Answer: A hydrogen bonding is interaction between lone pair and hydrogen atom. An Ion-Dipole interaction is the interaction between an ion formed and a dipole. Dipole forms because of the electronegativity difference between two atom participating in the bond formation, and an ion is formed when an atom gains or lose electron. This ion-dipole interaction is strongest interaction.
Therefore, The right choice is (B)
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.
If light moves at a speed of about 3.00 x 108 m/s, how long will it take light to travel the distance of a marathon (about 42.2 km)? Express your answer in microseconds using the correct number of significant figures. Do not enter your answer using scientific notation.
Light travels a distance of 3.69 meters within the given time frame of 12.3 nanoseconds.
Speed is the measure of how quickly an object changes its position concerning a specified frame of reference. It's a scalar quantity, indicating both magnitude and direction.
Common units for speed include meters per second (m/s) or kilometers per hour (km/h). This fundamental concept plays a pivotal role in various domains like travel, sports, and transportation.
If light travels at a constant speed of [tex]3.0 * 10^8 m/s[/tex] and continues for 12.3 nanoseconds, the distance covered can be calculated:
Distance = Speed × Time
Distance = [tex]3.0 * 10^8 m/s* 12.3 * 10^{-9} s[/tex]
Distance = 3.69 meters
In this instance, light travels a distance of 3.69 meters within the given time frame of 12.3 nanoseconds.
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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 .
A plastic rectangle has dimensions of 1.0 cm, 2.0 cm and 3.0 cm respectively. It masses 5.5 g when placed on an electronic mass balance. What is its density? Would it float in water? Density of water is 1.0 g/mL.
Answer:
0.917g/cm^3
Explanation:
Density = Mass/ Volume.
Given,
Mass = 5.5g
Volume of the plastic = L*W*H
= 1.0 x 2.0 x 3.0 = 6.0cm^3
Density = Mass/volume
= 5.5g/6.0cm^3
= 0.9166666666
= 0.917g/cm^3
Since the density of water is greater than that of the plastic, it means that the plastic will float in water.
Answer:
Sinking of the block in water will not happen.
Explanation:
An object’s (block in this case) density can be found out by dividing the mass of that object from its volume. In the given question the mass of rectangular block is given as 5.5 gram and the sides of the rectangular block is [tex]1, 2 and 3 cm[/tex] so volume will be [tex]6 cm^3[/tex]
[tex]Density = Mass/ Volume[/tex]
Now to calculate the density [tex]5.5/ 6= 0.91g/ml[/tex]
This block will float on water since the block’s density is less than water.
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.
Tell why you may be able to survive a bitter cold day in snow-cave."
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.
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).
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!
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.
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.
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.
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 .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
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.
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.
Just as the depletion of stratospheric ozone threatens life on Earth today, its accumulation was one of the crucial processes that allowed life to develop in prehistoric times: 3O2(g) → 2O3(g) At a given instant, the reaction rate in terms of [O2] is 2.17 × 10−5 mol/L·s. What is it in terms of [O3]? Enter your answer in scientific notation.
Answer:
Rate in terms of formation of [tex]O_{3}[/tex] is [tex]1.45\times 10^{-5}mol/L.s[/tex]
Explanation:
According to law of mass action for this reaction: Rate = [tex]-\frac{1}{3}\frac{\Delta [O_{2}]}{\Delta t}=\frac{1}{2}\frac{\Delta [O_{3}]}{\Delta t}[/tex][tex]-\frac{\Delta [O_{2}]}{\Delta t}[/tex] represents rate of disappearance of [tex]O_{2}[/tex] and [tex]\frac{\Delta [O_{3}]}{\Delta t}[/tex] represents rate of formation of [tex]O_{3}[/tex]Here, [tex]-\frac{\Delta [O_{2}]}{\Delta t}=2.17\times 10^{-5}mol/L.s[/tex]So, [tex]\frac{\Delta [O_{3}]}{\Delta t}=\frac{2}{3}\times -\frac{\Delta [O_{2}]}{\Delta t}=\frac{2}{3}\times (2.17\times 10^{-5}mol/L.s)=1.45\times 10^{-5}mol/L.s[/tex]Hence rate in terms of formation of [tex]O_{3}[/tex] is [tex]1.45\times 10^{-5}mol/L.s[/tex]The rate of formation is the time taken by the reaction to yield the product by the chemical change in the reactants. The rate of the formation of the ozone is [tex]1.45 \times 10^{-5} \;\rm mol/Ls[/tex].
What is the law of mass action?The law of mass action states that the rate of the reaction is proportional to the product of the reactant masses.
Rate according to the law of mass action:
[tex]\rm -\dfrac{1}{3}\dfrac {\Delta[O_{2}]}{\Delta t} = \rm \dfrac{1}{2}\dfrac{\Delta [O_{3}]}{\Delta t}[/tex]
Here,
Rate of disappearance of oxygen [tex](\rm -\dfrac {\Delta[O_{2}]}{\Delta t} ) = 2.17 \times 10^{-5} \;\rm mol/Ls[/tex] Rate of formation of ozone =[tex]\rm \dfrac{\Delta [O_{3}]}{\Delta t}[/tex]Substituting values in the above equation:
[tex]\begin{aligned}\rm \dfrac{\Delta [O_{3}]}{\Delta t} &= \dfrac{2}{3}\times - \rm \dfrac{\Delta [O_{2}]}{\Delta t}\\\\&= \dfrac{2}{3} \times (2.17 \times 10^{-5})\\\\&= 1.45 \times 10^{-5}\;\rm mol/Ls\end{aligned}[/tex]
Therefore, the rate of formation in the terms of ozone is [tex]1.45 \times 10^{-5} \;\rm mol/Ls.[/tex]
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Air at 293 K and 750 mm Hg pressure has a relative humidity of 80%. What is its percent humidity? The vapour pressure of water at 293 K is 17.5 mm Hg. (a) 80.38 (b) 80 (c) 79.62 (d) 78.51
Answer : The correct option is, (c) 79.62
Explanation :
The formula used for percent humidity is:
[tex]\text{Percent humidity}=\text{Relative humidity}\times \frac{p-p^o_v}{p-p_v}\times 100[/tex] ..........(1)
The formula used for relative humidity is:
[tex]\text{Relative humidity}=\frac{p_v}{p^o_v}[/tex] ...........(2)
where,
[tex]p_v[/tex] = partial pressure of water vapor
[tex]p^o_v[/tex] = vapor pressure of water
p = total pressure
First we have to calculate the partial pressure of water vapor by using equation 2.
Given:
[tex]p=750mmHg[/tex]
[tex]p^o_v=17.5mmHg[/tex]
Relative humidity = 80 % = 0.80
Now put all the given values in equation 2, we get:
[tex]0.80=\frac{p_v}{17.5mmHg}[/tex]
[tex]p_v=14mmHg[/tex]
Now we have to calculate the percent humidity by using equation 1.
[tex]\text{Percent humidity}=0.80\times \frac{750-17.5}{750-14}\times 100[/tex]
[tex]\text{Percent humidity}=79.62\%[/tex]
Therefore, the percent humidity is 79.62 %
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 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 gPart B Compute 1240.64/64. Round the answer appropriately Express your answer numerically using the proper number of significant figures. View Available Hint(s) Submit
Answer:
19
Explanation:
1240.64/64
This is a simple division that can easily be evaluated using any form of calculator.
Here, the main challenge is in expressing the answer in the proper number of significant figure.
The approach here is to express the answer to the which is least precise. This is usually the number with the lower significant values.
Here, 64 has two significant figures and it is of a lower rank. Our answer should be expressed this way:
[tex]\frac{1240.64}{64}[/tex] = 19.385 = 19
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.
You are performing a chemical reaction in a test tube. The test tube gets colder as the reaction takes place. This chemical reaction is... O a.a double replacement reaction. O b.endothermic Oc.a single replacement reaction. O d.a decomposition reaction. e. exothermic
Answer: Option (b) is the correct answer.
Explanation:
A chemical reaction in which heat energy is absorbed by the reactant molecules is known as an endothermic reaction.
Therefore, upon completion of this reaction the container in which reaction is carried out becomes colder.
A chemical reaction in which heat energy is released by the reactant molecules is known as an exothermic reaction.
Therefore, upon completion or during this type of reaction the container in which reaction is carried out becomes hot.
Thus, we can conclude that when the test tube gets colder as the reaction takes place then it means this chemical reaction is endothermic reaction.
Your patient gets a prescription for 62.5 mcg (micrograms, mg) of digoxin in liquid form. The label reads 0.0250 mg/mL. How many milliliters of digoxin should you give?
Answer:
2.5 militers (mL) of digoxin solution
Explanation:
if 1 microgram = 0.001 miligram
then 62.5 micrograms = X miligrams
X = (62.5 × 0.001) / 1 = 0.0625 miligrams
the we calculate the number of militers of digoxin needed by the patient:
if we have 0.0250 miligrams of digoxin in 1 mililiter of solution
then we have 0.0625 miligrams of digoxin in X mililiters of solution
X = (0.0625 × 1) / 0.0250 = 2.5 militers (mL) of digoxin solution
To calculate the amount of digoxin in milliliters, divide the amount of digoxin in milligrams by the concentration of the liquid form in milligrams per milliliter. In this case, the prescription is for 62.5 mcg of digoxin, and the concentration of the liquid form is 0.0250 mg/mL. The calculated volume of digoxin to be given is 2.5 mL.
Explanation:To calculate the amount of digoxin in milliliters that should be given, we need to use the given information and convert units. The prescription is for 62.5 mcg (micrograms) of digoxin, and the concentration of the liquid form is 0.0250 mg/mL. To convert micrograms to milligrams, we divide by 1000. Then, we can use the formula: Volume (in mL) = Amount (in mg) / Concentration (in mg/mL).
Amount of digoxin in mg = 62.5 mcg / 1000 = 0.0625 mg
Volume of digoxin in mL = 0.0625 mg / 0.0250 mg/mL = 2.5 mL
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.
How to dispose chemical waste? If unsure what should you do?
Sketch and label a covalent bond and an ionic bond (you do not need to draw the metallic bond or intermolecular force). Describe the role of electrons in covalent, ionic, and metallic bonds (you do not need to describe the intermolecular force), and describe the strength of these bonds. Also explain (no sketch) how differences in bonds cause diamond and graphite to have very different properties (2.11A).
Answer:
See attached image
Explanation:
An ionic bond is a type of chemical bond in which occurs an electrons transfer, where one of the atoms act as a donor and the other act as the acceptor. Compound with ionic bonds are very stables and strong because of the attraction of the opposite charges. These attractions help ions stay together forming solid nets.
A covalent bond is a type of chemical bond in which the electrons are shared. Unlike the ionic bond, the atoms act at the same time as donors and acceptors sharing their electrons. The covalent bond keeps the atoms together because an electrostatic force, thus this bond is weaker than the ionic bond.
A metallic bond is a force which keeps together metallic atoms. In this type of bond, there is no transfer or sharing of electrons. The atoms are kept together because of an electrostatic force. Since there is no electron exchange involved a metallic bond is weaker than the other two.
In diamond, every carbon atom shared four electrons with other carbon atoms between a covalent bond. These form a regular tetrahedron. On the other side, graphite has a layer structure, in which every layer hold carbon atoms wich shared electrons with other tree atoms. All the layers stay together because of the Van der Waals force. These difference in bonds cause to have different properties, diamond is one of the hardest materials and graphite it is not.
Name the following ternary acids. a. H2 SO4 e. HMnO4 i. HNO2 m. HClO4 b. H2 SO3 f. H2 CrO4 j. H2 CO3 n. HClO3 c. H2 S2 O3 g. H3 BO3 k. H2 C2 O4 o. HClO2 d. H3 PO4 h. HNO3 l. CH3 COOH p. HClO
Answer:
H2 SO4 (Sulfuric acid) - HMnO4 (Permanganic acid) - HNO2 (Nitrous acid) HClO4 (Perchloric acid) - H2 SO3 (Sulphurous acid) - H2CrO4 (Chromic acid) H2CO3 (Carbonic acid) - HClO3 (Chloric acid) - H3BO3 (boric acid) - HClO2 (Chlorous acid) H3PO4 (Phosphoric acid) HNO3 (Nitric acid) - HClO (Hypochlorous acid) - CH3 COOH (acetic acid) - H2S2O3 (Thiosulfuric acid)- H2C2O4 (oxalic acid)
Explanation:
To name the ternary acid, they have to obbey the following formula
Hx - Non metal - Oy
where the oxidation state in H and O are +1 y -2 respectively.
When the oxidation state of the central Non metal is odd, the atomicity of H is 1 and in the O is deduced in such a way that the sum of oxidation states is 0.
When the oxidation state of the central No metal is even, the atomicity of H is 2 and in the O is deduced in such a way that the sum of oxidation states is 0.
IV V VI VII
- - 1 Hypo ……. Ous
- 3 4 3 … ous
4 5 6 5 … ic
7 Per …… Ous
The oxalic acid is an organic compound of two carboxyl groups, so it is also called ethanedioc acid; It is actually known as oxalic by some plants of the genus oxalis. (prefix et (2 carbons))
The acetic acid that comprises a carboxylic group and a methyl group is also an organic compound. It is popularly known as acetic acid but it is actually called methylcarboxylic acid or having two carbons, ethanoic acid.
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]
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.
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]