Answer : The volume of [tex]CO_2[/tex] gas is 1.00 L
Explanation :
First we have to determine the moles of [tex]C_3H_4PO_7^{3-}[/tex].
Molar mass of [tex]C_3H_4PO_7^{3-}[/tex] = 182.9 g/mole
[tex]\text{ Moles of }C_3H_4PO_7^{3-}=\frac{\text{ Mass of }C_3H_4PO_7^{3-}}{\text{ Molar mass of }C_3H_4PO_7^{3-}}=\frac{15.0g}{182.9g/mole}=0.0820moles[/tex]
Now we have to calculate the moles of [tex]CO_2[/tex].
The given balanced chemical reaction is:
[tex]C_5H_8P_2O_{11}^{4-}(aq)+H_2O(aq)+CO_2(g)\rightarrow 2C_3H_4PO_7^{3-}(aq)+2H^+(aq)[/tex]
From the reaction we conclude that,
As, 2 moles of [tex]C_3H_4PO_7^{3-}[/tex] produce from 1 mole of [tex]CO_2[/tex]
So, 0.0820 moles of [tex]C_3H_4PO_7^{3-}[/tex] produce from [tex]\frac{0.0820}{2}=0.041moles[/tex] of [tex]CO_2[/tex]
Now we have to calculate the volume of [tex]CO_2[/tex] gas.
Using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = pressure of gas = 1.00 atm
V = volume of gas = ?
T = temperature of gas = 298 K
n = number of moles of gas = 0.041 mole
R = gas constant = 0.0821 L.atm/mole.K
Now put all the given values in the ideal gas equation, we get:
[tex](1.00atm)\times V=(0.041mole)\times (0.0821L.atmK^{-1}mol^{-1})\times (298K)[/tex]
[tex]V=1.00L[/tex]
Therefore, the volume of [tex]CO_2[/tex] gas is 1.00 L
What masses of monobasic and dibasic sodium phosphate will you use to make 250 mL of 0.1 M sodium phosphate buffer, pH = 7?
Answer:
Mass of monobasic sodium phosphate = 1.857 g Mass of dibasic sodium phosphate = 1.352 gExplanation:
The equilibrium that takes place is:
H₂PO₄⁻ ↔ HPO₄⁻² + H⁺ pka= 7.21 (we know this from literature)
To solve this problem we use the Henderson–Hasselbalch (H-H) equation:
pH = pka + [tex]log\frac{[A^{-} ]}{[HA]}[/tex]
In this case [A⁻] is [HPO₄⁻²], [HA] is [H₂PO₄⁻], pH=7.0, and pka = 7.21
If we use put data in the H-H equation, and solve for [HPO₄⁻²], we're left with:
[tex]7.0=7.21+log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\ -0.21=log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\\\10^{-0.21} =\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\0.616 * [H2PO4^{-}] = [HPO4^{-2}][/tex]
From the problem, we know that [HPO₄⁻²] + [H₂PO₄⁻] = 0.1 M
We replace the value of [HPO₄⁻²] in this equation:
0.616 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.1 M
1.616 * [H₂PO₄⁻] = 0.1 M
[H₂PO₄⁻] = 0.0619 M
With the value of [H₂PO₄⁻] we can calculate [HPO₄⁻²]:
[HPO₄⁻²] + 0.0619 M = 0.1 M
[HPO₄⁻²] = 0.0381 M
With the concentrations, the volume and the molecular weights, we can calculate the masses:
Molecular weight of monobasic sodium phosphate (NaH₂PO₄)= 120 g/mol.Molecular weight of dibasic sodium phosphate (Na₂HPO₄)= 142 g/mol.mass of NaH₂PO₄ = 0.0619 M * 0.250 L * 120 g/mol = 1.857 gmass of Na₂HPO₄ = 0.0381 M * 0.250 L * 142 g/mol = 1.352 gA solution is prepared by mixing 30 ml of a 0.2 M hydrochloric acid (HCl) with 20 ml of a 0.3 M potassium hydroxide solution (KOH), what will be the final pH? (Show your calculation to get full credit.) Hint: Determine whether hydrochloric acid and potassium hydroxide are strong or weak acids or bases.
Answer:
The final pH is 7
Explanation:
KOH + HCl ------> KCl + H2O
KOH ----> K+ + OH-
0,3 M 0,3 M + 0,3 M
HCl -----> H+ + Cl-
0,2 M 0,2M + 0,2M
In both cases they are strong acid and base because they dissociate completely; The initial molarity is the same, at the end of the dissociation.
As we have an acid next to a strong base, the final solution will have a neutralization reaction where the OH- of the base react with the H + of the acid to give water. The excess of OH- or H + defines whether the solution will have an acid or alkaline pH.
Molarity = moles / volume (L)
Moles of OH- = 0,3 M x 0,020 L = 0,006 moles
(remember volume in L, so 20mL = 0,020 L)
Moles of H+ = 0,2 M x 0,030 L = 0,006 moles
As we have the same quantity for OH- and H+, the solution will have neutral pH.
Consider the following reaction: A+B C +D a) What is the correct expression for the equilibrium constant (k)? b) The k value for this reaction is 45000. What does this tell you about the reaction? c) If the Ris 45000 are the reactants more stable than the products?
Answer:
a) k = [tex]\frac{[C][D]}{[A][B]}[/tex]
b) A value of 45000 means that tendency of your reaction is to have more products in a ratio of 45000:1
c) No, the products are more stable than reactants.
Explanation:
a) For a reaction:
A + B → C + D
The equlibrium constant (k) is:
k = [tex]\frac{[C][D]}{[A][B]}[/tex]
Where [x] is the molar concentration of x. Always the expression of equilibrium constant is molar concentration of products over molar concentration of reagents.
b) Having in mind the expression of equilibrium constant when k>1 the concentration of products is higher than concentration of reagents. Thus, when k<1 concentration of reagents is higher than concentration of products.
A value of 45000 means that tendency of your reaction is to have more products in a ratio of 45000:1
c) Again, a value of 45000 means that tendency of your reactants is react to produce products. Thus, the products are more stable than reactants.
I hope it helps!
For the reaction A+B+C=>D+E, the initial reaction rate was measured for various initial concentrations of reactants. The following data were collected:
Trial A(M) --------B(M)------ C(M) ------ Initial Rate(M/s)
1 ------- 0.40 -------- 0.40 ------ 0.40 --------- 1.2 x 10^-4
2 --------0.40 --------- 0.40 -------1.20 -----------3.6 x 10^-4
3------- 0.80 ----------0.40------- 0.40 ---------- 4.8 x 10^-4
4 ------- 0.80 ----------0.80 -------0.40 ----------4.8 x 10^-4
What is the value of the rate constant k for this reaction?
Answer : The value of the rate constant 'k' for this reaction is [tex]1.9375\times 10^{-3}M^{-2}s^{-1}[/tex]
Explanation :
Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
For the given chemical equation:
[tex]A+B+C\rightarrow D+E[/tex]
Rate law expression for the reaction:
[tex]\text{Rate}=k[A]^a[B]^b[C]^c[/tex]
where,
a = order with respect to A
b = order with respect to B
c = order with respect to C
Expression for rate law for first observation:
[tex]1.24\times 10^{-4}=k(0.40)^a(0.40)^b(0.40)^c[/tex] ....(1)
Expression for rate law for second observation:
[tex]3.6\times 10^{-4}=k(0.40)^a(0.40)^b(1.20)^c[/tex] ....(2)
Expression for rate law for third observation:
[tex]4.8\times 10^{-4}=k(0.80)^a(0.40)^b(0.40)^c[/tex] ....(3)
Expression for rate law for fourth observation:
[tex]4.8\times 10^{-4}=k(0.80)^a(0.80)^b(0.40)^c[/tex] ....(4)
Dividing 1 from 2, we get:
[tex]\frac{3.6\times 10^{-4}}{1.24\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(1.20)^c}{k(0.40)^a(0.40)^b(0.40)^c}\\\\3=3^c\\c=1[/tex]
Dividing 1 from 3, we get:
[tex]\frac{4.8\times 10^{-4}}{1.24\times 10^{-4}}=\frac{k(0.80)^a(0.40)^b(0.40)^c}{k(0.40)^a(0.40)^b(0.40)^c}\\\\4=2^a\\a=2[/tex]
Dividing 3 from 4, we get:
[tex]\frac{4.8\times 10^{-4}}{4.8\times 10^{-4}}=\frac{k(0.80)^a(0.40)^b(0.40)^c}{k(0.80)^a(0.80)^b(0.40)^c}\\\\1=2^b\\b=0[/tex]
Thus, the rate law becomes:
[tex]\text{Rate}=k[A]^2[B]^0[C]^1[/tex]
Now, calculating the value of 'k' by using any expression.
Putting values in equation 1, we get:
[tex]1.24\times 10^{-4}=k(0.40)^2(0.40)^0(0.40)^1[/tex]
[tex]k=1.9375\times 10^{-3}M^{-2}s^{-1}[/tex]
Hence, the value of the rate constant 'k' for this reaction is [tex]1.9375\times 10^{-3}M^{-2}s^{-1}[/tex]
What is the unit of catalytic reaction rate?
a) Mole/time
b) Mole * kg * time
c) Mole * kg / time
d) other
Answer:
The answer is the option "A"
Explanation:
The reaction rate is the amount of substance formed or transformed per unit of time. The catalytic action allows to increase the reaction rate or the selectivity. The units are moles / time (mol / s)
mol = moles
s = seconds
So the correct answer is "A"
What is inaccurate about describing a solution as containing or using free hydrogen atoms?
Answer: An aqueous solution will contains both hydrogen ion and hydroxide ion.An here we are talking about free hydrogen ions, which means the description will be related to the pH. If the ratio of ion are equal then the pH will be 7 i.e. neutral. But if there are free hydrogen atoms in the solution we have, the pH of solution will be less than 7 and the solution will be described ad acidic. So the inaccurate description for a solution containing free hydrogen ion is that it is basic in nature.
Which of the following reactions is a double displacement reaction? A) HCI (9) - Hz (g) + Cl2 (9) B) HCI (aq) + NaOH (aq) - H2O (1) + Naci (aq) C) Mg () + 2HCl (aq) - MgCl2 (aq) + H2(g) D) HCI (9) + C-H10 (9) -CsH..CI (9)
Answer:
HCI (aq) + NaOH (aq) --> H2O (l) + NaCl (aq)
Explanation:
A double displacement reaction is a type of reaction where two reactants exchange ions to form two new compounds. Here we can see that both the acid and the base exchanged an ion each to for salt and water, 2 new products.
Consider four small molecules, A–D, which have the following binding affinities for a specific enzyme (these numbers are the equilibrium constants Kd for the dissociation of the enzyme/molecule complex). Which binds most tightly to the enzyme? Which binds least tightly?
A) 4.5 μM
B) 13 nM
C) 8.2 pM
D) 6.9 mM
Answer:
Binding affinity measures the strength of the interaction between a molecule to its ligand; it is expressed in terms of the equilibrium dissociation constant; and the higher value of this constant, the more weaker the binding between the molecule and the ligand is. On the other hand, small constans means that the interaction is tight. So "C" binds most tightly to the enzyme and "D" binds least tightly.
The exhaust gas from an automobile contains 1.5 percent by volume of carbon monoxide. What is the concentration of CO in mg/m' at 25°C and 1 atm pressure? What is the concentration in mg/m' in the exhaust pipe if it is at 200°C and 1.1 atm of pressure?
Answer:
(a) 17,178 mg/m3
(b) 11,625 mg/m3
Explanation:
The concentration of CO in mg/m3 can be calculated as
[tex]C (mg/m3) =(P/RT)*MW*C_{ppm}[/tex]
For standard conditions (1 atm and 25°C), P/RT is 0.0409.
Concentration of 1.5% percent by volume of CO is equivalent to 1.5*10,000 ppm= 15,000 ppm CO.
The molecular weigth of CO is 28 g/mol.
(1) For 25°C and 1 atm conditions
[tex]C=(P/RT)*MW*C_{ppm}\\\\C=0.0409*28*15,000=17,178[/tex]
(b) For 200°C and 1.1 atm,
[tex]P/RT=0.0409*(P/P_{std})*(T_{std}/T)\\P/RT=0.0409*(1.1atm/1atm)*(273+15K/273+200K)=0.0277[/tex]
Then the concentration in mg/m3 is
[tex]C=(P/RT)*MW*C_{ppm}\\\\C=0.0277*28*15,000=11,625[/tex]
The National Institute for Occupational Safety and Health (NIOSH) sets standards for CCl4 in air at 12.6 mg/m^3 of air (a time average over 40 hr). The CCl4 found in a sample is 4800 ppb. Does the sample meet the NIOSH standard?
Answer:
No, it doesn't.
Explanation:
To convert ppb to mg/m³ we first need to convert to ppm, by just divide the amount by 1,000, so the concentration in the sample is 4.8 ppm.
mg/m³ = (ppm x molar mass)/molar volume
Using the molar mass in gram and the molar volume in liters, multiplying by the parts per million, we will get the concentration in mg/m³.
Molar mass of C = 12 g/mol; molar mass of Cl = 35.5 g/mol
Molar mass of CCl4 = 12 + 4x35.5 = 154 g/mol
Assuming, 25ºC and 1 atm, the molar volume of an ideal gas is 24.45 L, so:
mg/m³ = (4.8 x 154)/24.45
mg/m³ = 30.2
Which is higher than the limit of 12.6 mg/m³
Answer:
Sample is greater than the standard (32.96 mg/m³ > 12.6 mg/m³)
Explanation:
Please look at the solution in the attached Word file
How to dispose chemical waste? If unsure what should you do?
A well-insulated tank contains concentrated HCl solution in water. If more water is added to this tank the temperature of the system will __________.
Select one:
a. increase
b. decrease
c. no change
d. cannot tell
Answer: Option (a) is the correct answer.
Explanation:
As the given situation shows that water is being added to water. This will liberate heat into the surround as the reaction will be exothermic in nature.
It is known that chemical reactions in which heat is released are known as exothermic reactions.
Hence, when more water is added to this tank the temperature of the system will increase due to release of heat.
Thus, we can conclude that if more water is added to this tank the temperature of the system will increase.
Calculate the cell potential for a cell operating with the following reaction at 25 degrees Celsius, in which [MnO4^1-] = .01M, [Br^1-] = .01M, [Mn^2+] = .15M, and [H^1+] = 1M. The reaction is 2 MnO4^1-(aq) + 10 Br^1-(aq) + 16 H^1+(aq) --> 2 Mn^2+(aq) + 5 Br2(l) + 8 H2O(l)
Answer: The cell potential of the cell is 0.31 V
Explanation:
We know that:
[tex]E^o_{(Br_2/2Br^-)}=1.07V\\E^o_{(MnO_4^-/Mn^{2+})}=1.51V[/tex]
The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction. Here, [tex]MnO_4^-[/tex] will undergo reduction reaction will get reduced. And, bromine will get oxidized.
Oxidation half reaction: [tex]2Br^-(aq.)+2e^-\rightarrow Br_2(l);E^o_{(Br_2^/2Br^-)}=1.07V[/tex] ( × 5)
Reduction half reaction: [tex]MnO_4^-(aq.)+8H^+(aq.)+5e^-\rightarrow Mn^{2+}(aq.)+4H_2O(l);E^o_{(MnO_4^-/Mn^{2+})}=1.51V[/tex] ( × 2)
The net reaction follows:
[tex]2MnO_4^-(aq.)+10Br^-(aq.)+16H^+(aq.)\rightarrow 2Mn^{2+}(aq.)+5Br_2(l)+8H_2O(l)[/tex]
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
Putting values in above equation, we get:
[tex]E^o_{cell}=1.51-(1.07)=0.44V[/tex]
To calculate the EMF of the cell, we use the Nernst equation, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]^2}{[MnO_4^{-}]^2\times [Br^-]^{10}\times [H^+]^{16}}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = ?V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = 0.44 V
n = number of electrons exchanged = 10
[tex][H^{+}]=1M[/tex]
[tex][Mn^{2+}]=0.15M[/tex]
[tex][MnO_4^{-}]=0.01M[/tex]
[tex][Br^{-}]=0.01M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=0.44-\frac{0.059}{10}\times \log(\frac{(0.15)^2}{(0.01)^2\times (0.01)^{10}\times (1)^{16}})\\\\E_{cell}=0.31V[/tex]
Hence, the cell potential of the cell is 0.31 V
In a laboratory experiment, a fermenting aqueous solution of glucose and yeast produces carbon dioxide gas and ethanol. The solution was heated by burning natural gas in a Bunsen burner to distill the ethanol that formed in the flask. During the distillation, the ethanol evaporated and then condensed in the receiving flask. The flame of the burner was kept too close to the bottom of the flask and some of the glucose decomposed into a black carbon deposit on the inside of the flask. During this experiment the following changes occurred. Which of these changes involved a physical change and not a chemical change? Check all that apply. Check all that apply.
1-condensation of ethanol
2-evaporation of ethanol
3- formation of carbon dioxide gas from glucose burning of natural gas
4-formation of ethanol from glucose by yeast
5-formation of a carbon deposit inside the flask
Answer:
1-condensation of ethanol
2-evaporation of ethanol
Explanation:
According to the question ,
As the compound condenses , it is a basic process of the change in the state , and hence , it is only a physical change . Then , the ethanol is changing its state, hence , again a physical change is observed . The carbon dioxide gas is formed from the glucose would be a chemical change since a new product is formed i.e carbon dioxide , from the reaction between the glucose and yeast . The formation of a carbon deposit was due to the heat which altered the chemical make - up of the substance and is therefore a chemical change .An object with a mass of 100 kg is dropped from a height of 20 m. If the velocity of the object before hitting the ground is 15 m/s, is there a loss of energy in the form of heat? If so, how much? Assume, g= 9.8 m/s^2
Answer:
Yes, there is a loss of 8350 J of energy in the form of heat
Explanation:
The principle of energy conservation is described mathematically
as the energy conservation equation as follows:
ΔK+ΔU=Q+W
where:
ΔK: changes in kinetics energy [J]ΔU: changes in potential energy [J]Q: gain or loss of heat[J]W: work done by the system or in the system[J]Analyzing each term of the equation:
[tex]ΔK=\frac{1}{2}mv_{f} ^{2} - \x]frac{1}{2}mv_{i} ^{2}=\frac{1}{2}*100*15^{2} - \frac{1}{2}*100*0^{2}=11250 J[/tex]
ΔU=[tex]mgh_{f}-mgh_{i}=100*9,8*0-100*9,8*20=-19600[J][/tex]
Q=?
W=0 [J]
Replacing in the main equation:
11250-19600=Q+0
Q= -8350 [J]
So, the answer is YES, there is a loss of 8350 J of energy in the form of heat.
Reynolds number E. What is the mean velocity u. (ft/s) and the Reynolds number Re = pu., D/ for 35 gpm (gallons per minute) of water flowing in a 1.05- in. ID. pipe if its density is p = 62.3 lb/ft and its viscosity is = 1.2 cP? What are the units of the Reynolds number?
Answer:
The mean velocity is 13 ft/s.
The Reynolds number is 88,583 and it is dimensionless.
Explanation:
We have water flowing in a pipe of 1.05 in diameter.
The density is ρ=62.3 lb/ft and the viscosity is 1.2 cP.
The mean velocity can be calculated as
[tex]u=\frac{Q}{A}=\frac{Q}{\pi*D^2/4}=\frac{35gpm }{3.14*(1.05in)^2/4}\\\\ u=\frac{35}{0.865}*\frac{gal}{min}\frac{1}{in^2}*\frac{231in^3}{1gal}*\frac{1}{60s} \\\\ u=156\,in/s=13\,ft/s[/tex]
The Reynolds number now can be calculated for this flow as
[tex]Re=\frac{\rho*u*D}{\mu}[/tex]
being ρ: density, u: mean velocity of the fluid, D: internal diameter of the pipe and μ the dynamic viscosity.
To simplify the calculation, we can first make all the variables have coherent units.
Viscosity
[tex]\mu=1.2cP=\frac{1.2}{100}\frac{g}{cm*s}*\frac{1lb}{453.6g}*\frac{30.48cm}{1ft}= 0.0008\frac{lb}{ft*s}[/tex]
Diameter
[tex]D=1.05in*(\frac{1ft}{12in} )=0.0875ft[/tex]
Then the Reynolds number is
[tex]Re=\frac{\rho*u*D}{\mu}\\\\Re=62.3\frac{lb}{ft^3}*13\frac{ft}{s} *0.0875ft*\frac{1}{0.0008}*\frac{ft*s}{lb}\\\\Re=88,583[/tex]
The Reynolds number (Re) and mean velocity (u) for water flowing through a pipe can be calculated using the flow rate, pipe dimensions, fluid density, and viscosity. The Reynolds number is a dimensionless quantity that helps predict the flow pattern in the pipe.
Explanation:The question relates to the calculation of the Reynolds number and the mean velocity (u) for a given flow rate of water through a pipe. First, to find the mean velocity u, the flow rate needs to be converted to cubic feet per second (ft³/s) and then divided by the cross-sectional area of the pipe. The Reynolds number Re is a dimensionless number used to predict flow patterns in different fluid flow situations. It is calculated using the formula Re = ρuD/μ, where ρ is the fluid density, u is the mean velocity, D is the pipe diameter, and μ is the dynamic viscosity of the fluid.
To proceed with the calculation, the given flow rate of 35 gallons per minute (gpm) is converted to cubic feet per second, the pipe's internal diameter is converted to feet, the density of water (62.3 lb/ft³) is used, and the viscosity (1.2 cP) is converted to lb/(ft·s). The mean velocity u is then calculated, and subsequently, the Reynolds number Re is determined.
The units of the Reynolds number are indeed unitless, as demonstrated by the cancellation of units in its definition, ensuring it is a dimensionless quantity.
What is the wavelength of radiation that has a frequency
of6.912 x 10-14 s-1?
Answer: The wavelength of radiation is [tex]4.34\times 10^{21}m[/tex]
Explanation:
The relationship between wavelength of light and frequency of light is given by the equation:
[tex]\nu=\frac{c}{\lambda}[/tex]
where,
[tex]\nu[/tex] = frequency of radiation = [tex]6.912\times 10^{-14}s^{-1}[/tex]
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\lambda[/tex] = wavelength of radiation = ?
Putting values in above equation, we get:
[tex]6.912\times 10^{-14}s^{-1}=\frac{3\times 10^8m/s}{\lambda}\\\\\lambda=4.34\times 10^{21}m[/tex]
Hence, the wavelength of radiation is [tex]4.34\times 10^{21}m[/tex]
how long would it take to administer exactly 500 mL of fluid through an IV with a drop factor of 30 gtt/mL if the drip rate is 60 gtt/min?
Answer : The time taken will be 25 min.
Explanation :
First we have to determine the amount of fluid.
As, 1 mL contains 30 gtt
So, 500 mL contains [tex]\frac{500mL}{1mL}\times 30gtt=1500gtt[/tex]
Now we have to determine the time taken.
As, 60 gtt takes time = 1 min
So, 1500 gtt takes time = [tex]\frac{1500gtt}{60gtt}\times 1min=25min[/tex]
Therefore, the time taken will be 25 min.
Final answer:
To administer 500 mL of fluid with a drop factor of 30 gtt/mL at a drip rate of 60 gtt/min, the calculation reveals it will take 250 minutes.
Explanation:
To calculate the time required to administer exactly 500 mL of fluid through an IV with a drop factor of 30 gtt/mL at a drip rate of 60 gtt/min, we must first understand the given terms. The drop factor (gtt/mL) is a measure indicating the number of drops (gtt) that make up 1 mL of fluid. The drip rate (gtt/min) specifies how many drops of fluid are administered per minute.
First, we calculate the total number of drops in 500 mL of fluid, using the drop factor:
Total drops = Volume (mL) × Drop factor (gtt/mL)= 500 mL × 30 gtt/mL= 15000 gttNext, to find out how long it will take to administer these 15000 drops at a rate of 60 drops per minute, we divide the total number of drops by the drip rate:
Time (min) = Total drops ÷ Drip rate (gtt/min)= 15000 gtt ÷ 60 gtt/min= 250 minutesTherefore, it will take 250 minutes to administer exactly 500 mL of fluid through the IV at the specified conditions.
Determine the hybridization around the central atom for each of the following molecules.
a) HCN
b) C(CH3)4
c) H3O+
d) - CH3
Answer:
a) HCN - hybridization sp
b) C(CH₃)₄ - hybridization sp³
c) H₃O⁺ - hybridization sp³
d) - CH₃ - hybridization sp³
Explanation:
Hybridization occurs to allow an atom to make more covalent bonds than the original electronic distribution would allow or to allocate ligands in an energetically stable geometry.
Carbon can have thre hybridization states: sp³ , sp² and sp.
Oxygen usualluy has an sp³ hybridization.
In order to determine the hybridization, we need to consider the number of atoms attached to the central atom and the number of lone pairs.
The figure attached shows the species and the hybridization of their central atoms.
A medium-sized carrot weighs 64 g and contains 7.0 g of carbohydrate.
What percent, by mass, of the carrot is carbohydrate? Express your answer using two significant figures.
Answer:
10,93% of a carrot is carbohydrate.
Explanation:
In order to solve this you just have to create a rule of three where 64g is the 100% and you want to calculate how much is 7g in percentage:
[tex]\frac{64}{100}= \frac{7}{x}\\ x=\frac{7*100}{64}\\ x=10,9375[/tex]
So there are 10.9375 % carbohydrates in a piece of carrot.
The percent, by mass, of carbohydrate in a medium-sized carrot is approximately 10.9%.
Explanation:To find the percent, by mass, of carbohydrate in a medium-sized carrot, we can use the following formula:
Percent mass of carbohydrate = (mass of carbohydrate / total mass) × 100
In this case, the mass of carbohydrate is given as 7.0 g and the total mass of the carrot is 64 g. Plugging these values into the formula:
Percent mass of carbohydrate = (7.0 g / 64 g) × 100 = 10.9%
Therefore, approximately 10.9% of the carrot's mass is carbohydrate.
A 36.5 lb child has a Streptococcus infection. Amoxicillin is prescribed at a dosage of 25 mg per kg of body.How many hours should pass between each administration?number of hours:Amoxicillin should be stored between 0 °C and 20 °C. Should the amoxicillin be stored in the freezer or the refrigerator?
Answer:
36.5 lbs weight of child able to take 413.91 mg/ day dose
Amoxicillin should be stored in temperature 0°C to 20°C, therefore it must be stores in refrigerator as it provided temperature only between 0°C to 5°C.
Explanation:
Given data:
Dosage of Amoxicillin as prescribed is 25 mg/kg-day
Weight of the Child weight = 36.5 lbs
As We know 1 lbs = 0.4536 kg
therefore, the weight of Child is 36.5\times 0.4536 kg = 16.5564 kg
From the information about dosage,
1 kg of body takes = 25 mg/day
so, for 16.556 kg body [tex]16.5564\ kg \ body\ takes = 25\times 16.5564 = 413.91 mg/day[/tex]
Therefore 36.5 lbs weight of child able to take 413.91 mg/ day dose
Amoxicillin should be stored in temperature 0°C to 20°C, therefore it must be stored in refrigerator as it provided temperature only between 0°C to 5°C.
Final answer:
To calculate the amoxicillin dosage for a child, convert the child's weight to kilograms, multiply by the prescribed mg/kg dosage, then divide by the tablet strength. Amoxicillin should be stored in a refrigerator, not a freezer.
Explanation:
The question relates to the prescription and administration of amoxicillin dosage based on a patient's weight. To determine the number of tablets to administer, you will need to convert the child's weight from pounds to kilograms (1 pound = 0.453592 kg), multiply the child's weight in kilograms by the prescribed dosage per kilogram and then divide the total dosage by the amount of medicine per tablet.
Using the information provided, if a doctor prescribes amoxicillin at 30mg/kg to a child weighing 73.5 lbs, first convert the weight: 73.5 lbs × 0.453592 = 33.3 kg approximately. Next, calculate the total dosage: 33.3 kg × 30 mg/kg = 999 mg. Since amoxicillin is available in 500 mg tablets, divide the total dosage by the tablet strength: 999 mg / 500 mg/tablet = about 2 tablets (always round to the nearest whole number when it involves whole tablets).
Amoxicillin should be stored between 0 °C and 20 °C, which is typically within the temperature range of a refrigerator, not a freezer. Therefore, amoxicillin should be stored in the refrigerator to maintain its efficacy.
What is the chemistry principle for silly putty and candle
waxas it pertains to hydrophilic/hydrophobic, polar/nonpolar
andsoluble/non-soluble and binding/non-binding?
Explanation:
Silly Putty -
It is an inorganic polymer , in which the molecules are linked via covalent bonding . And the bonds can easily be broken , by the application of even a small amount of pressure .
The composition of silly putty is , the Elmer's glue solution and the sodium borate in the ration , 4 : 1 .
silly putty is polar , hydrophilic and binding in nature .
Candle Wax -
The Wax of the candle is an ester of the ethylene glycol and fatty acids , and the chemical properties of was is , it is hydrophobic in nature and is insoluble in water as well .
The principles affecting the properties of silly putty and candle wax involve their hydrophobic, nonpolar nature, making them insoluble in water and more soluble in nonpolar solvents like heptane. Solubility depends on the balance of intermolecular forces and molecule polarity, with polar and ionic substances being hydrophilic. Heat transfer during solvation is dictated by the relative strength of solute-solvent interactions compared to original solute-solute and solvent-solvent interactions.
The chemistry principle for substances such as silly putty and candle wax as it relates to hydrophilic/hydrophobic, polar/nonpolar, and soluble/non-soluble properties is rooted in their intermolecular forces and molecular structure. Silly putty, which is a silicone polymer, is primarily hydrophobic and non-polar, making it insoluble in water and more soluble in nonpolar solvents. On the other hand, candle wax, typically composed of long-chain hydrocarbons, is also nonpolar and hydrophobic, displaying similar solubility characteristics. Things that are polar and can dissolve in water are referred to as being hydrophilic, which allows them to form hydrogen bonds with water. This property is not present in nonpolar substances like silly putty and candle wax, which resist mixing with water.
In terms of solubility predictions for various substances:
Vegetable oil is nonpolar and more soluble in a nonpolar solvent like heptane.
Isopropyl alcohol is polar due to the presence of an -OH group and more soluble in water.
Potassium bromide is ionic and highly soluble in water which is a polar solvent.
When solutions form, heat may be released or absorbed depending on the intermolecular forces involved. In solvation where the solute-solvent interactions are stronger than the solute-solute and solvent-solvent interactions, heat is released (exothermic). Conversely, if breaking the original interactions requires more energy than is released during new interaction formation, the process is endothermic and absorbs heat.
Which one of the following is not a valid expression for the rate of the reaction below? 4NH3 + 7O2 → 4NO2 + 6H2O Which one of the following is not a valid expression for the rate of the reaction below? 4NH3 + 7O2 4NO2 + 6H2O 16 Δ[H2O]Δt 14 Δ[NO2]Δt - 14 Δ[NH3]Δt - 17 Δ[O2]Δt All of the above are valid expressions of the reaction rate.
Answer : All of the above are valid expressions of the reaction rate.
Explanation :
The given rate of reaction is,
[tex]4NH_3+7O_2\rightarrow 4NO_2+6H_2O[/tex]
The expression for rate of reaction for the reactant :
[tex]\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}[/tex]
[tex]\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}[/tex]
The expression for rate of reaction for the product :
[tex]\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}[/tex]
[tex]\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}[/tex]
From this we conclude that, all the options are correct.
Based on the equation of the reaction, all of the given expressions are valid expressions of the reaction rate.
What is the rate of a chemical reaction?The rate of a chemical reaction is the rate at which reactants are used up or the rate at which products are formed.
The rate of a reaction is obtained from the balanced equation of the reaction.
The equation of the reaction is given below:
4NH3 + 7O2 → 4NO2 + 6H2ORate of disappearance of NH3 = - 1/4 × Δ[NH3]/Δt
Rate of disappearance of O2 = - 1/7 × Δ[O2]/Δt
Rate of formation of NO2 = 1/4 × Δ[NO2]/Δt
Rate of formation of H2O = 1/6 × Δ[H2O]/Δt
Therefore, all of the above are valid expressions of the reaction rate.
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At 1.00 atmosphere pressure, a certain mass of a gas has a temperature of 100oC. What will be the temperature at 1.13 atmosphere pressure if the volume remains constant?
Answer: Final temperature of the gas will be 330 K.
Explanation:
Gay-Lussac's Law: This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.
[tex]P\propto T[/tex] (At constant volume and number of moles)
[tex]{P_1\times T_1}={P_2\times T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 1.00 atm
[tex]P_2[/tex] = final pressure of gas = 1.13 atm
[tex]T_1[/tex] = initial temperature of gas = [tex]100^0C=(100+273)K=373K[/tex] K
[tex]T_2[/tex] = final temperature of gas = ?
[tex]{1.00\times 373}={1.13\times T_2}[/tex]
[tex]T_2=330K[/tex]
Therefore, the final temperature of the gas will be 330 K.
The questions are: Find the equations for the reaction of
HCLwith MgO and HCL and Mg. How the reactions different?
Answer:
[tex]Mg(s) + HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)[/tex]
[tex]MgO(s) + HCl(aq) \rightarrow MgCl_2(aq) + H_2O(l)[/tex]
Explanation:
When Mg reacts with HCl, magnesium chloride and hydrogen is formed. Mg is an active element and displaces hydrogen from HCl. So, this is a type of single displacement reaction.
[tex]Mg(s) + HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)[/tex]
When magnesium oxide (MgO) reacts with HCl, magnesium chloride and water is formed. This reaction is a type of neutralization reaction. MgO is a water insoluble base and HCl is acid. So. in this reaction, acid reacts with base to form salt [tex]MgCl_2[/tex]
[tex]MgO(s) + HCl(aq) \rightarrow MgCl_2(aq) + H_2O(l)[/tex]
The equation for the reaction between HCl and MgO is 2HCl(aq) + MgO(s) → MgCl2(aq) + H2O(l). The equation for the reaction between HCl and Mg is 2HCl(aq) + Mg(s) → MgCl2(aq) + H2(g). The reactions differ in terms of the reactants involved.
Explanation:The equation for the reaction between HCl and MgO is:
2HCl(aq) + MgO(s) → MgCl2(aq) + H2O(l)
The equation for the reaction between HCl and Mg is:
2HCl(aq) + Mg(s) → MgCl2(aq) + H2(g)
The reactions differ in terms of the reactants involved. In the reaction with MgO, the reactant is an oxide compound (MgO) while in the reaction with Mg, the reactant is a pure metal (Mg).
Solve for x: LaTeX: \frac{1}{x}=\:66.54. Report your answer to 4 significant figures.
Answer:
x = 0.01503 = 1.503 × 10⁻² , four significant figures
Explanation:
Significant figures refer to the digits of a number that have meaning and contribute to the precision of the given number.
Since the given equation is:
[tex]\frac{1}{x}=\:66.54[/tex]
Cross-multiplying to solve for the value of x:
[tex]x \times 66.54 = 1[/tex]
⇒ [tex]x = \frac{1}{66.54}[/tex]
⇒ x = 0.01503 = 1.503 × 10⁻² , has four significant figures.
Calculate the molarity of a solution made by
dissolving0.145mol Na2SO4 in enough water to form exactly 750mL
ofsolution.
Answer:
0.1933 M
Explanation:
Data provided in the question:
Number of Moles of Na₂SO₄ initially taken = 0.145 mol
Volume of solution, V = 750 mL = 0.750 L
Now, the molarity is given as:
Molarity = [tex]\frac{\textup{moles of solute}}{\textup{volume of solution in liter}}[/tex]
or
Molarity = [tex]\frac{0.145}{0.750}[/tex]
or
Molarity = 0.1933 M
The molarity of the solution is 0.1933 M.
Explanation:To calculate the molarity of a solution, you need to know the moles of solute and the volume of solution. In this case, we have 0.145 mol of Na2SO4 and a volume of 750 mL. First, convert the volume to liters by dividing by 1000: 750 mL = 0.75 L. Now, calculate the molarity using the formula: Molarity (M) = moles of solute / volume of solution in liters.
So, Molarity = 0.145 mol / 0.75 L = 0.1933 M
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In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.379 L flask at 1191 K. At equilibrium, the flask contains 0.145 mol of CO gas, 0.218 mol of H2 gas, and 0.25 mol of methane. What is the water concentration at equilibrium (Kc = 0.30 for this process at 1191 K)?
Enter to 4 decimal places.
HINT: Look at sample problem 17.7 in the 8th ed Silberberg book. Write a balanced chemical equation. Write the Kc expression. Calculate the equilibrium concentrations of all the species given (moles/liter). Put values into Kc expression, solve for the unknown.
Answer: The concentration of water at equilibrium is 0.3677 mol/L
Explanation:
Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric coefficients. It is represented by [tex]K_{c}[/tex]
For a general chemical reaction:
[tex]aA+bB\rightarrow cC+dD[/tex]
The [tex]K_{c}[/tex] is written as:
[tex]K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]
The chemical equation for the conversion of methane to carbon monoxide and hydrogen gas follows:
[tex]CH_4+H_2O\rightleftharpoons 3H_2+CO[/tex]
The [tex]K_{c}[/tex] is represented as:
[tex]K_{c}=\frac{[H_2]^3[CO]}{[CH_4][H_2O]}[/tex] ....(1)
To calculate the concentration, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
We are given:
[tex][CO]=\frac{0.145mol}{0.379L}=0.383mol/L[/tex]
[tex][H_2]=\frac{0.218mol}{0.379L}=0.575mol/L[/tex]
[tex][CH_4]=\frac{0.25mol}{0.379L}=0.660mol/L[/tex]
[tex]K_c=0.30[/tex]
Putting values in equation 1, we get:
[tex]0.30=\frac{(0.575)^3\times 0.383}{0.660\times [H_2O]}[/tex]
[tex][H_2O]=\frac{(0.575)^3\times 0.383}{0.660\times 0.30}=0.3677[/tex]
Hence, the concentration of water at equilibrium is 0.3677 mol/L
A blood sample with a known glucose concentration of 102.0 mg/dL is used to test a new at home glucose monitor. The device is used to measure the glucose concentration in the blood sample five times. The measured glucose concentrations are 98.4 , 104.3 , 97.4 , 106.7 , and 93.0 mg/dL. Calculate the absolute error and relative error for each measurement made by the glucose monitor.
Explanation:
Actual value = 102.0 mg/dL
Calculation of absolute error and relative error for the measured value, 98.4 mg/dL
Absolute error = |Actual value - measured value|
= | 102.0 - 98.4 | = 3.6
Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]
= [tex]\frac{3.6}{102.0} = 0.03529[/tex]
Calculation of absolute error and relative error for the measured value, 104.3 mg/dL
Absolute error = |Actual value - measured value|
= | 102.0 - 104.3 | = 2.3
Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]
= [tex]\frac{2.3}{102.0} = 0.02255[/tex]
Calculation of absolute error and relative error for the measured value, 97.4 mg/dL
Absolute error = |Actual value - measured value|
= | 102.0 - 97.4 | = 4.6
Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]
= [tex]\frac{4.6}{102.0} = 0.04509[/tex]
Calculation of absolute error and relative error for the measured value, 106.7 mg/dL
Absolute error = |Actual value - measured value|
= | 102.0 - 106.7 | = 4.7
Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]
= [tex]\frac{4.7}{102.0} = 0.04607[/tex]
Calculation of absolute error and relative error for the measured value, 93.0 mg/dL
Absolute error = |Actual value - measured value|
= | 102.0 - 93 | = 9
Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]
= [tex]\frac{9}{102.0} = 0.08823[/tex]
Carlos Santana is a popular Latin-rock and jazz musician. O True ○ False
Answer:
True
Explanation:
Carlos Santana
Carlos Santana is a famous american - mexico musician and even a guitarist with some unique music categories like , jazz , Latin , salsa . He started off in late late 1960's with the brand name Santana , which consists of a fusion of rock and roll and even Latin American jazz . And then after his career grows and he became famous .
Hence , the statement is correct .