Answer:
First you need to know that the warm air is less dense than the colder air and thus it tends to rise; so snow-caves are constructed in some way that the tunnel entrance is below the main space, and so the warm air can be retained into the cave. In a typical snow-cave, you can reach temperatures over 0°C even when the temperature outside are under zero degrees celsius.
Ranjit titrates a sample 10.00 mL of Ba(OH)2 solution to the endpoint using 12.58 mL of 0.1023 M H2SO4.
Based on this data, calculate the concentration of the barium hydroxide solution.
[Ba(OH)2] = ___ M
Answer: The concentration of [tex]Ba(OH)_2[/tex] comes out to be 0.129 M.
Explanation:
To calculate the concentration of base, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ba(OH)_2[/tex]
We are given:
[tex]n_1=2\\M_1=0.1023M\\V_1=12.58mL\\n_2=2\\M_2=?M\\V_2=10.00mL[/tex]
Putting values in above equation, we get:
[tex]2\times 0.1023\times 12.58=2\times M_2\times 10.00\\\\M_2=0.129M[/tex]
Hence, the concentration of [tex]Ba(OH)_2[/tex] comes out to be 0.129 M.
Assume a pill has a dosage of 350 mg of medication. How much medication is this in grams?
Answer:
0.350 grams
Explanation:
The given mass of the pill that has to be taken as a dosage for the medication = 350 mg
The medication has to be determined in grams which means that milligram has to be converted to grams.
The conversion of mg to g is shown below:
[tex]1\ milligram=10^{-3}\ gram[/tex]
So,
[tex]350\ milligram=350\times 10^{-3}\ gram[/tex]
[tex]350\ milligram=0.350\ gram[/tex]
The medication required in grams = 0.350 grams
Final answer:
To convert 350 mg to grams, divide by 1000, resulting in 0.35 grams of medication.
Explanation:
To convert the dosage of medication from milligrams to grams, you need to know the conversion factor between these two units.
There are 1000 milligrams in one gram.
Therefore, you can find the amount in grams by dividing the milligram dosage by 1000.
For the pill with a dosage of 350 mg of medication, the conversion to grams would be:
Total given grams/1000
= 350 mg ÷ 1000
= 0.35 grams
If you mix 10 mL of a 0.1 M HCl solution with 8 mL of
a0.2 M NaOH solution, what will be the resulting pH?
Answer:
The pH of the resulting solution is 12.52.
Explanation:
[tex]Molarity=\frac{n}{V}[/tex]
n = number of moles
V = volume of the solution in Liters
1)1 mol of HCl gives 1 mol of hydrogen ion.
[tex][HCl]=[H^+]=0.1 m[/tex]
Concentration of the hydrogen ion = 0.1 M
Volume of the solution = 10 mL = 0.010 L
[tex]0.1 M=\frac{n}{0.010L}[/tex]
Moles of hydrogen ions = = 0.001 mol
2) 1 mol of NaOH gives 1 mol of hydroxide ion.
[tex][NaOH]=[OH^-]=0.2 M m[/tex]
Concentration of the Hydroxide ions = 0.2 M
Volume of the solution ,V'= 8 mL = 0.008 L
[tex]0.2=\frac{n'}{V'}[/tex]
Moles of hydroxide ions ,n ' = 0.0016
1 mol of HCl neutralizes 1 mol of NaOH ,then 0.001 mol of HCl will neutralize 0.001 mol NaOH.
So left over moles of hydroxide ions in the solution will effect the pH of the solution:
Left over moles of hydroxide ions in the solution = 0.0016 mol - 0.0010 mol = 0.0006 mol
Left over concentration of hydroxide ions:
[tex][OH^-]'=\frac{0.0006 mol}{0.010 L+0.008 L}=0.0333 mol/L[/tex]
[tex]pOH=-\log[OH^-]=-\log[0.03333 M]=1.48[/tex]
pH +pOH = 14
pH = 14 - 1.48 = 12.52
The pH of the resulting solution is 12.52.
If the half-life of 37Rb is 4.7x101 years, how long would it take for 0.5 grams of a 2 gram sample to radioactively decay?
Answer: The time required will be 19.18 years
Explanation:
All the radioactive reactions follows first order kinetics.
The equation used to calculate half life for first order kinetics:
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
We are given:
[tex]t_{1/2}=4.7\times 10^1yrs[/tex]
Putting values in above equation, we get:
[tex]k=\frac{0.693}{4.7\times 10^1yr}=0.015yr^{-1}[/tex]
Rate law expression for first order kinetics is given by the equation:
[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]
where,
k = rate constant = [tex]0.015yr^{-1}[/tex]
t = time taken for decay process = ?
[tex][A_o][/tex] = initial amount of the reactant = 2 g
[A] = amount left after decay process = (2 - 0.5) = 1.5 g
Putting values in above equation, we get:
[tex]0.015yr^{-1}=\frac{2.303}{t}\log\frac{2}{1.5}\\\\t=19.18yrs[/tex]
Hence, the time required will be 19.18 years
Two scientists work together on an experiment, but they have different hypotheses. When the scientists look at the experimental results, they interpret the data in different ways and come to different conclusions. Which of the following does this situation best illustrate?
Answer:
When experiments are carried out or research is done in a certain field of knowledge the scientist at first hypothesise certain knowledge or make theoretical hypotheses about that field of knowledge.
And then conduct the experiment or research to derive certain conclusions and get answers which they can apply on the hypothesis made or on the previous knowledge they have and thereby confirm or negate the hypothesis.
When two scientists are working on similar experiment and they tend to differ in the conclusions drawn by the result, as they get from experiment then it is called as confirmation bias among the scientists.
Silver Nitrate.
Hi experts, can someone give me some real word applications beyond the chemistry laboratory about silver nitrate?
what is it used for in real life?
Explanation:
Except for the use in the chemistry laboratory , were it is used to synthesize many useful products, silver nitrate is also has biolofical and medical relevance.
Silver nitrate is commonly used for silver staining, for demonstrating the reticular fibers, the proteins and the nucleic acids. It is also used as stain in the scanning electron microscopy.
Silver Nitrate is also used for the bone ulcers as well as the burns and the acute wounds.
Verona dissolves 20. grams of NaCl with water to make a 100 ml solution. What is the molarity of the solution? There are 1,000 mL in 1 L O a. 3.4 M O b.0.34 M O G.58 M O d. 20. M
Answer:
b.0.34 M
Explanation:
Given that:
Mass of NaCl = 20 grams
Molar mass of NaCl = 58.44 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{20\ g}{58.44\ g/mol}[/tex]
[tex]Moles= 0.3422\ mol[/tex]
Given that volume = 100 mL
Also,
[tex]1\ mL=10^{-3}\ L[/tex]
So, Volume = 100 / 1000 L = 0.1 L
Considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
[tex]Molarity=\frac{0.3422}{0.1}[/tex]
Molarity = 0.34 M
If heat flows into a system and the system does work on the surroundings, what will be the signs on q and w? Select the correct answer below O positive q, positive w O positive q, negative w O negative q, positive w O negative q, negative w
Answer:
q = Positive
w = Negative
Explanation:
As per first law of thermodymanics,
ΔE = q + w
Where,
ΔE = Change in internal energy
q = Heat absorbed or heat released by the system
w = Work done
Sign conventions are used for heat transfer and work done during a thermodynamics process.
Sign convention for Heat transfer
q is positive when heat is added to the system or heat absorbed by the system this is because energy of the system is increased.q is negative when heat is withdrawn from the system or heat released by the system.Sign convention for Work done
w is positive if work is done on the system or work is done by the surroundingsw is negative if work is done on the system or work is done on the surrounding.In the given question, work is done on the surroundings so, w is negative.
Heat flows into a system or in other word heat is added to the system,
So q is positive.
What is the ground state electron configuration of a
calciumatom?
Answer:
[tex]1s^22s^22p^63s^23p^64s^2[/tex]
Explanation:
Calcium is the chemical element with symbol Ca and the atomic number equal to 20. As alkaline earth metal, the element, calcium is reactive metal. IT lies in the second group and forth period of the periodic table.
The number of the valence electrons of the calcium element is 2 and thus primarily denotes these electrons and forms ionic bond.
The ground state- electron configuration for calcium is: [tex]1s^22s^22p^63s^23p^64s^2[/tex]
Air containing 0.06% carbon dioxide is pumped into a room whose volume is 12,000 ft3. The air is pumped in at a rate of 3,000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.3% carbon dioxide, determine the subsequent amount A(t), in ft3, in the room at time t.
Answer:
[tex]A(t)=1.8+34.2*e^{-t}[/tex]
Explanation:
The concentration of CO2 in the room will be the amount of CO2 in the room at time t, divided by the volume of the room.
Let A(t) be the amount of CO2 in the room, in ft3 CO2.
The air entering the room is 3000 ft3/min with 0.06% concentrarion of CO2. That can be expressed as (3000*0.06/100)=1.8 ft3 CO2/min.
The mixture leaves at 3000 ft3/min but with concentration A(t)/V. We can express the amount of CO2 leaving the room at any time is A(t).
We can write this as a differential equation
[tex]dA/dt=v_i-v_o=1.8-A[/tex]
We can rearrange and integrate
[tex]dA/dt=v_i-v_o=1.8-A\\\\dA/(A-1.8)=-dt\\\\\int(dA/(A-1.8) = -\int dt\\\\ln(A-1.8)=-t+C\\\\A-1.8=e^{-t}* e^{C}=C*e^{-t}\\\\A=1.8+C*e^{-t}[/tex]
We also know that A(0) = 12000*(0.3/100)=36 ft3 CO2.
[tex]A(0)=1.8+C*e^{-0}\\36=1.8+C*1\\C=34.2[/tex]
Then we have the amount A(t) as
[tex]A(t)=1.8+34.2*e^{-t}[/tex]
The subsequent amount A(t) of carbon dioxide in the room at time t is determined by solving the differential equation that models the problem as a tank mixing problem, accounting for the rate of air being pumped in and out of the room.
Explanation:To determine the subsequent amount A(t), in ft³, of carbon dioxide in the room at any given time t, we need to set up a differential equation that accounts for the rate of air being pumped in and out of the room. This situation is analogous to a classic problem in differential equations and mathematical modeling called the tank mixing problem.
The rate at which carbon dioxide enters the room is constant at 0.06% of the 3,000 ft³/min being pumped in, which equals 1.8 ft³/min of CO₂. The rate at which it leaves the room is proportional to the concentration of carbon dioxide in the room at time t. The change in the amount of carbon dioxide at any moment is then given by the rate in minus the rate out.
Let C0 be the initial concentration of CO₂ (0.3%), CV be the volume of the room (12,000 ft³), and r be the rate of air exchange (3,000 ft³/min). The differential equation modeling this situation is:
dA(t)/dt = r * (0.06% * V) - r * (A(t) / V)with initial condition A(0) = C0 * V. This equation can be solved using separation of variables and integrating both sides,
A(t) = V * (C0 - 0.0006) * e^(-rt/V) + 0.0006VUpon inserting the values for V, C0, and r, we can find the expression for A(t).
A specific brand of gourmet chocolate candy contains 7.00 g of dietary fat in each 22.7-gpiece. How many kilograms of dietary fat are in a box containing 1.00 lb of candy?
Express your answer numerically in kilograms.
Answer: 0.14 kg
Explanation:
Gourmet chocolate candy contains 7.00 g of dietary fat in each 22.7-g piece
That is 1 piece of candy weighs 22.7 g and contains 7.00 g of dietary fat
Converting the mass in pounds to kg
1 lb = 0.45 kg = 450 grams (1kg=1000g)
Number of pieces = [tex]\frac{450}{22.7g}=20pieces[/tex]
1 piece contains = 7 g of dietary fat
Thus 30 pieces would contain =[tex]\frac{7}{1}\times 20=140g[/tex] of dietary fat
1 g = 0.001 kg
Thus 140 grams =[tex]\frac{0.001}{1}\times 140=0.14kg[/tex]
Thus 0.14 kg of dietary fat are in a box containing 1.00 lb of candy.
What is the pH if 1mL of 0.1M HCl is added to 99mL of pure water?
Now if instead of pure water a buffer is used: HPO4-2/H2PO4- pKa = 7.2 Assume the initial pH of this buffer is 7 (like the pure water example). -First you must you must use the Henderson-Hasselbalch equation to determine the ratio of HPO4-2/H2PO4- , which is 0.063M to .1M. Using the same amount of HCl added (.001M), determine the change in pH that occurs to the buffer when the HCl is added.
(i already answered the first part, I just need the second part. Show and explain your work please!)
Answer:
pH of buffer after addition of 1 mL of 0,1 M HCl = 7,0
Explanation:
It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:
pH = pka + log₁₀
Where A⁻ is conjugate base and HA is conjugate acid
The equilibrium of phosphate buffer is:
H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺ Kₐ₂ = 6,20x10⁻⁸; pka=7,2
Thus, Henderson–Hasselbalch equation for 7,00 phosphate buffer is:
7,0 = 7,2 + log₁₀ [tex]\frac{[HPO4^{2-}] }{[H2PO4^{-}]}[/tex]
Ratio obtained is:
0,63 = [tex]\frac{[HPO4^{2-}] }{[H2PO4^{-}]}[/tex]
As the problem said you can assume [H₂PO₄⁻] = 0,1 M and [HPO4²⁻] = 0,063M
As the amount added of HCl is 0,001 M the concentrations in equilibrium are:
H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺
0,1 M +x 0,063M -x 0,001M -x -because the addition of H⁺ displaces the equilibrium to the left-
Knowing the equation of equilibrium is:
[tex]K_{a} = \frac{[HPO_{4}^{2-}][H^{+}]}{[H_{2} PO_{4}^{-}]}[/tex]
Replacing:
6,20x10⁻⁸ = [tex]\frac{[0,063-x][0,001-x]}{[0,1+x]}[/tex]
You will obtain:
x² -0,064 x + 6,29938x10⁻⁵ = 0
Thus:
x = 0,063 → No physical sense
x = 0,00099990
Thus, [H⁺] in equilibrium is:
0,001 M - 0,00099990 = 1x10⁻⁷
Thus, pH of buffer after addition of 1 mL of 0,1 M HCl =
-log₁₀ [1x10⁻⁷] = 7,0
A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. In this example you can see its effect!
I hope it helps!
An ice cube at 0.00 ˚C with a mass of 8.32 g is placed into 55 g of water, initially at 25 ˚C. If no heat is lost to the surroundings, what is the final temperature of the entire water sample after all the ice is melted (report your answer to three significant figures)?
Answer:
The final temperature of the entire water sample after all the ice is melted, is 12,9°C. We should realize that if there is no loss of heat in our system, the sum of lost or gained heat is 0. It is logical to say that the temperature has decreased because the ice gave the water "heat" and cooled it
Explanation:
This is the begin:
Q1 = Q which is gained from the ice to be melted
Q2 = Q which is lost from the water to melt the ice
Q1 + Q2 = 0
We are informed that the ice is at 0 ° so we have to start calculating how many J, do we need to melt it completely. If the ice had been at a lower temperature, it should be brought to 0 ° with the formula
Q = mass. specific heat. (ΔT)
and then make the change of state by the latent heat of fusion.
The heat of fusion for water at 0 °C is approximately 334 joules per gram.
So Q = Hf . mass
Q1 = 334 J/g . 8.32 g = 2778,88 J
For water we should use this:
Q = mass. specific heat. (ΔT)
Q2 = 55g . 4180 J/kg. K (Tfinal -T initial)
Q2 = 55g . 4180 J/kg. K (Tfinal -T initial)
(notice we have kg, so we have to convert 55 g, to kg, 0,055kg so units can be cancelled)
Q2 = 0,055kg . 4180 J/kg. K (Tfinal (The unknown) -25°)
T° should be in K for the units of Specific heat but it is the same. The difference is the same, in K either in °C
25°C = 298K
Q2 = 0,055kg . 4180 J/kg. K (Tfinal -298K)
Now the end
Q1 + Q2 = 0
334 J/g . 8.32 g + 0,055kg . 4180 J/kg. K (Tfinal -298K)
2778,88 J + 229,9 J/K (Tfinal - 298 K) = 0
2778,88 J + 229,9 J/K x Tfinal - 68510,2 J = 0
229,9 J/K x Tfinal = 68510,2 J - 2778,88 J
229,9 J/K x Tfinal = 65731,4 J
Tfinal = 65731,4 J / 229,9 K/J
Tfinal = 285,9 K
Tfinal = 285,9 K - 273K = 12,9 °C
The final temperature of the water after an 8.32 g ice cube at 0.00 °C is placed into 55 g of water at 25 °C and all the ice melts is 12.959 °C, using the conservation of energy and assuming no heat loss to the surroundings.
When an 8.32 g ice cube at 0.00 °C is placed into 55 g of water at 25 °C, we can find the final temperature after all the ice is melted by applying the principle of conservation of energy. The heat gained by the ice melting must be equal to the heat lost by the water cooling down. We will assume that no heat is lost to the surroundings in this perfectly insulated system.
We need to calculate the heat required to melt the ice cube (Qmelt) using the heat of fusion of water (which is 79.9 cal/g or 334 J/g), and the heat lost by the water as it cools down (Qwater) using the specific heat of water (which is 4.184 J/g°C).
First, calculate the heat necessary to melt the ice:
Qmelt = mass of ice * heat of fusion
Qmelt = 8.32 g * 334 J/g
Qmelt = 2778.08 J
Next, set up the equation based on the conservation of energy, where the heat lost by the water (Qwater) equals the heat gained by the ice (Qmelt):
Qwater = mass of water * specific heat of water * temperature change
Qwater = Qmelt
55 g * 4.184 J/g°C * (25 °C - final temperature) = 2778.08 J
Solving for the final temperature:
230.62 J/°C * (25 °C - final temperature) = 2778.08 J
25 °C - final temperature = 2778.08 J / 230.62 J/°C
25 °C - final temperature = 12.041 °C
Final temperature = 25 °C - 12.041 °C
Final temperature = 12.959 °C
Therefore, the final temperature of the water after all the ice has melted is 12.959 °C (rounded to three significant figures).
A piece of an unknown metal has a volume of 19.8 ml and a mass of 210.0 grams. The density of the metal is g/mL A piece of the same metal with a mass of 86.0 grams would have a volume of mL.
Answer: The density of the metal is 10.60 g/mL and the volume occupied by 86.0 grams is 8.11 mL
Explanation:
To calculate the density of unknown metal, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex] ......(1)
Volume of unknown metal = 19.8 mL
Mass of unknown metal = 210.0 g
Putting values in equation 1, we get:
[tex]\text{Density of unknown metal}=\frac{210.0g}{19.8mL}\\\\\text{Density of unknown metal}=10.60g/mL[/tex]
The density of the metal remains the same.
Now, calculating the volume of unknown metal, using equation 1, we get:
Density of unknown metal = 11.45 /mL
Mass of unknown metal = 86.0 g
Putting values in above equation, we get:
[tex]10.60g/mL=\frac{86.0g}{\text{Volume of unknown metal}}\\\\\text{Volume of unknown metal}=8.11mL[/tex]
Hence, the density of the metal is 10.60 g/mL and the volume occupied by 86.0 grams is 8.11 mL
The Prandtl number, Pr, is a dimensionless group important in heat transfer. It is defined as Pr Cp*mu/k = where Cp is the heat capacity of a fluid, mu is the fluid viscosity, and k is the fluid thermal conductivity. For a given fluid, Cp 0.5 J/(g * deg C), k 0.2 W/(m * deg C), and mu 2200 lbm (ft* h}. Determine the value of the Prandtl number for this fluid. Please keep two significant figures in your final answer
Answer:
Pr = 2273.58
Explanation:
Pr = Cp*μ/κ∴ Cp = 0.5 J/g.°C
∴ κ = 0.2 W/m.°C * ( J/s / W ) = 0.2 J/s.m.°C
∴ μ = 2200 Lbm/ft.h * ( 453.592 g/Lbm ) * ( ft / 0.3048 m ) * ( h/3600 s )
⇒ μ = 909.433 g/m.s
⇒ Pr = ((0.5 J/g.°C )*( 909.433 g/m.s )) / 0.2 J/s.m.°C
⇒ Pr = 2273.58
Which is the largest scale? Subatomic Miniscopic Atomic Macroscopic Microscopic
Answer:
Macroscopic scale
Explanation:
Subatomic scale is the scale at which atomic constituents, such as nucleus which contains protons and neutrons, and electrons, which orbit in the elliptical paths around nucleus exists.
Miniscopic scale is a reference scale and is not a standard scale for measurement. Usually, this refers to minute objects.
Atomic scale is the scale which is at size of the atoms.
Macroscopic scale is length scale on which the objects or the phenomena are enough large to be visible with naked eye, without magnifying the optical instruments. It is the largest scale.
Microscopic scale is scale of the objects that require microscope to see them.
A liquid of mass 10 kg is enclosed in a cylinder of radius 1 m and length 5 m, what is the density of liquid? a) 0.63 kg/m3 b) 0.44 kg/m3 c) 0.54 kg/m3 d) 0.83 kg/m3
Answer:
a) [tex]0.63\frac{kg}{m^{3}}[/tex]
Explanation:
Density is given by the expression [tex]d=\frac{m}{V}[/tex], where m is the mass of the substance and V is the volume occupied by the substance.
As the problem says that the liquid is enclosed in a cylinder, you should find the volume of that cylinder that will be the same volume of the liquid, so:
For a cylinder the volume is given by V=[tex]2\pi r^{2}h[/tex]
Replacing the values given, we have:
[tex]V=2\pi (1m)^{2}(5m)[/tex]
[tex]V=15.708m^{3}[/tex]
Replacing the values of m and V in the equation of density, we have:
[tex]d=\frac{10kg}{15.708m^{3}}[/tex]
[tex]d=0.63\frac{kg}{m^{3}}[/tex]
A sealed can with an internal pressure of 721 mmHg at
25degrees C is thrown into an incinerator operating at 755 degrees
C.What will be the pressure inside the heated can, assuming
thecontainer remains intact during incineration?
Answer:
2486 mmHg
Explanation:
Gay-Lussac's Law states the pressure varies directly with temperature when volume remains constant:
P₁/T₁ = P₂/T₂
Where P₁ and T₁ are initial pressure and temperature and P₂ and T₂ are final pressure and temperatue.
The problem says initial pressure is 721 mmHg, initial temperature is 25°C and final temperature is 755°C. The question is final pressure.
°C must be converted to absolute temperature (K), thus:
25°C + 273,15 = 298,15 K
755°C + 273,15 = 1028,15 K
Thus, pressure P₂ is:
(T₂·P₁) / T₁ = P₂
1028,15K · 721mmHg / 298,15 K = 2486 mmHg
I hope it helps!
As accurately as possible find the residual molar volume of saturated liquid water at 83 bar.
Answer:
v = 2.512 E-5 m³/mol
Explanation:
∴ P = 80 bar → V = 0.001384 m³/Kg......sat. liq water table
∴ P = 85 bar → V = 0.0014013 m³/Kg
⇒ P = 83 bar → V = ?
specific volume ( V ):
⇒ V = 0.001384 + (( 83 - 80 ) / ( 85 - 80 ))*( 0.0014013 - 0.001384 )
⇒ V = 0.00139438 m³/Kg
molar volume ( v ):
∴ Mw water = 18.01528 g/mol
⇒ v = 0.00139438 m³/Kg * ( Kg/1000g ) * ( 18.01528 g/mol )
⇒ v = 2.512 E-5 m³/mol
Drugs.com contains information on all the following categories except: a. Pill Identification b. Product d. Manufacturers C. Approval date b. Drugs A-Z c. Interaction Checker e. Calculators
Answer: Drugs.com is the site which have all we need regarding medicines and uses of medicines. It give the facility to search A-Z drugs, once you have searched the desired medicine by this feature of site, it further gives information about identification of pills, approval date, and interaction between drug-drug and drug- the food you eating, the manufacturing date of medicines. It doesn't give information about the calculators.
Therefore, (e) is the correct option here.
Drugs.com is a resourceful website for information about various drugs. It provides details like Pill Identification, Drugs A-Z, Interaction Checker, and more. However, it does not provide any calculator-type feature.
Explanation:Drugs.com provides comprehensive information related to different types of medications. It covers areas such as Pill Identification, Product, Manufacturers, Drug Approval Date, Drugs A-Z, and Interaction Checker. However, Drugs.com does not offer any feature or functionality related to 'Calculators' pertaining to drug calculations or dosage calculations. Use of the site could enable someone to determine whether they have a substance use disorder, but it does not replace professional medical advice.
Learn more about Drugs.com here:https://brainly.com/question/13264475
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Calculate the freezing point of the solution.After mixing these 2 bottles together, set Kf of water = 1.86 ° C / m.
Bottle 1 contained 0.3 grams of glucose in 1000 grams of water.
The 2nd bottle contains 0.5 mol fructose in 1000 grams of water.
Answer : The freezing point of solution is 273.467 K
Explanation : Given,
Mass of glucose (solute) = 0.3 g
Mass of water (solvent) = 1000 g = 1 kg
Moles of fructose (solute) = 0.5 mol
Mass of water (solvent) = 1000 g = 1 kg
Molar mass of glucose = 180 g/mole
First we have to calculate the moles of glucose.
[tex]\text{Moles of glucose}=\frac{\text{Mass of glucose}}{\text{Molar mass of glucose}}=\frac{0.3g}{180g/mole}=0.00167mole[/tex]
Now we have to calculate the total moles after mixing.
[tex]\text{Total moles}=\text{Moles of glucose}+\text{Moles of fructose}[/tex]
[tex]\text{Total moles}=0.00167+0.5=0.502moles[/tex]
Now we have to calculate the molality.
[tex]\text{Molality}=\frac{\text{Total moles}}{\text{Mass of water(solvent in kg)}}[/tex]
[tex]\text{Molality}=\frac{0.502mole}{(1+1)kg}=0.251mole/kg[/tex]
Now we have to calculate the freezing point of solution.
As we know that the depression in freezing point is a colligative property that means it depends on the amount of solute.
Formula used :
[tex]\Delta T_f=K_f\times m[/tex]
[tex]T^o_f-T_f=K_f\times m[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point
[tex]T_f^o[/tex] = temperature of pure water = [tex]0^oC[/tex]
[tex]T_f[/tex] = temperature of solution = ?
[tex]K_f[/tex] = freezing point constant of water = [tex]1.86^oC/m[/tex]
m = molality = 0.251 mole/kg
Now put all the given values in this formula, we get
[tex]0^oC-T_f=1.86^oC/m\times 0.251mole/kg[/tex]
[tex]T_f=-0.467^oC=273.467K[/tex]
conversion used : [tex]K=273+^oC[/tex]
Therefore, the freezing point of solution is 273.467 K
If the caffeine concentration in a particular brand of soda is 2.13 mg/oz, drinking how many cans of soda would be lethal? Assume that 10.0 g of caffeine is a lethal dose, and there are 12 oz in a can. cans of soda:
Answer:
400 cans of soda would be lethal.
Explanation:
In a can of soda, there is (2.13 mg/oz * 12 oz) 25 mg caffeine.
25 mg * (1g / 1000 mg) = 0.025 g
If in a can of soda there is 0.025 g of caffeine, a lethal dose of caffeine will be ingested after drinking (10.0 g * (1 can / 0.025 g)) 400 cans of soda.
What is the transition interval for phenol red? 24 a. pH 3.1-4.4 b. pH 6.4-8.0 c. pH 6.2-7.6 d. pH 8.0-10.0
Answer:
The correct option is: b. pH 6.4-8.0
Explanation:
Phenol red is a weak acid that is used as a pH indicator and exists in the form of stable red crystals.
The color of the phenol red solution changes from yellow to red when the change in pH is observed. The color of phenol red transitions from yellow to red when the pH is 6.8 - 8.2 or 6.4 - 8.0
Above the pH of 8.2, the phenol red solution turns a bright pink in color.
The units of density are kg/m2. If the density of a liquid is 760.0 kg/m' what is the specific volume? a) 1.316 x 10 m2/kg b) 1.316 x 10 m3/kg c) without the molecular weight of the liquid it is impossible to determine the specific volume d) none of the above are correct
Answer:
The correct answer is: 1.316 . 10⁻³ m³/kg.
Explanation:
The density (ρ) of a substance is the ratio of its mass (m) to its volume (V). At constant temperature and pressure, its value is constant and it is an intrinsic property of materials. The units of density are kg/m³.
[tex]\rho = \frac{m}{V}[/tex]
The specific volume (ν) of a substance is the ratio of its mass to its volume. We can see that it is the reciprocal of density and an intrinsic property of matter as well. Therefore, the units of specific volume are m³/kg.
[tex]\nu = \frac{V}{m}=\frac{1}{\rho }[/tex]
Given we know the density of the liquid, we can use this relationship to find out its specific volume:
[tex]\nu =\frac{1}{\rho }=\frac{1}{760.0kg/m^{3} } =1.316 .10^{-3} m^{3} /kg[/tex]
The volume of a sphere is given by V-(4/3)r where r is the radius. The density of magnesium is 1.74 g/cm What is the mass of a magnesium sphere with a radius of 0.80 cm? • V= x2 m= TEMPERAURE CONVERTIONS show formulas used Convert 38.0 °F 10 °C = 3.33 (38-32)x(5/6)=3333°C Convert 23.5 °C 10 °F - 245 (22.5%6/5)+32=72.5°F SPD.
Answer:
The answer to your question is:
mass = 3.74 g
Explanation:
Data
V = (4/3) πr³
density = 1.74 g/cm³
radius = r = 0.80 cm
Process
V = (4/3) π(0.8)³ Substitution
V = 2.1446 cm³
mass = density x volume
mass = 1.74 x 2.1446 Substitution
mass = 3.74 g
I don't understand if the second section is also a question.
A wall is constructed of a section of stainless steel (k = 16 W/m -°C) 40 mm thick with identical layers of plastic on both sides of the steel The overall heat-transfer coefficient, considering convection on both sides of the plastic, is 120 W/m2 °C If the overall temperature difference across the arrangement is 60°C, calculate the temperature difference across the stainless steel The area of the wall is one meter square
Answer:
The temperature difference across the stainless steel is 18°C
Explanation:
The heat flows through the plastic layer by convection and then through the steel layer by conduction and then through the plastic layer on the other side.
The heat flux q/A can be expressed as:
[tex]q/A = h*\Delta T[/tex]
It can also be expressed as
[tex]q/A=h*(T_A-T_B) = h_1(T_A-T_1)=k/\Delta x*(T_1-T_2)=h_2(T_2-T_B)[/tex]
being Δx/k*(T1-T2) the conductive heat flux through the steel.
If we want to know the temperatur difference across the stainless steel (T1-T2) we can write:
[tex](k/\Delta x)*(T_1-T_2)=h*(T_A-T_B)\\\\(T_1-T_2)=(\Delta x/k)*h*(T_A-T_B)\\\\(T_1-T_2)=(0.04/16)*120*60=18 \, ^{\circ} C[/tex]
People tend to speak more quietly in restaurants than they do when they are having an ordinary conversation Restaurant conversation is about 45 dB. If ordinary conversation is 100 times greater than restaurant conversation, how loud is ordinary conversation?
Answer:
A loud ordinary conversation following the supplied information in the question is about 4500 dB. But, in the official decibel system measure a loud conversation does not overcome 100 dB.
Explanation:
Using the supplied data of the exercise, we say that in a restaurant conversation the value is 45 dB. If we multiply this by 100 we will have a value for a laud ordinary conversation.
45×100 = 4500 dB.
but as I mentioned in the answer, in the official decibel system measure a loud conversation between 2 man reaches a maximal of 100 dB.
What typically occurs in a substance where hydrogen bonding exists when compared to the same substance without H-bonds?
Question options:
A) Decrease in boiling point and decrease in vapor pressure
B) Increase in boiling point and decrease in vapor pressure
C) Increase in boiling point and increase in vapor pressure
E) Decrease in boiling point and increase in vapor pressure
F) There is no difference
Answer:
B) Increase in boiling point and decrease in vapor pressure
Explanation:
Vapor pressure is inversely related to the Boiling point , as
higher the boiling point, lower the vapor pressure. and
Lower the boiling point, higher the vapor pressure.
Hydrogen bonding.
The electrostatic attraction between Hydrogen , bonded to electronegative atom like F, O, N and the more electronegative atom is called as Hydrogen bonding.
For example -
In alcohols, - OH group has Hydrogen that is bonded to more electronegative atom O.
As ,
Extra energy is required to break Hydrogen bonds.
because the substance which exhibits Hydrogen bonding have lower vapor pressure than that of the substance with out Hydrogen bonding.
Hence , the substance with Hydrogen bonding , has higher boiling point,.
Hence , the correct option is Increase in boiling point and decrease in vapor pressure .
Copper sulfate is a blue solid that is used to control algae growth. Solutions of copper sulfate that come in contact with the surface of galvanized (zinc-plated) steel pails undergo the following reaction that forms copper metal on the zinc surface. How many grams of zinc would react with 454 g (1 lb) of copper sulfate (160 g/mol)?
Answer:
185.5156 g
Explanation:
The reaction of copper sulfate with zinc is shown below as:
[tex]CuSO_4+Zn\rightarrow ZnSO_4+Cu[/tex]
Given that :
Amount of copper sulfate = 454 g
Molar mass of copper sulfate = 160 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus, moles are:
[tex]moles= \frac{454\ g}{160\ g/mol}[/tex]
[tex]moles\ of\ copper\ sulfate= 2.8375\ mol[/tex]
From the reaction,
1 mole of copper sulfate reacts with 1 mole of zinc
Thus,
2.8375 moles of copper sulfate reacts with 2.8375 moles of zinc
Moles of Zinc that should react = 2.8375 moles
Mass of zinc= moles×Molar mass
Molar mass of zinc = 65.38 g/mol
Mass of zinc = 2.8375 ×65.38 g = 185.5156 g
Final answer:
185.595 grams of zinc would react with 454 g of copper sulfate, based on the one-to-one mole ratio between copper sulfate and zinc and the molar mass of zinc (65.38 g/mol).
Explanation:
The student is asking how many grams of zinc would react with 454 g of copper sulfate, which has a molar mass of 160 g/mol. According to the provided reaction, CuSO4 (aq) + Zn(s) → Cu(s) + ZnSO4 (aq), there is a one-to-one mole ratio between copper sulfate and zinc. Given the molar mass of copper sulfate (160 g/mol), we first find the number of moles of copper sulfate in 454 g.
Number of moles of CuSO4 = (454 g) / (160 g/mol) = 2.8375 mol
Since the mole ratio between copper sulfate and zinc is 1:1, an equal number of moles of zinc will react with the copper sulfate. We then need to find the molar mass of zinc to convert moles of zinc to grams. Zinc has a molar mass of approximately 65.38 g/mol.
Mass of zinc = Number of moles of Zn × molar mass of Zn = 2.8375 mol × 65.38 g/mol = 185.595 g
Therefore, 185.595 grams of zinc would react with 454 g of copper sulfate.
In the reaction C zH4 + H2 - e) +4 CHo the carbon atoms are a) oxidized b) reduced c) cannot be determined
Answer: Option (b) is the correct answer.
Explanation:
Reduction is defined as the process in which there occurs gain of hydrogen. Whereas oxidation is defined as the process in which there occurs loss of hydrogen.
As the given reaction is as follows.
[tex]C_{2}H_{4} + H_{2} \rightarrow C_{2}H_{6}[/tex]
Since, hydrogen is being added in this chemical reaction. It means that reduction is taking place and carbon atom is reduced.
Thus, we can conclude that in the given reaction carbon atoms are reduced.