The weak acid HA is 2% ionized (dissociated) in a 0.20 M solution.
1. What is Ka for this acid?
2. What is the pH of this solution?
Answer:
1. Ka = 8.16x10⁻⁵
2. pH = 2.40
Explanation:
1. The dissociation of a weak acid in water occurs as follows:
HA ⇄ H⁺ + A⁻
Ka = [H⁺] [A⁻] / [HA]
As 2% of the 0.20M solution is dissociated:
[H⁺] = [A⁻] = 0.20M * 2% = 0.004M -As H⁺ and A⁻ comes from the same reaction, their concentrations are the same
[HA] = 0.20M * 98% = 0.196M
Ka = (0.004)² / (0.196M) = 8.16x10⁻⁵
2. pH = -log [H⁺] = -log [0.004M]
pH = 2.40100 POINTS & BRAINLIEST !!!! PLEASE HELP three qualitative and three quantitative observations, and one inference
Answer: 3 qualitative: the word pharmacology is written in blue, the stairs are gray, the tile on the first floor is yellow. 3 quantitative: there are four people wearing yellow, there are Four pink boxes, there are two floors.
Inference: there are many people performing experiments so it’s a lab.
Explanation:
Answer:
Qualitive - The bars are yellow, The floor is green, and the floor has pharmacology on it.
Quantitative - There are 8 support bars, There is 3 gray coats, and there is 27 pills that I can see on that picture.
Inference - There are many people doing experiments because the word pharmacology is on the floor.
Explanation:
This picture tells you that pharmacology is a form of study of medicine and drugs.
The speed of light in a vacuum is 2.998×108 m/s 2.998 × 10 8 m / s . Calculate its speed in miles per hour (miles/h m i l e s / h ).
Answer:
The speed of light in a vacuum is 6.69 * 10⁸ [tex]\frac{miles}{h}[/tex]
Explanation:
Two magnitudes are directly proportional when increasing one quantity increases the other in the same proportion or when decreasing one quantity decreases the other in the same proportion.
The rule of three allows the resolution of problems that are related to the proportionality of three known values and a fourth value that is always an unknown. In other words, it is useful to establish the proportionality between 2 values a and b through the knowledge of a third value c in order to calculate a fourth value x. In the case of direct margins, the rule of three between a, b and c and the unknown x is:
a ⇒ b
c ⇒ x
So: [tex]x=\frac{c*b}{a}[/tex]
In this case, knowing that 1 meter is equal to 0.000621 miles, 1 second is equal to 0.000278 hours, the simple rule of three is applied as follows: if 1 meter is equal to 0.000621 miles, 2.998 * 10⁸ meters are equal to how many miles?
[tex]miles=\frac{2.998*10^{8} meters*0.000621 miles}{1 meter}[/tex]
miles=186,175.8
[tex]2.998*10^{8} \frac{m}{s} =\frac{2.998*10^{8}m}{s}=\frac{186,175.8miles}{s}[/tex]
Replacing the seconds by their equivalent in meters:
[tex]\frac{186,175.8miles}{s} =\frac{186,175.8miles}{0.000278 hours}= 669,697,122.3 \frac{miles}{h}[/tex]
Then:
2.998*10⁸ [tex]\frac{m}{s}[/tex] = 669,697,122.3 [tex]\frac{miles}{h}[/tex] ≅ 6.69 * 10⁸ [tex]\frac{miles}{h}[/tex]
The speed of light in a vacuum is 6.69 * 10⁸ [tex]\frac{miles}{h}[/tex]
How many ml of 0.213-M Na3PO4 are required to deliver 66.4 mmol Na1+ ions?
Given :
Molarity of [tex]Na_3PO_4[/tex] , M = 0.213 M .
To Find :
How many ml of 0.213 M of [tex]Na_3PO_4[/tex] are required to deliver 66.4 mmol [tex]Na^+[/tex].
Solution :
Volume required :
[tex]V=[\text{ 66.4 mmol of }Na^+] + [\dfrac{\text{1 mm of }Na_3PO_4}{\text{3 mmol of }Na^+}]+[\dfrac{\text{1 ml }Na_3PO_4}{\text{0.213 mmol of }Na_3PO_4}][/tex]
So ,
[tex]V=\dfrac{66.4}{3\times 0.213}\ ml\\\\V=103.91\ ml[/tex]
Therefore , volume of [tex]Na_3PO_4[/tex] required is 103.91 ml .
Hence , this is the required solution .
A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 84 g of water (with an initial temperature of 20 °C) in a calorimeter. The final temperature of the system is 52.1°C. The specific heat of water is 4.184 J/g*⁰C. What is the specific heat of the metal? questions below A) 0.171 B) 0.343 C) 1.717 D) 3.433
want to talk
Explanation:
What are electrolytes?
Answer:
Electrolytes are minerals in your body that have an electric charge. They are in your blood, urine, tissues, and other body fluids. Electrolytes are important because they help. Balance the amount of water in your body
Explanation:
The work needed to move a mass of 2.12 kg 47.1 m against a resistive force of 21.4 N is ___ J.
Answer:
1007.94 J
Explanation:
Work done: This can be defined as the product of force and distance, along the direction of the force. The S.I unit of work is Joules (J) .
Form the question above,
W = F×d........................... Equation 1
Where W = Work done, F = Resistive force, d = distance.
Given: F = 21.4 N, d = 47.1 m
Substitute these values into equation 1
W = 21.4(47.1)
W = 1007.94 J
Which of the following is NOT a "weak" interaction?
a. hydrogen bonds b. van der Waals forces c. disulfide bonds d. ionic interactions e. hydrophobic interactions
Answer:
D.ionic interaction is a strong interaction
When atoms come close enough to one another, an ionic connection is created. Atoms can interact with one another ionically, and this distance is greater than the bonding distance. Therefore, option D is correct.
What is ionic interaction ?The main interaction taking place in ionic compounds is ionic bonding, a form of chemical bonding that includes the electrostatic attraction between two atoms or ions with dramatically differing electronegativities.
Ions with opposing charges are drawn to one another through ion-ion interactions. They are also known as ionic bonds, because they are what keep ionic compounds together. opposing charges attract each other whereas like charges repel each other.
The whole transfer of valence electrons between atoms is referred to as an ionic bond. It is a kind of chemical connection that produces two ions with opposing charges. In ionic bonding, the nonmetal receives the electrons to become a negatively charged anion while the metal gives them up to become a positively charged cation.
Thus, option D is correct.
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What is a observation
Isotopes are atoms with the SAME number of protons, but DIFFERENT numbers of neutrons. Question 8 options: True False
Answer: The correct answer would be : True
I hope that this helps you !
In Europe, gasoline efficiency is measured in km/L. If your car’s gas mileage is 25.0 mi/gal, how many liters of gasoline would need to buy to complete a 250 km trip to Europe? Use the following conversions: 1 km= 0.6214 mi and 1 gal = 3.78L
Answer:
[tex]V=23.5L[/tex]
Explanation:
Hello,
In this case, it is convenient to compute the car's mileage in km/L as follows:
[tex]25.0\frac{mi}{gal}*\frac{1km}{0.6214mi}*\frac{1gal}{3.78L}=10.64\frac{km}{L}[/tex]
In such a way, since the distance is measured to be 250 km, the volume requirement is:
[tex]V=\frac{250km}{10.64kg/L}\\ \\V=23.5L[/tex]
Regards.
Dr. Winters decides not to tell the people in his experiment about the potential risks. He
.
A) is within the limits of ethical behavior
B) is being noncompliant
C) is making an honest mistake
D) has a lot of integrity
Answer: sorry I cant help you I need the same answer
Explanation:
Answer:
b
Explanation:
What causes blood cells to shrink?
Two particles of the element iodine are chemically bound. The two particles compose what?
Group of answer choices
a compound
a molecule
an isotope
an atom
Answer:
compound and elements.
Explanation:
A sample of mass 6.814 grams is added to another sample weighing 0.08753 grams.
The subsequent mixture is then divided into exactly 3 equal parts. One of those parts has the yield multiplied by 7.6335 times; what is the final mass?
Answer:
17.5609grams
Explanation:
According to the question, a sample of mass 6.814 grams is added to another sample weighing 0.08753 grams i.e.
= 6.814 + 0.08753
= 6.90153 grams
The result of this mixture i.e. 6.90153 grams is divided into three equal parts i.e.
= 6.90153 ÷ 3
= 2.30051 grams
This means that each equal part weighs 2.30051 grams
One of the yield (2.30051 grams) is multiplied by 7.6335 times i.e.
= 2.30051 × 7.6335
= 17.56094309
Hence, the final mass is 17.5609grams
Which of the following substances has the lowest density?
A) A mass of 1.5 kg and a volume of 1.2 L
B) A mass of 25 g and a volume of 20 mL
C) A mass of 750 g and a volume of 70 dL
D) A mass of 5 mg and a volume of 25 UL (mcL)
Answer:
C) A mass of 750 g and a volume of 70 dL .
Explanation:
Hello,
In this case, for substantiating the substance having the lowest density we need to compute it in the same units for each case as shown below:
[tex]\rho=\frac{m}{V}[/tex]
A) [tex]\rho =\frac{1.5kg}{1.2L}*\frac{1000g}{1kg} *\frac{1L}{1000mL}=1.25g/mL[/tex]
B) [tex]\rho =\frac{25g}{20mL}*=1.25g/mL[/tex]
C) [tex]\rho =\frac{750g}{70dL}*\frac{10dL}{1L}*\frac{1L}{1000mL} =0.107g/mL[/tex]
D) [tex]\rho =\frac{15mg}{25\mu L}*\frac{1g}{1000mg} *\frac{1000\mu L}{1mL}=0.6g/mL[/tex]
Therefore, the lowest density corresponds to C) A mass of 750 g and a volume of 70 dL
Regards.
Select the statements that correctly describe buffers.
A) A buffer is made up of a strong acid and a strong base.
B) The pH of a buffer solution does not change significantly when a small amount of acid is added.
C) The pH of a buffer solution is determined by the ratio of the concentration of acid to the concentration of base.
D) The K_a of a buffer decreases slightly when a small amount of acid is added to the buffer solution.
E) A strong acid added to the buffer solution reacts with the weak acid of the buffer.
Answer:
B) The pH of a buffer solution does not change significantly when a small amount of acid is added.
C) The pH of a buffer solution is determined by the ratio of the concentration of acid to the concentration of base.
Explanation:
Buffer solution is useful in maintaining the pH of a solution constant . It is made of a weak acid and its salt or a weak base and its salt . Ph of a buffer is given by the following relation .
pH = pKa + log [ A⁻ ] / [ HA ]
A- is a base and HA is a weak acid .
When we add acid , it reacts with A⁻ or base and gets neutralised .
When we add base , it reacts with acid HA and gets neutralised . In this way it maintains the pH of the solution .
Correct statements are as follows:
B) When a tiny amount of solvent is introduced to a buffer, the pH doesn't really alter considerably.
C) The ratio of acids to base concentration determines the pH of a phosphate buffer.
So, options B) and C) are correct.
Define pH.ph scale used in science to measure the acidity or alkalinity of an aqueous solution. pH originally stood for "potential of hydrogen." The pH of acidic solutions is lower than that of basic or alkaline solutions.
Correct statements are as follows:
B) When a tiny amount of solvent is introduced to a buffer, the pH doesn't really alter considerably.
C) The ratio of acids to base concentration determines the pH of a phosphate buffer.
So, options B) and C) are correct.
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A 2 mole sample of F2(g) reacts with excess NaOH(aq) according to the equation above. If the reaction is repeated with excess NaOH(aq) but with 1 mole of F2(g), which of the following is correct?
Group of answer choices
The amount of OF2(g) produced is doubled.
The amount of OF2(g) produced is halved.
The amount of NaF(aq) produced remains the same.
The amount of NaF(aq) produced is doubled.
Answer:
B - The amount of OF2(g) produced is halved.
Explanation:
Because its right.
When instead of 2 moles of F₂, 1 mole of F₂ reacts with excess NaOH, the amount of OF₂(g) produced is halved.
Let's consider the balanced reaction that occurs when F₂ reacts with NaOH.
2 F₂(g) + 2 NaOH(aq) → OF₂(g) + 2 NaF(aq) + H₂O(l)
First, let's see the moles of OF₂ and NaF obtained from 2 moles of F₂, using the molar ratios derived from the balanced chemical equation.
[tex]2 mol F_2 \times \frac{1 molOF_2}{2mol F_2} = 1 mol OF_2\\\\2 mol F_2 \times \frac{2 molNaF}{2mol F_2} = 2 mol NaF[/tex]
Now, let's compare with the moles of OF₂ and NaF obtained from 1 mol of F₂, again using the same molar ratios derived from the balanced chemical equation.
[tex]1 mol F_2 \times \frac{1 molOF_2}{2mol F_2} = 0.5 mol OF_2\\\\1 mol F_2 \times \frac{2 molNaF}{2mol F_2} = 1 mol NaF[/tex]
As we can see, since we have half the amount of F₂, we obtain half the amount of the products. Then, the only right option is: The amount of OF₂(g) produced is halved.
When instead of 2 moles of F₂, 1 mole of F₂ reacts with excess NaOH, the amount of OF₂(g) produced is halved.
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6. What effect does an increase in blood hydrogen ion levels have on the pH of the blood?
Answer: the two parameters change inversely; as hydrogen ion concentration increases, pH falls. due to the logarithmic relationship, a large change in hydrogen ion concentration is actually a small change in pH. For example, doubling the hydrogen ion concentration causes pH to fall by just 0.3.
A 2.600×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created by dissolving a sample of C3H8O3 in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 998.9 mL . The density of water at 20.0∘C is 0.9982 g/mL.Part ACalculate the concentration of the glycerol solution in percent by mass.Express your answer to four significant figures and include the appropriate units.Part BCalculate the concentration of the glycerol solution in parts per million.Express your answer as an integer to four significant figures and include the appropriate units.
Answer:
A. 0.2395 w/w %
B. 2394ppm
Explanation:
A. To find concentrationin percent by mass of the solution we need to calculate mass of glycerol and mass of water. The formula is:
Mass glycerol / Total mass * 100
Mass glycerol:
The solution is 2.6x10⁻²moles / L. As there is 1L of solution there are 2.6x10⁻² moles of glycerol. In mass (Using molar mass glycerol: 92.09g/mol):
2.6x10⁻² moles of glycerol * (92.09g / mol) = 2.394g glycerol
Mass of water:
998.9mL and density = 0.9982g/mL:
998.9mL * (0.9982g/mL) = 997.1g of water.
That means percent by mass is:
% by mass: 2.394g / (997.1g + 2.394g) * 100 = 0.2395 w/w %
B. Parts per million are mg of glycerol per L of solution. As in 1L there are 2.394g. In mg:
2.394g * (1000mg / 1g) = 2394mg:
Parts per million: 2394mg / L = 2394ppm
Considering the definition of percent by mass
A) the concentration of the glycerol solution in percent by mass is 0.2395%.
B) the concentration of the glycerol solution is 2394.34 ppm.
Concentration of the glycerol solution in percent by mass
A) The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.
The mass percentage of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.
The mass percentage is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:
[tex]mass percentage=\frac{mass of solute}{mass of solution} x100[/tex]
In this case, you have a 2.600×10⁻² M solution of glycerol (C₃H₈O₃) in water. The sample was created by dissolving a sample of C₃H₈O₃ in water and then bringing the volume up to 1.000 L.
So, being the molarity the number of moles of solute that are dissolved in a certain volume, the number of moles of glycerol can be calculated as:
number of moles of glycerol= 2.600×10⁻² M× 1 L
number of moles of glycerol= 2.600×10⁻² moles
Being the molar mass glycerol 92.09 [tex]\frac{g}{mole}[/tex], the mass of glycerol can be calculated as:
2.600×10⁻² moles×92.09 [tex]\frac{g}{mole}[/tex]= 2.39434 grams of glycerol
On the other side, the volume of water needed was 998.9 mL and the density of water at 20.0∘C is 0.9982 [tex]\frac{g}{mL}[/tex]. So, the mass of water needed can be calculated as:
998.9 mL×0.9982 [tex]\frac{g}{mL}[/tex]= 997.1 grams of water
Finally, the mass percentage of the solution can be calculated as:
[tex]mass percentage=\frac{mass of glycerol}{mass of glycerol + mass of water} x100[/tex]
Solving:
[tex]mass percentage=\frac{2.39434 grams}{2.39434 grams+ 997.1 grams} x100[/tex]
[tex]mass percentage=\frac{2.39434 grams}{999.49434 grams} x100[/tex]
mass percentaje= 0.2395 %
In summary, the concentration of the glycerol solution in percent by mass is 0.2395%.
Parts per million (ppm)B) Parts per million (ppm) is a unit of measurement of concentration that measures the number of units of substance in each million units of the whole. In this case, the concentration measurement refers to mg of glycerol per L of solution.
Being the mass of glycerol 2.39434 grams equal to 2394.34 mg (1 g=1000mg), the concentration is:
[tex]concentration=\frac{2394.34 mg}{1L}[/tex]
concentration= 2394.34 ppm
In summary, the concentration of the glycerol solution is 2394.34 ppm.
Learn more about:
mass percentage: brainly.com/question/19168984?referrer=searchResults brainly.com/question/18646836?referrer=searchResultsbrainly.com/question/24201923?referrer=searchResults brainly.com/question/9779410?referrer=searchResults brainly.com/question/17030163?referrer=searchResults ppmhttps://brainly.com/question/16727593?referrer=searchResultshttps://brainly.com/question/13565240?referrer=searchResultsILL GIVE YOU BRAINLIST !!! HAVE TO GET IT RIGHT !!
1.
Explain.
Identify Which are the quantitative data in the example above?
Answer: The 3rd and 6th bullet point is the quantitative data.
Explanation: Quantitative data is expressed by NUMBERS and Qualitative data is expressed by WORDS. The 3rd and 6th one is correct because they both use numbers to compare how much time hummingbirds spent feeding on nectar.
There are 360 degrees in a circle. How many arcminutes will be in 86% of a circle?
Answer:
There are 18572.9 arcminutes in 86% of a circle.
Explanation:
The degrees in 86% of a circle is:
[tex] degrees = \frac{86}{100}*360 = 309.6 ^{\circ} [/tex]
Now, we need to find the number of arcminutes in 309.6°:
[tex] arcminute = \frac{59.99 arcminute}{1 degree}*309.6 degrees = 18572.9 [/tex]
Therefore, there are 18572.9 arcminutes in 86% of a circle.
I hope it helps you!
Look at the image of the solar system,
According to Kepler, which planet travels the fastest?
Neptune
Our Solar System
O Earth
Saturn
O Jupiter
O Mercury
Neptune
Mars
Uranus
Asteroid Bilt.
The Sun
Mercury
Earth
Venus
Jupiter
Mark this and return
Save and Exit
Nex
Submit
Answer:
Mercury
Explanation:
CH3
/
CH3----C-----CH3
/
CH3
Spell out the full name of compound.
Answer:
A. Neopentane.
B. 2,2-dimethyl-propane.
Explanation:
Hello,
In this case, for the given compound we can spell out two possible names:
A. Neopentane: it is, because the molecule has five carbon atoms wherein there is a central one, bonded to the other four.
B. 2,2-dimethyl-propane: it is, because we find two methyl radicals at the second carbon.
Regards.
During studies of the following reaction (i), a chemical engineer measured a less-than-expected yield of N2 and discovered that the following side reaction (ii) occurs. (i) N2O4(l) 2 N2H4(l) 3 N2(g) 4 H2O(g) (ii) 2 N2O4(l) N2H4(l) 6 NO(g) 2 H2O(g) In one experiment 12.7 g of NO formed when 101.1 g of each reactant was used. What is the highest percent yield of N2 that can be expected
Answer:
Maximum expected yield = 87.2%
Explanation:
Equations of reactions:
Main reaction: N₂O₄(l) + 2N₂H₄(l) ---> 3N₂(g) + 4H₂O(g)
Side reaction: 2N₂O₄(l) + N₂H₄(l) ----> 6NO(g) + 2H₂O(g)
Molar mass of N₂O₄ = 92 g/mol; molar mass of N₂H₄ = 32 g/mol; molar mass of N₂ = 28 g/mol; molar mass of of NO = 30 g/mol; molar mass of water = 18 g/mol
In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂.
101.1 g of N₂O₄ will react with 2 * 32 * 101.1 / 92 g of N₂H₄ = 70.33 g of N₂H₄
N₂O₄ is the limiting reactant
101.1 g of N₂O₄ will react to produce 3 * 14 * 101.1 / 92 g of N₂ = 46.15 g of N₂
In the side reaction, (6 * 30 g) of NO is produced from (2 * 92 g) of N₂O₄ and 32 g of N₂H₄
12.7 g of N₂O₄ will be produced from ( 2 * 92 * 12.7/180 g) of N₂O₄ and (32 * 12.7/180) g of N₂H₄ to produce
mass of N₂O₄ used = 12.98 g
mass of N₂H₄ used = 2.26 g
mass of N₂O₄ left for main reaction = 101.1 - 12.98 = 88.12 g
mass of N₂H₄ left for main reaction = 101.1 - 2.26 = 98.84 g
In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂
88.12 g of N₂O₄ will react with 2 * 32 * 88.12 / 92 g of N₂H₄ = 61.30 g of N₂H₄
N₂O₄ is the limiting reactant.
88.12 g of N₂O₄ will to react produce 3 * 14 * 88.12 / 92 g of N₂ = 40.23 g of N₂
Percentage yield = (theoretical yield/actual yield) * 100%
Percentage yield = (40.23/46.15) * 100% = 87.2%
Therefore, maximum expected yield = 87.2%
Some SO2 and O2 are mixed together in a flask at 1100 K in such a way ,that at the instant of mixing, their partial pressures are, respectively, 1.00 atm and 0.500 atm. When the system comes to equilibrium at 1100 K, the total pressure in the flask is found to be 1.35 atm. Given: 2SO2(g) + O2(g) ⇌ 2SO3(g); ΔrH = − 198.2 kJ. mol-1 1.1 Calculate Kp at 1100 K
Answer:
The answer is "[tex]\bold{0.525\ \ atm^{-1}}[/tex]"
Explanation:
Given equation:
[tex]2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)[/tex]
Given value:
[tex]\Delta rH =-198.2 \ \ \frac{KJ}{mol}[/tex]
[tex]Kp=1100 \ K[/tex]
[tex]\Delta x = 2-(2+1)\\\\[/tex]
[tex]= 2-(2+1)\\\\= 2-(3)\\\\= -1[/tex]
[tex]\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm) &1-2x&0.5-x&2x\end{array}\right[/tex]
calculating the total pressure on equilibrium= [tex](1-2x)+(0.5-x)+2x \ atm\\\\[/tex]
[tex]= 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x\\[/tex]
[tex]\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to x= 0.15\\\\[/tex]
calculating the pressure in [tex]So_2[/tex]:
[tex]= (1-2 \times 0.15)[/tex]
[tex]= 1-0.30 \\\\ =0.70 \ atm[/tex]
calculating the pressure in [tex]O_2[/tex]:
[tex]= (0.5- 0.15)\\\\= 0.35 \ atm \\[/tex]
calculating the pressure in [tex]So_3[/tex]:
[tex]= (2 \times 0.15)\\\\= (.30) \ atm \\\\[/tex]
Calculating the Kp at 1100 K:
[tex]= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\= 0.5247 \ \ or \ \ 0.525 \ \ atm^{-1} \\\\[/tex]
Does the density of a liquid depend on its volume? Write your answer as a CER.
Answer: Yes
Explanation:
Density of a liquid depend on its volume. This is because Density is mass of liquid divided by volume.
Density is inversely proportional to volume.
As density increases, volume decreases and vice versa. The density for water is 1g/ milliliter but it changes with changes in temperature or there are impurities dissolved in it. Ice is less dense that liquid water and it's the major reason it's float because it's volume is inversely proportional to it's density.
A patient needs 40.0 mg of an antibiotic per kilogram of body weight each day. If the patient weighs 55 kilograms, how much antibiotic, in milligrams, should the patient receive each day for optimal therapy?
Answer:
2,200 milligrams
Explanation:
40 mg per 1 kilogram means 40 times 55 which is 2,200 mg
The boiling point of another member of this homologous series was found to be 309 KK. What is the likely molecular formula for this compound?
Answer: Pentane C5H12
Explanation:
The boiling point of a substance is simply defined as the temperature whereby a liquid's vapor pressure is equal to the pressure that is surrounding the liquid and hence, the liquid will changes into vapor.
The likely molecular formula for this compound is Pentane i.e C5H12 due to the fact that its boiling point is between Butane with formula C4H10 and Hexane with formula C6H14 boiling points.
The molecular formula of the of the substance is [tex]\bold {C_5H_1_0}[/tex] (pentane).
Boiling point:
It is the temperature at which liquid's vapor pressure is equal to the pressure that is surrounding the liquid and hence, the liquid start to into vaporize.
The boiling point of the given compound is 309 K which is between Butane and Hexane.
Therefore, the molecular formula of the of the substance is [tex]\bold {C_5H_1_0}[/tex] (pentane).
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Differentiate between a pure substance and a mixture and provide examples
Explanation:
pure substances are those substances which are made up of one elements
mixture that those substances which are made by more than two or three elements
Answer:
Brainliest is appreciated
Explanation:
Substances which have a specific composition and cannot be separated into any constituents are called pure substances. Pure substances are further divided into elements and compounds. The combination of two or more pure substances is called a mixture. Mixtures can be classified into two types viz. heterogeneous and homogeneous mixtures.
EXAMPLES
pure substance
gold, pure water, hydrogen gas
mixture substance
oil and water, sand and sugar