gas
d = Density of the gas
M = molar mass of the gas
R = universal gas constant
Density at point-1 = [tex]d_1[/tex]
[tex]P_1 = 10,000 Pa=0.0986 atm[/tex]
[tex]1 Pa=9.86923\times 10^{-6} atm[/tex]
[tex]T_1 = 300^oC = 573.15 K[/tex]
M = 28.97 g/mol
[tex]d_1=\frac{PM}{RT}=\frac{0.0986 atm\times 28.97 g/mol}{0.0821 atm L/ mol K\times 573.15 K}[/tex]
[tex]d_1 =0.0607 g/ml[/tex]
1 g = 0.001 kg
[tex]1 mL = 10^{-6} m^3[/tex]
[tex]d_1=\frac{0.0607\times 0.001 kg}{10^{-6} m^3}=60.7 kg/m^3[/tex]
Density at point-2 = [tex]d_2[/tex]
[tex]P_2 = 575,000 Pa=5.67 atm[/tex]
[tex]T_2 = 700^oC = 973.15 K[/tex]
M = 28.97 g/mol
[tex]d_2=\frac{PM}{RT}=\frac{5.67 atm\times 28.97 g/mol}{0.0821 atm L/ mol K\times 973.15 K}[/tex]
[tex]d_2 =2.06 g/ml=2060 kg/m^3[/tex]
Define "Triose", "Tetrose", "Pentose", and "Hexose" and identify which of these is/are most abundant
Answer:
Monosaccharides are the simplest form of carbohydrates that cannot be hydrolyzed to smaller compounds. Monosaccharides are the basic units of carbohydrates and are also known as simple sugars.
The monosaccharides are classified on the basis of number of carbon atoms present.
Triose is a type of monosaccharide molecule, which is composed of 3 carbon atoms.
Tetrose is a type of monosaccharide molecule, which is composed of 4 carbon atoms.
Pentose is a type of monosaccharide molecule, which is composed of 5 carbon atoms.
Hexose is a type of monosaccharide molecule, which is composed of 6 carbon atoms.
D-glucose is a hexose sugar and it is the most abundant monosaccharide in the nature.
Identify which two compounds below are constitutional isomers
(CH3)3COCH3
(CH3)2CHOCH3
(CH3)2CHOCH2CH3
Answer:
(CH₃)₃COCH3₃ and (CH₃)₂CHOCH₂CH₃
Explanation:
Isomers are compounds which have the same molecular formula. Constitutional isomers have different connectivity; the atoms are connected in different ways.
1. (CH₃)₃COCH₃
2. (CH₃)₂CHOCH3₃
3. (CH₃)₂CHOCH₂CH₃
Molecules 1 and 3 have the same formula (C₅H₁₂O) and are isomers. Molecule 2 is not an isomer. From the structural formula, it is clear that Molecules 1 and 3 have different connectivity.
Calculate the total number of days therapy available within a 300 ml bottle of ranitidine (as hydrochloride) 75 mg/5 ml. oral solution, when it is prescribed at a dose of 300 mg at night?
Answer: This therapy is available for 15 days.
Explanation:
We are given:
Oral solution dosage = 75 mg/5 mL
To calculate the volume of oral situation for single dose per, we use unitary method:
The volume required for 75 mg of solution is 5 mL
So, the volume required for 300 mg of solution will be = [tex]\frac{5mL}{75mg}\times 300mg=20mL[/tex]
The total volume of the ranitidine bottle = 300 mL
To calculate the number of days, we divide the total volume of the bottle by the volume of dose taken per night, we get:
[tex]\text{Number of days}=\frac{\text{Total volume}}{\text{Volume of dose taken per night}}=\frac{300mL}{20mL}=15[/tex]
Hence, this therapy is available for 15 days.
Two moles of ideal He gas are contained at a pressure of 1 atm and a temperature of 300 K. 34166 J of heat are transferred to the gas, as a result of which the gas expands and does 1216 J of work against its surroundings. The process is reversible. (Note: C = 1.5R) Please calculate the final temperature of the gas
Explanation:
The given data is as follows.
n = 2 mol, P = 1 atm, T = 300 K
Q = +34166 J, W= -1216 J (work done against surrounding)
[tex]C_{v}[/tex] = [tex]\frac{3R}{2}[/tex]
Relation between internal energy, work and heat is as follows.
Change in internal energy ([tex]\Delta U[/tex]) = Q + W
= [34166 + (-1216)] J
= 32950 J
Also, [tex]\Delta U = n \times C_{v} \times \Delta T[/tex]
= [tex]3R \times (T_{2} - T_{1})[/tex]
32950 J = [tex]3 \times 8.314 J/mol K \times (T_{2} - 300 K)[/tex]
[tex]\frac{32950}{24.942} = T_{2} - 300 K[/tex]
1321.06 K + 300 K = [tex]T_{2}[/tex]
[tex]T_{2}[/tex] = 1621.06 K
Thus, we can conclude that the final temperature of the gas is 1621.06 K.
4. Each time that you prepare a diluted bleach solution you will want it to have a [OH-] = 0.02. Calculate how much 1.00 M NaOH you need to add to d.1. water to get every 20 mL of this concentration of OH- ion. Express your answer in drops of 1.00 M NaOH. (Assume that there are 20 drops in 1 m.)
Answer:
8 drops of 1.00 M NaOH will be needed.
Explanation:
Concentration of [tex][OH^-][/tex] in bleach solution = 0.02 M
[tex]NaOH\rightarrow OH^-+Na^+[/tex]
[tex]NaOH=[OH^-]=0.02M[/tex]
Concentration of bleach solution we want ,[tex]M_1[/tex] = 0.02 M
Volume of the bleach solution,[tex]V_1[/tex] = 20 ml
Concentration of NaOH solution,[tex]M_2[/tex] = 1.00 M
Volume of the NaOH solution required ,[tex]V_2[/tex] = ?
[tex]M_1V_1=M_2V_2[/tex]
[tex]0.02 M\times 20 mL=1.00 M\times V_2[/tex]
[tex]V_2=\frac{0.02 M\times 20 mL}{1.00 M}=0.4 mL[/tex]
1 mL = 20 drops
0.4 mL = 0.4 × 20 drops = 8 drops
8 drops of 1.00 M NaOH will be needed.
Consider the titration of 100 mL of 0.200 M HCHO, with 1.00 M NaOH. The pK, of HCHO2 is 3.75. a) What is the pH before ANY NaOH is added? b) What is the pH after 5.00 mL of NaOH are added? c) After 10 mL of NaOH are added? d) What is the pH when 20 mL of NaOH have been added? What is this point in the titration called?
Answer:
a) pH = 2,23
b) pH = 3,26
c) pH = 3,74
d) pH = 7,98. Here we have the equivalence point of the titration
Explanation:
In a titration of a strong base (NaOH) with a weak acid (HCOOH) the reaction is:
HCOOH + NaOH → HCOONa + H₂O
a) Here you have just HCOOH, thus:
HCOOH ⇄ HCOO⁻ + H⁺ where ka =1,8x10⁻⁴ and pka = 3,74
When this reaction is in equilibrium:
[HCOOH] = 0,200 -x
[HCOO⁻] = x
[H⁺] = x
Thus, equilibrium equation is:
1,8x10⁻⁴ = [tex]\frac{[x][x] }{[0,200-x]}[/tex]
The equation you will obtain is:
x² + 1,8x10⁻⁴x - 3,6x10⁻⁵ = 0
Solving:
x = -0,006090675 ⇒ No physical sense. There are not negative concentrations
x = 0,005910674
As x = [H⁺] and pH = - log [H⁺]
pH = 2,23
b) Here, it is possible to use:
HCOOH + NaOH → HCOONa + H₂O
With adition of 5,00 mL of 1,00M NaOH solution the initial moles are:
HCOOH = [tex]0,100 L.\frac{0,200 mol}{L} =[/tex] = 2,0x10⁻² mol
NaOH = [tex]0,005 L.\frac{1,00 mol}{L} =[/tex] = 5,0x10⁻³ mol
HCOO⁻ = 0.
In equilibrium:
HCOOH = 2,0x10⁻² mol - 5,0x10⁻³ mol = 1,5x10⁻² mol
NaOH = 0 mol
HCOO⁻ = 5,0x10⁻³ mol
Now, you can use Henderson–Hasselbalch equation:
pH = 3,74 + log [tex]\frac{5,0x10^{-3} }{1,5x10^{-2} }[/tex]
pH = 3,26
c) With adition of 10 mL of 1,00M NaOH solution the initial moles are:
HCOOH = [tex]0,100 L.\frac{0,200 mol}{L} =[/tex] = 2,0x10⁻² mol
NaOH = [tex]0,010 L.\frac{1,00 mol}{L} =[/tex] = 1,0x10⁻² mol
HCOO⁻ = 0.
In equilibrium:
HCOOH = 2,0x10⁻² mol - 1,0x10⁻² mol = 1,0x10⁻² mol
NaOH = 0 mol
HCOO⁻ = 1,0x10⁻² mol
Now, you can use Henderson–Hasselbalch equation:
pH = 3,74 + log [tex]\frac{1,0x10^{-2} }{1,0x10^{-2} }[/tex]
pH = 3,74
d) With adition of 20 mL of 1,00M NaOH solution the initial moles are:
HCOOH = [tex]0,100 L.\frac{0,200 mol}{L} =[/tex] = 2,0x10⁻² mol
NaOH = [tex]0,020 L.\frac{1,00 mol}{L} =[/tex] = 2,0x10⁻² mol
HCOO⁻ = 0.
Here we have the equivalence point of the titration, thus, the equilibrium is:
HCOO⁻ + H₂O ⇄ HCOOH + OH⁻ kb = kw/ka where kw is equilibrium constant of water = 1,0x10⁻¹⁴; kb = 5,56x10⁻¹¹
Concentrations is equilibrium are:
[HCOOH] = x
[HCOO⁻] = 0,1667-x
[OH⁻] = x
Thus, equilibrium equation is:
5,56x10⁻¹¹ = [tex]\frac{[x][x] }{[0,01667-x]}[/tex]
The equation you will obtain is:
x² + 5,56x10⁻¹¹x - 9,27x10⁻¹³ = 0
Solving:
x = -9,628361x10⁻⁷⇒ No physical sense. There are not negative concentrations
x = 9,627806x10⁻⁷
As x = [OH⁻] and pOH = - log [OH⁻]; pH = 14 - pOH
pOH = 6,02
pH = 7,98
I hope it helps!
An aqueous solution of sulfuric acid has a composition of 25.0 wt% sulfuric acid and a specific gravity of 1.22. Calculate the volume of solution that contains 245 kg of sulfuric acid.
Answer: The volume of solution is [tex]8.03\times 10^5mL[/tex]
Explanation:
The relationship between specific gravity and density of a substance is given as:
[tex]\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}[/tex]
Specific gravity of sulfuric acid solution = 1.22
Density of water = 1.00 g/mL
Putting values in above equation we get:
[tex]1.22=\frac{\text{Density of sulfuric acid solution}}{1.00g/mL}\\\\\text{Density of sulfuric acid solution}=(1.22\times 1.00g/mL)=1.22g/mL[/tex]
We are given:
25% (m/m) sulfuric acid solution. This means that 25 g of sulfuric acid is present in 100 g of solution
Conversion factor: 1 kg = 1000 g
Mass of solution having 254 kg or 245000 g of sulfuric acid is calculated by using unitary method:
If 25 grams of sulfuric acid is present in 100 g of solution.
So, 245000 grams of sulfuric acid will be present in = [tex]\frac{100}{25}\times 245000=980000g[/tex]
To calculate volume of a substance, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of solution = 1.22 g/mL
Mass of Solution = 980000 g
Putting values in above equation, we get:
[tex]1.22g/mL=\frac{980000g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{980000g}{1.22g/mL}=8.03\times 10^5mL[/tex]
Hence, the volume of solution is [tex]8.03\times 10^5mL[/tex]
Calculating molarity using solute moles A chemist prepares a solution of sodium chloride (NaCl) by measuring out 6.89 mol of sodium chloride into a 200 ml volumetric flask and filling the the mark with water. Calculate the concentration in mol/L of the chemist's sodium chloride solution. Round your answer to 3 significant digits. X 5 ? Explanation Check
Answer: Concentration of the chemist's sodium chloride solution is 34.4 mol/L.
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n= moles of solute
[tex]V_s[/tex] = volume of solution in ml
Given : moles of [tex]NaCl[/tex] = 6.89
volume of solution = 200 ml
Putting in the values we get:
[tex]Molarity=\frac{6.89\times 1000}{200}=34.4mol/L[/tex]
Thus the concentration of the chemist's sodium chloride solution is 34.4 mol/L.
A sample of nitrogen gas, stored in a 2.96-L container at 32.0°C, exerts a pressure of 4.27 atm. Calculate the number of moles of nitrogen gas in the container. Enter your answer in the box provided. mol
Answer: The number of moles of nitrogen gas is 0.505 moles.
Explanation:
To calculate the number of moles of nitrogen gas, we use ideal gas equation, which is:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 4.27 atm
V = Volume of the gas = 2.96 L
T = Temperature of the gas = [tex]32.0^oC=[32.0+273]K=305K[/tex]
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of gas = ?
Putting values in above equation, we get:
[tex]4.27atm\times 2.96L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 305K\\n=0.505mol[/tex]
Hence, the number of moles of nitrogen gas is 0.505 moles.
Explanation:
The given data is as follows.
Volume = 2.96 L, Temperature = [tex]32.0^{o}C[/tex] = (32 + 273) K = 305 K,
Pressure = 4.27 atm, n = ?
And, according to ideal gas equation, PV = nRT
[tex]4.27atm \times 2.96 L= n \times 0.0821 Latm/mol K \times 305K[/tex]
n = [tex]\frac{12.64 atm L}{25.04 Latm/mol}[/tex]
= 0.505 mol
Thus, we can conclude that the number of moles of nitrogen gas in the container is 0.505 mol.
Based on sulubility rules, which of the
followingwill occur when solutions of ZnSO4(aq)
andMgCl2(aq) are mixed? Write the NIE if aprecipitation
is considered likely.
A. ZnCl2 wil
precipitate;Mg2andSO42_ will be
spectatior ions.
B. ZnSO4 will precipitate;Mg2+ andCl-
will be spectator ions.
C. MgSO4 will precipitate; Zn2+and Cl-
will be spectator ions.
D. MgCl2 will precipitate; Zn2+
andSO42- will be spectator ions.
E. No precipitate will form.
Answer:
The correct option is: E. No precipitate will form.
Explanation:
A solubility chart refers to the list of solubility of various ionic compounds. It shows the solubility of the various compounds in water at room temperature and 1 atm pressure.
Also, according to the solubility rules, the salts of chlorides, bromides and iodides are generally soluble and mostly all salts of sulfate are soluble.
Since, all the compounds formed in this double replacement reaction are soluble in water. Therefore, no precipitate will be formed.
ZnSO₄ (aq) + MgCl₂ (aq) → ZnCl₂ (aq) + MgSO₄ (aq)
A certain rock has a mass of 125 g. The rock is gently lowered into a graduated cylinder whose water level is 250 mL. When the rock is completely submerged, the water level rises to 300 mL What is the volume of the rock?
Answer:
50 mL
Explanation:
The equilibrium reaction below has the Kc = 3.93. If the volume of the system at equilibrium is increased from 2.00 liters to 8.00 liters, how and for what reason will the equilibrium shift? Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer.
co (g) + 3H2(g) -CH4(g) + H2O (g)
Answer:
the equilibium changes since Q > Kc, by increasing the volume, therefore, the reaction will try to use some of the excess product and favor the reverse reaction to reach equilibrium.
Explanation:
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
∴ Kc = [ H2O ] * [ CH4 ] / [ H2 ]³ * [ CO ] = 3.93.....equilibrium, V = 2.00L
PV = nRT; assuming P,T a standart conditions ( 1 atm, 298 K )
⇒ n / V = P / RT
∴ V = 8.00L
∴ R = 0.082 atm.L/K.mol
⇒ mol CO(g) = 0.327 mol = mol H2O = mol CH4
⇒ mol H2(g) = 0.327 mol CO * ( 3mol H2 / mol CO) = 0.981 mol
⇒ [ CO ] = 0.041 = [ CH4 ] = [ H2O]
⇒ [ H2 ] = 0.123 M
∴ Q = [ H2O ] * [ CH4 ] / [ H2 ]³ * [ CO ]
⇒ Q = ( 0.041² ) / (( 0.123³ ) * ( 0.041 ))
⇒ Q = 22.03
Q > Kc⇒ we have more product present than we would have in the equilibrium.
Answer:
Q = 62.9
Q > Kc, so the reaction will proceed to the left so that Q takes the value of Kc.
Explanation:
Let's consider the following reaction.
CO(g) + 3 H₂(g) ⇄ CH₄(g) + H₂O(g)
The equilibrium constant (Kc) is:
[tex]Kc= 3.93 =\frac{[CH_{4}].[H_{2}O]}{[CO].[H_{2}]^{3} }[/tex]
If the volume is multiplied by 4 (2.00 L → 8.00 L), the concentrations will be divided by 4. The reaction quotient (Q) is:
[tex]Q=\frac{(0.25[CH_{4}]).(0.25[H_{2}O])}{(0.25[CO]).(0.25[H_{2}])^{3} }=16\frac{[CH_{4}].[H_{2}O]}{[CO].[H_{2}]^{3} }=16Kc=16 \times 3.93 = 62.9[/tex]
Q > Kc, so the reaction will proceed to the left so that Q takes the value of Kc.
Calculate the freezing temperature of the following solution of 0.50 M glucose (a covalent compound). Assume that the molality of the solution is 0.50 m. (The molar and molal concentrations of dilute aqueous solutions are often identical to two significant figures.) Enter your answer in the provided box. 0.50 m glucose (a covalent compound) °C
Answer:
-0.93 °C
Explanation:
Hello,
The freezing-point depression is given by:
[tex]T_f-T_f^*=-iK_{solvent}m_{solute}[/tex]
Whereas [tex]T_f[/tex] is the freezing temperature of the solution, [tex]T_f^*[/tex] is the freezing temperature of the pure solvent (0 °C since it is water), [tex]i[/tex] the Van't Hoff factor (1 since the solute is covalent), [tex]K_{f,solvent}[/tex] the solvent's freezing point depression point constant (in this case [tex]1.86 C\frac{kg}{mol}[/tex]) and [tex]m_{solute}[/tex] the molality of the glucose.
As long as the unknown is [tex]T_f[/tex], solving for it:
[tex]T_f=T_f^*-iK_fm\\T_f=0C-1*1.86C\frac{kg}{mol}*0.5\frac{mol}{kg} \\T_f=-0.93C[/tex]
Best regards.
AG' for the isomerization reaction glucose-1-phosphate (GIP) $ glucose-6-phosphate (G6P) is -7.1 kJ/mol. Calculate the equilibrium ratio of [G1P] to (G6P) at 25°C. Read this carefully to make sure you solve for the correct ratio! SHOW WORK! [G1P][G6P] =
Answer:
The ratio [G1P]/[G6P] = 5.7 . 10⁻².
Explanation:
Let us consider the reaction G1P ⇄ G6P, with ΔG° = -7.1 kJ/mol. According to Hess's Law, we can write the inverse reaction, and Gibbs free energy would have an opposite sign.
G6P ⇄ G1P ΔG° = 7.1 kJ/mol
This is the reaction for which we want to find the equilibrium constant (the equilibrium ratio of [G1P] to [G6P]):
[tex]Kc=\frac{[G1P]}{[G6P]}[/tex]
The equilibrium constant and Gibbs free energy are related by the following expression:
[tex]Kc=e^{-\Delta G\si{\textdegree}/R.T } } =e^{-7.1kJ/mol/8.314.10^{-3}kJ/mol.K.298K} } }=5.7.10^{-2}[/tex]
where,
R is the ideal gas constant (8.314 . 10⁻3 kJ/mol.K)
T is the absolute temperature (in kelvins)
Final answer:
To calculate the equilibrium ratio of [G1P] to [G6P], the Gibbs free energy equation is used with ΔG', universal gas constant R, and the temperature T substituted. The result is that the concentration of G6P is approximately 1.331 times that of G1P at equilibrium and at 25°C.
Explanation:
The student is asking about the equilibrium ratio of concentrations of glucose-1-phosphate ([G1P]) to glucose-6-phosphate ([G6P]) at 25°C when the standard free energy change (ΔG') for the isomerization reaction is given as -7.1 kJ/mol. To calculate this ratio, we can use the Gibbs free energy equation for the equilibrium constant (Keq):
ΔG' = -RT ln(Keq)
Where ΔG' is the standard free energy change, R is the universal gas constant (8.314 J/mol K), T is the temperature in Kelvin (25°C + 273.15 = 298.15 K), and Keq is the equilibrium constant which for this reaction is [G6P]/[G1P].
Substituting the values into the equation we get:
-7100 J/mol = -(8.314 J/mol K)(298.15 K) ln([G6P]/[G1P])
Now, we solve for ln([G6P]/[G1P]):
ln([G6P]/[G1P]) = ΔG' / (-R * T)
ln([G6P]/[G1P]) = -7100 J/mol / (-(8.314 J/mol K)(298.15 K))
ln([G6P]/[G1P]) = 0.286
Exponentiating both sides to remove the natural logarithm, we get:
[G6P]/[G1P] = e0.286 = 1.331
Therefore, at equilibrium and at 25°C, the concentration of G6P is approximately 1.331 times that of G1P.
A pellet of Zn of mass 10.0g is dropped into a flaskcontaining
dilute H2SO4 at a pressure of P=1.00 bar and
temperature of 298K. What is the reaction thatoccures? Calculate w
for the process.
Answer: Work done for the process is -390 J
Explanation:
The chemical equation for the reaction of zinc metal with sulfuric acid follows:
[tex]Zn+H_2SO_4\rightarrow ZnSO_4+H_2[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of zinc = 10.0 g
Molar mass of zinc = 65.38 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of zinc}=\frac{10.0g}{65.38g/mol}=0.153mol[/tex]
The equation given by ideal gas follows:
[tex]P\Delta V=nRT[/tex]
where, P = pressure of the gas
[tex]\Delta V[/tex] = Change in volume of the gas
T = Temperature of the gas = 298 K
R = Gas constant = 8.314 J/mol.K
n = number of moles of gas = 0.153 mol
Putting values in above equation, we get:
[tex]P\Delta V=0.153mol\times 8.314J/mol.K\times 298K\\\\P\Delta V=397J[/tex]
To calculate the work done, we use the equation:
[tex]\text{Work done}=-P\Delta V\\\\W=-390J[/tex]
Hence, work done for the process is -390 J
How many significant figures are represented in each of the following numbers? a) 7.1 x 10^-5 b) 0.00677 c) 750
Answer:
a) 7.1 x 10⁻⁵ : 2 significant figures
b) 0.00677 : 3 significant figures
c) 750 : 2 significant figure
Explanation:
The significant digits or figures refers to the digits of a given number that carry meaning and also contributes to precision of the given number.
a) 7.1 x 10⁻⁵ = 0.000071 : 2 significant figures, leading zeros are not significant.
b) 0.00677 : 3 significant figures, leading zeros are not significant.
c) 750 : 2 significant figure, trailing zeros are not significant.
Assume that the NO, concentration in a house with a gas stove is 150 pg/m°. Calculate the equivalent concentration in ppm at STP.
Explanation:
It is known that for [tex]NO_{2}[/tex], ppm present in 1 [tex]mg/m^{3}[/tex] are as follows.
1 [tex]\frac{mg}{m^{3}}[/tex] = 0.494 ppm
So, 150 [tex]pg/m^{3}[/tex] = [tex]\frac{150}{1000} mg/m^{3}[/tex]
= 0.15 [tex]mg/m^{3}[/tex]
Therefore, calculate the equivalent concentration in ppm as follows.
[tex]0.15 \times 0.494 ppm[/tex]
= 0.074 ppm
Thus, we can conclude that the equivalent concentration in ppm at STP is 0.074 ppm.
Write Huckel's rule below and determine how many electrons are required to make an aromatic ring with n = 0, 1, and 2.
Answer: The number of electrons for n = 0, 1 and 2 are 2, 6 and 10 respectively.
Explanation:
Huckel's rule is used to determine the aromaticity in a compound. The number of delocalized [tex]\pi-[/tex] electrons are calculated by using the equation:
[tex]\text{Number of delocalized }\pi-\text{ electrons}=4n+2[/tex]
where,
n = 0 or any whole number
Calculating the value of electrons for n = 0Putting values in above equation, we get:
[tex]\text{Number of delocalized }\pi-\text{ electrons}=4(0)+2=2[/tex]
Calculating the value of electrons for n = 1Putting values in above equation, we get:
[tex]\text{Number of delocalized }\pi-\text{ electrons}=4(1)+2=6[/tex]
Calculating the value of electrons for n = 2Putting values in above equation, we get:
[tex]\text{Number of delocalized }\pi-\text{ electrons}=4(2)+2=10[/tex]
Hence, the number of electrons for n = 0, 1 and 2 are 2, 6 and 10 respectively.
Enter your answer in the provided box. Calculate the number of g of CO2 produced from the combustion of 5.24 mol of CzHg. The balanced equation is: C3H2(g) + 502(g) → 3CO2(g) + 4H2O(g). g CO2
Answer:
691.84g
Explanation:
I'm assuming that by CzHg, you mean C3H2
First, use the mole ratio in the equation to find the moles of CO2
n (CO2)= n ( C3H2) × 3
= 5.24 × 3
= 15.72
To find the mass of CO2 produced in grams, complete the following calculation
m= n × MM
where
m = mass
n= moles
MM= molecular mass
m= 15.72 × (12.01 +( 16×2))
m =691.8372
m= 691.84g
Define the terms "Aldonic Acid", "uronic Acid", and "Aldaric Acid"
Explanation:
Aldonic Acid:
Aldonic acids are suger acids.
General formula of aldonic acid = [tex]HOOC-(CHOH)_n-CH_2OH[/tex]
Aldonic acids are obtained by the oxidation of aldehydic group of suger.
So, aldonic acids have hydroxyl group at one terminal and carboxylic group at another terminal.
Gluconic acid is an example of aldonic acid.
Uronic Acid:
It is also a type of suger acid having carbonyl functional group at one terminal and carboxylic group at other terminal.
It is obtained by oxidation of hydroxyl group of the sugar.
Aldaric Acid:
Aldaric acid is also a type of sugar acid having carboxylic acid functional group at both the ends.
Both the hydroxyl group and aldehydic group are oxidized to form class of compound, called aldaric acid.
A 2.60 gram sample of a compound know to contain only indium and chlorine is dissolved in 50.0 g of tin(IV) chloride (Kb = 9.43oC kg mol-1). The normal boiling point is raised from 114.1oC for pure SnCl4 to 116.3oC for the solution. What is the molecular weight and probable molecular formula of the solute?
Answer: 1. The molecular weight of the compound is 222.8 g/mol
2. The probable molecular formula of the solute is [tex]InCl_3[/tex]
Explanation:
Elevation in boiling point is given by:
[tex]\Delta T_b=i\times K_f\times m[/tex]
[tex]\Delta T_b=T_b-T_b^0=(116.3-114.1)^0C=2.2^0C[/tex] = elevation in boiling point
i= vant hoff factor = 1 (for non electrolyte)
[tex]K_b[/tex] = boiling point constant = [tex]9.43^0Ckg/mol[/tex]
m= molality
[tex]\Delta T_b=i\times K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent (tin chloride)= 50.0 g =0.05 kg
Molar mass of unknown solute = M g/mol
Mass of unknown solute = 2.6 g
[tex]2.2=1\times 9.43\times \frac{2.6g}{M g/mol\times 0.05kg}[/tex]
[tex]M=222.8g/mol[/tex]
The possible formula for the compound would be [tex]InCl_3[/tex] as indium has valency of 3 and chlorine has valency of 1 has molecular mass almost equal to 222.8.
when doing a problem like .3065g/138.03ml is the answer 2.220532X10(^3) g/ml or is it written differently?
Answer:
It is written differently.
Explanation:
The answer as given is not written correctly.
First, we must consider significant figures. Any non-zero number is a significant figure, and zeroes between significant figures are also significant, so 0.3065 has four significant figures and 138.03 has five significant figures. (Leading zeroes before the decimal point are not significant)
When two numbers are multiplied or divided, the answer has the same number of significant figures as the number that had the least number of significant figures. So in this case, the result will have four significant figures.
The result is written in scientific notation:
(0.3065)/(138.03) = 2.221 x 10⁻³
Show that 1 kJ/kg = 1000 m2/S2
Answer:
1000m2 / s2
Explanation:
Hello! In order to verify this, we have to do unit conversion. We also have to know that J (Joule) = kg * m2 / s2
Then we can start with the test.
1kJ / kg * (1000J / 1kJ) = 1000J / kg
1000J / kg = 1000kg * m2 / kg * s2
In this step we can simplify "kg".
So the result is
1000m2 / s2
Final answer:
To demonstrate that 1 kJ/kg equals 1000 m²/s², we recognize that the joule (J) is defined as kg-m²/s², and by converting kJ to J and canceling out the kg units, we affirm the equality.
Explanation:
To show that 1 kJ/kg is equal to 1000 m²/s², we start by recognizing that the unit of energy, the joule (J), is defined as 1 kilogram-meter²/second² (kg-m²/s²). Therefore, when we talk about energy per unit mass, we are effectively dividing energy by mass, leaving us with units of m²/s².
Given that 1 joule is 1 kg·m²/s², 1 kJ is 1000 joules (since the prefix 'kilo' means 1000). So when we have 1 kJ/kg, it's the same as saying 1000 J/kg. When we divide each term (kg·m²/s²) by kg, the kilograms cancel out, leaving us with m²/s². Thus, 1 kJ/kg is indeed equivalent to 1000 m²/s².
Activity of the bacterial enzyme acetoacetate decarboxylase sharply declines below pH 6, suggesting there is an ionizable catalytic residue with pKa ~ 6. Using mutagenesis, this critical catalytic residue was determined to be Lys-115. a) What is the typical pKa of a lysine side chain and what ratio of deprotonated to protonated lysine would be expected at pH 7.5 (standard pH inside a bacterium)
Answer:
The typical pka of a lysine side chain is 10.5At pH=7-5 the ratio of deprotonated to protonated lysine is 0.001Explanation:
From literature, we know that the typical pka of a lysine side is 10.3.
Then we use the Henderson-Hasselbalch equation:
[tex]pH=pka+log\frac{[A^{-} ]}{[HA]}[/tex]
In this case, [A⁻] is the concentration of deprotonated lysine, and [HA] is the concentration of protonated lysine.
We put the data from the problem in the equation and calculate the ratio of deprotonated to protonated lysine:
[tex]7.5=10.5+log\frac{[A^{-} ]}{[HA]}\\-3.0=log\frac{[A^{-} ]}{[HA]}\\10^{-3.0}=\frac{[A^{-} ]}{[HA]}\\0.001=\frac{[A^{-} ]}{[HA]}[/tex]
How many neutrons and protons are there in the nuclei of the following atoms. a. Li-7 b. O-16 c. Th-232 d. Pu-239
Answer:
a.
P = 3 , N = 4
b.
P = 8 , N = 8
c.
P = 90 , N = 142
d.
P = 94 , N = 145
Explanation:
Mass number = Number of protons + Number of neutrons
Also,
Atomic number = Number of protons
Li - 7
Given, Mass number of Lithium = 7
For lithium, atomic number = 3
So,
Number of protons = 3
Number of neutrons = Mass number - Number of protons = 7 - 3 = 4
O - 16
Given, Mass number of oxygen = 16
For oxygen, atomic number = 8
So,
Number of protons = 8
Number of neutrons = Mass number - Number of protons = 16 - 8 = 8
Th - 232
Given, Mass number of Thorium = 232
For Thorium, atomic number = 90
So,
Number of protons = 90
Number of neutrons = Mass number - Number of protons = 232 - 90 = 142
Pu - 239
Given, Mass number of Plutonium = 239
For Plutonium, atomic number = 94
So,
Number of protons = 94
Number of neutrons = Mass number - Number of protons = 239 - 94 = 145
A pipe of diameter 10 cm carries water at a velocity of 5 m/s. Determine the volumetric flow rate in m'/min (2 pts) a. b. the mass flow rate in kg/min (use the density of water on Thatcher's sheet)
Explanation:
It is given that diameter of the pipe is 10 cm which is also equal to [tex]10 \times 10^{-2}m[/tex].
Velocity of water = 5 m/s
(a) Formula to calculate volumetric flow rate is as follows.
Q = Area of the pipe (A) × Velocity of water (V)
= [tex]\frac{\pi}{4} \times 10 \times 10^{-2} \times 5 m^{3}/sec[/tex]
= 0.039 [/tex]m^{3}/sec[/tex]
= [tex]\frac{0.039 m^{3}/sec \times 60 sec}{1 min}[/tex]
= 2.36 [tex]m^{3} min^{-1}[/tex]
Hence, the volumetric flow rate is 2.36 [tex]m^{3} min^{-1}[/tex].
(b) Formula to calculate mass flow rate is as follows.
[tex]Q \times \rho[/tex]
= [tex]2.36 m^{3} min^{-1} \times 1000 kg m^{-3}[/tex]
= 2356.19 kg/min
Therefore, the mass flow rate is 2356.19 kg/min.
The specific reaction rate of a reaction at 492k is 2.46 second inverse and at 528k 47.5 second inverse.find activation energy and ko.
Answer:
Ea = 177x10³ J/mol
ko = [tex]1.52x10^{19}[/tex] J/mol
Explanation:
The specific reaction rate can be calculated by Arrhenius equation:
[tex]k = koxe^{-Ea/RT}[/tex]
Where k0 is a constant, Ea is the activation energy, R is the gas constant, and T the temperature in Kelvin.
k depends on the temperature, so, we can divide the k of two different temperatures:
[tex]\frac{k1}{k2} = \frac{koxe^{-Ea/RT1}}{koxe^{-Ea/RT2}}[/tex]
[tex]\frac{k1}{k2} = e^{-Ea/RT1 + Ea/RT2}[/tex]
Applying natural logathim in both sides of the equations:
ln(k1/k2) = Ea/RT2 - Ea/RT1
ln(k1/k2) = (Ea/R)x(1/T2 - 1/T1)
R = 8.314 J/mol.K
ln(2.46/47.5) = (Ea/8.314)x(1/528 - 1/492)
ln(0.052) = (Ea/8.314)x(-1.38x[tex]10^{-4}[/tex]
-1.67x[tex]10^{-5}[/tex]xEa = -2.95
Ea = 177x10³ J/mol
To find ko, we just need to substitute Ea in one of the specific reaction rate equation:
[tex]k1 = koxe^{-Ea/RT1}[/tex]
[tex]2.46 = koxe^{-177x10^3/8.314x492}[/tex]
[tex]1.61x10^{-19}ko = 2.46[/tex]
ko = [tex]1.52x10^{19}[/tex] J/mol
Dinitrogen monoxide gas is collected at-3.0 °C in an evacuated flask with a measured volume of 5.0 L. When all the gas has been collected, the press the flask is measured to be 0.100 atm . Calculate the mass and number of moles of dinitrogen monoxide gas that were collected. Round your answer to 2 significant digits. mass: mole: nol Explanation Check 2019 McGraw-Hill Education All Rights Reserved
Answer:
0.971 grams
Explanation:
Given:
Temperature = 3.0° C = 3 + 273 = 276 K
Volume, V = 5.0 L
Pressure, P = 0.100 atm
Now, from the relation
PV = nRT
where,
n is the number of moles,
R is the ideal gas constant = 0.082057 L atm/mol.K
thus,
0.1 × 5 = n × 0.082057 × 276
or
n = 0.022 moles
Also,
Molar mass of the Dinitrogen monoxide gas (N₂O)
= 2 × Molar mass of nitrogen + 1 × Molar mass of oxygen
= 2 × 14 + 16 = 44 grams/mol
Therefore, Mass of 0.022 moles of N₂O = 0.022 × 44 = 0.971 grams
A 1.24g sample of a hydrocarbon, when completely burned in an excess of O2 yields 4.04g Co2 and 1.24g H20. Draw plausible structure for the hydrocarbon molecule
Answer:
Plausible structure has been given below
Explanation:
Molar mass of [tex]CO_{2}[/tex] is 44 g/mol and molar mass of [tex]H_{2}O[/tex] is 18 g/molNumber of mole = (mass/molar mass)4.04 g of [tex]CO_{2}[/tex] = [tex]\frac{4.04}{44}moles[/tex] [tex]CO_{2}[/tex] = 0.0918 moles of [tex]CO_{2}[/tex]
1 mol of [tex]CO_{2}[/tex] contains 1 mol of C atom
So, 0.0918 moles of [tex]CO_{2}[/tex] contains 0.0918 moles of C atom
1.24 g of [tex]H_{2}O[/tex] = [tex]\frac{1.24}{18}moles[/tex] [tex]H_{2}O[/tex] = 0.0689 moles of [tex]H_{2}O[/tex]
1 mol of [tex]H_{2}O[/tex] contain 2 moles of H atom
So, 0.0689 moles of [tex]H_{2}O[/tex] contain [tex](2\times 0.0689)moles[/tex] of [tex]H_{2}O[/tex] or 0.138 moles of [tex]H_{2}O[/tex]
Moles of C : moles of H = 0.0918 : 0.138 = 2 : 3
Empirical formula of hydrocarbon is [tex]C_{2}H_{3}[/tex]
So, molecular formula of one of it's analog is [tex]C_{4}H_{6}[/tex]
Plausible structure of [tex]C_{4}H_{6}[/tex] has been given below.