Answer:
The master mix contains the following reagents: dNTPs, DNA Polymerase, PCR buffer and MgCl2.
Explanation:
The DNA templates are the gene fragments to amplify by PCR, thereby they have to be added separately in each tube. Moreover, the primer pairs are specific for each gene, thereby they have to be added separately in each tube.
Deoxynucleotide triphosphates (dNTPs) are the building blocks of the DNA molecules: dGTP, dATP, TTP, and dCTP.
The PCR buffer provides a suitable medium for the activity of the DNA polymerase, often it contains Tris-Hcl and KCl.
MgCl2 is a cofactor for the activity of the DNA Polymerase.
The DNA Polymerase is an enzyme that amplifies DNA by adding nucleotides to the 3' end.
Two samples of the same rainwater are tested using two indicators at an environrnental lab. The first indicator, Methyl Orange, reveals a distinct yellow color when added to the sample. The second indicator, Litmus, turns red when placed in contact with the water sample.
Required:
a. Identify a possible pH value for the rainwater.
b. Explain, in terms of hydronium ions and hydroxide ion concentrations, the pH value of the rainwater.
Answer:
A. The pH value of rainwater is acidic about 4.4
B. The molar concentrations of the Hydronium ions are more than that of the hydroxide ions. That is why the rainwater is acidic with a pH of less than 7
Explanation:
A. Methyl orange is an acid indicator that is used to detect acidic solutions which have pH values that fall within the range of about 4.4 to 7. The distinct yellow colour change that was shown by the methyl orange as it was added to the water shows that the pH value is acidic, with a value above 4.4. (it has to be like this before methyl orange changes to yellow colour)
B. The Hydronium ( H30+) ion concentrations and the hydroxide (OH-) ion concentrations are used to measure the pH values of substances.
We can tell that the Hydronium ( H30+) ion concentrations are more than the hydroxide (OH-) ion concentrations in the sample of rainwater tested. This can be detected from the colour change that both the methyl orange and the litmus paper gave. The indicators showed that the rainwater solution was indeed acidic. Hence, the pH value will be less than 7, but greater than 4.4.
A gas has volume of 800.0mL at -23.0°c and 300.0torr. What would the volume of the gas be at 227.0°c and 600.0torr of pressure
Answer:
Explanation:
use gas law eqation
P1 * V1 / T1 = P2 * V2 /T2
600*V1/227 = 300*800/23
V1 = 300*800*227 / 23*600 = ............ can you solve this and get the answer?
Determine if the following salt is neutral, acidic or basic. If acidic or basic, write the appropriate equilibrium equation for the acid or base that exists when the salt is dissolved in aqueous solution. If neutral, write only NR. LiNO₃
Answer:
It is neutral (NR)
Explanation:
Salts are formed when the ionizable hydrogens in an acid is replaced by metallic or ammonium ions from bases. The reaction is known as a neutralization reaction.
The nature of a salt formed from this reaction depends on the nature of the reacting acid and base.
If the reaction is between a strong acid and strong base, the salt produced is a neutral salt.
If the reaction occurs between a strong acid and a weak base, the salt produced is acidic.
If the reaction occurs between a strong base and a weak acid, the salt produced is a basic salt.
Considering the salt above, LiNO3.
On hydrolysis, addition of water, the following products are obtained:
LiNO3 + H2O ----> LiOH + HNO3
The products obtained, LiOH and HNO3 are a strong base and a strong acid respectively. Therefore, the salt, LiNO3, is a neutral salt.
The salt, LiNO₃ is a neutral, NR salt as it's a salt formed from the reaction of a strong acid and a strong base.
In neutralisation reactions, acids and bases react to form salt and water.
However, the salt formed may be acidic, basic or neutral. This is dependent on the type of acid and base which form the salt.
A strong acid and a strong base react to yield a neutral salt like, LiNO₃.
The equilibrium equation when LiNO₃ is dissolved in aqeous solution is;
LiNO₃ + H2O ==>. LiOH + HNO₃Evidently, LiOH and HNO₃ are an example strong base and acid respectively.
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6. To isolate benzoic acid from a bicarbonate solution, it is acidified with concen- trated hydrochloric acid, as in experiment 1. What volume of acid is needed to neutralize the bicarbonate
Answer:
For our assumed experiment; the expected volume of Hcl acid needed to neutralize the bicarbonate is 0.13 mL
Explanation:
We are going attempt this question experimentally.
We know that benzoic acid originate from the relationship between benzene and a carboxylic group. So basically , the functional group of a carboxylic acid (-COOH) joins with a benzene ring(C₆H₆) to form a simple aromatic carboxylic acid known as Benzoic acid. (C₇H₆O₂)
However, it is possible to isolate benzoic acid from a bicarbonate solution in the presence of an acidified concentrated hydrochloric acid.
Let assume that ;
0.20 g of benzoic acid was reacted with 2 mL of a 20% solution of NaHCO₃, the amount of the excess NaHCO₃ can be determined by subtracting the amount of benzoic acid from the amount of NaHCO₃.
Let first calculate the number of moles in 0.20 g of benzoic acid
we know that the standard molar mass of benzoic acid is 122.12 g/mol
number of moles of benzoic acid = mass of benzoic acid/molar mass of benzoic acid =
number of moles of benzoic acid = 0.20/ 122.12
number of moles of benzoic acid = 0.0016 mol
number of moles of bicarbonate solution = mass of bicarbonate solution/ molar mass of bicarbonate solution
number of moles of bicarbonate solution = 0.2/84.00654 g/mol
number of moles of bicarbonate solution = 0.00238 mol
∴
(0.00238 - 0.0016) mol
= 7.8 × 10⁻⁴ mol
Let assume that the concentrated HCl is 12 M
Also. HCl and NaHCO₃ react together at the ratio of 1:1; thus the volume of Hcl acid needed to neutralize the bicarbonate is:
[tex]= ( 7.8 * 10^{-4} \ \ mol )* ( \dfrac{2\ L}{ 12 \ M})*( \dfrac{10^3 \mL}{1 \ L})[/tex]
= 0.13 mL
Thus; for our assumed experiment; the expected volume of Hcl acid needed to neutralize the bicarbonate is 0.13 mL
Determine the amount of heat (in kJ) associated with the production of 5.71 × 104 g of ammonia according to the following equation. N2(g) + 3H2(g) 2NH3ΔH°rxn = −92.6 kJ Assume that the reaction takes place under standard-state conditions at 25°C.
Answer:
[tex]Q=-3.11x10^5kJ[/tex]
Explanation:
Hello,
In this case, for the given reaction, we are given the standard enthalpy of reaction per mole of ammonia that is -92.6 kJ, it means, that forming one mole of ammonia will release 92.6 kJ of energy. In such a way, for the formation of 5.71x10⁴ g of ammonia, the following amount of heat will be released:
[tex]Q=5.71x10^4gNH_3*\frac{1molNH_3}{17gNH_3}*-92.6\frac{kJ}{molNH_3}\\ \\Q=-3.11x10^5kJ[/tex]
Best regards.
The amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ
We'll begin by calculating the number of mole in 5.71×10⁴ g of NH₃
Mass of NH₃ = 5.71×10⁴ g
Molar mass of NH₃ = 14 + (3×1) = 17 g/mol
Mole of NH₃ =?Mole = mass / molar mass
Mole of NH₃ = 5.71×10⁴ / 17
Mole of NH₃ = 3358.82 moles Finally, we shall determine the heat required to produce 3358.82 moles (i.e 5.71×10⁴ g) of NH₃. This can be obtained as follow:N₂(g) + 3H₂(g) —> 2NH₃(g) ΔH°rxn = −92.6 kJ
Since reaction took place at standard conditions, it means:
1 moles of NH₃ required −92.6 kJ
Therefore,
3358.82 moles of NH₃ will require = 3358.82 × –92.6 = –311026.732 KJ
Thus, the amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ
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A geochemist in the field takes a 46.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X. He notes the temperature of the pool, 21°C, and caps the sample carefully. Back in the lab, the geochemist filters the sample and then evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 0.87 g.
Required:
Using only the information above, can you calculate the solubility of X in water at 21°C? If yes, calculate it.
Answer: The solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.
Explanation:
The given data is as follows.
Volume of sample water = 46 ml
Temperature = [tex]21^oC[/tex]
After vaporization, washes and then drying the weight of mineral X = 0.87 g
This means that 46.0 ml of water contains 0.87 g of X. Therefore, grams present in 1 ml of water will be calculated as follows.
1 ml of water = [tex]\frac{0.87 g}{46.0 ml}[/tex]
= [tex]1.891 \times 10^{-2}[/tex] g/ml
Therefore, we can conclude that solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.
How is excitation in spectroscopy brought about
Answer: the exciation of molecules is brount by absorption of energy in spectroscpy
Explanation:
C3H7-C(=O)-NH2 IUPAC NAME ?
Answer:
Amide
Explanation:
O=NH2 is the Amide group versus NH2, which is the amine group.
Answer:
Butamide
Explanation:
C3H7-C(=O)-NH2 IUPAC NAME
C4H9NO
H H H
H - C - C - C - C = O
H H H N - H
H
But amide
Amide because R-CO-NH2 ie C(=O)-NH2
But because 4 Cabon
The realization that guanine and thymine base pair to cysteine and adenine, respectively, through their keto-like tautomeric forms has ultimately led to which types of technologies seen in popular TV shows like CSI and Law and Order?
a. DNA sequencing and profiling
b. Flexible display monitors
c. Smart phones
d. Police cars
Answer:
The correct answer to the following question will be Option A (DNA sequencing and profiling).
Explanation:
It is indeed a procedure through which a specific DNA sequence named as the profile is chosen to take from either a sample or perhaps a survey. For humans, much more of the DNA sequence will be the same, although only different regions differ in sequencing, such locations are considered polymorphic. The disparity between individuals regardless of such a particular process is named as polymorphisms, and then this method is being used in the detection of murders, parental conflicts, etc.The other three situations aren't connected to the situation in question. That option 1 seems to be the right answer.
Pick the odd one out?
Ethanol
Hexane
Oil
Carbon tetrachloride
Answer: Ethanol is the odd one out.
Explanation:
A polar compound is defined as the compound which is formed when there is a difference of electronegativities between the atoms. It is also defined as the bond which is formed due to the unequal sharing of electrons between the atoms.
Non-polar compound is defined as the compound which is formed when there is no difference of electronegativities between the atoms or the polarities cancel out.
Hexane [tex](C_6H_{14}), Oil (mixture of hydrocarbons) and carbon tetrachloride [tex](CCl_4)[/tex] all are non polar whereas ethanol is polar due to electronegative difference between hydrogen and oxygen.
Which of the following is a conjugate acid-base pair in the reaction represented by the
equation below?
H2PO4 + H20 H3PO, + OH
H2PO, and H2O
b) H,PO, and OH
c) H2PO, and H3PO,
None of the above
Answer: [tex]H_2PO_4[/tex] and [tex]H_3PO_4[/tex]
Explanation:
According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.
For the given reaction:
[tex]H_2PO_4^-+H_2O\rightleftharpoons H_3PO_4+OH^-[/tex]
Here, [tex]H_2O[/tex] is loosing a proton, thus it is considered as an acid and after losing a proton, it forms [tex]OH^-[/tex] which is a conjugate base.
Similarly , [tex]H_2PO_4^-[/tex] is gaining a proton, thus it is considered as an base and after gaining a proton, it forms [tex]H_3PO_4[/tex] which is a conjugate acid.
Thus [tex]H_2PO_4[/tex] and [tex]H_3PO_4[/tex] is a conjugate acid-base pair in the reaction represented by the equation below
The Bronsted-Lowry conjugate acid-base hypothesis defines an acid as a substance that loses protons and donates them to another chemical to produce conjugate base, and a base as a substance that takes protons to generate conjugate acid.
Thus, a proton is being lost, making it an acid, and once a proton is lost, a conjugate base is formed. Similar to that, is gaining a proton, making it a base, and then it produces a conjugate acid after gaining a proton.
The Brnsted-Lowry hypothesis, often known as the proton theory of acids and bases, is an independent theory of acid-base reactions that was put forth in 1923 by Johannes Nicolaus Brnsted and Thomas Martin Lowry.
This theory's central idea is that when an acid and a base interact, the acid creates its conjugate base and the base creates its conjugate acid by exchanging a proton (the hydrogen cation, or H+).
Thus, The Bronsted-Lowry conjugate acid-base hypothesis defines an acid as a substance that loses protons and donates them to another chemical to produce conjugate base, and a base as a substance that takes protons to generate conjugate acid.
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How many moles of PC15 can be produced from 51.0 g of Cl2 (and excess P4)?
Express your answer to three significant figures and include the appropriate units.
LIT....ITS NOT .227 or .228!!!!
Answer:
0.287 mole of PCl5.
Explanation:
We'll begin by calculating the number of mole in 51g of Cl2. This is illustrated below:
Molar mass of Cl2 = 2 x 35.5 = 71g/mol
Mass of Cl2 = 51g
Number of mole of Cl2 =..?
Mole = Mass /Molar Mass
Number of mole of Cl2 = 51/71 = 0.718 mole
Next, we shall write the balanced equation for the reaction. This is given below:
P4 + 10Cl2 → 4PCl5
Finally, we determine the number of mole of PCl5 produced from the reaction as follow:
From the balanced equation above,
10 moles of Cl2 reacted to produce 4 moles of PCl5.
Therefore, 0.718 mole of Cl2 will react to produce = (0.718 x 4)/10 = 0.287 mole of PCl5.
Therefore, 0.287 mole of PCl5 is produced from the reaction.
What happens in a double replacement reaction
Answer: D
Explanation: The elements in two compunds switch places
Now construct a different electrochemical cell. You put a zinc metal anode in contact with a 0.052 M solution of zinc nitrate and a silver cathode in contact with a 0.0042 M solution of silver(I) nitrate. What is the value of the electric potential at the moment the reaction begins
Answer:
[tex]1.66~V[/tex]
Explanation:
We have to start with the half-reactions for both ions:
[tex]Zn^+^2~+2e^-~->Zn[/tex] V= -0.76
[tex]Ag^+~e^-~->~Ag[/tex] V= +0.80
If we want a spontaneous reaction (galvanic cell) we have to flip the first reaction, so:
[tex]Zn~->~Zn^+^2~+2e^-~[/tex] V= +0.76
[tex]Ag^+~+~e^-~->~Ag[/tex] V= +0.80
If we want to calculate ºE we have to add the two values, so:
ºE=0.76+0.80 = 1.56 V
Now, we have different concentrations. So, if we want to calculate E we have to use the nerts equation:
[tex]E=ºE~+~\frac{0.059}{n}LogQ[/tex]
On this case, Q is equal to:
[tex]Q=\frac{[Zn^+^2]}{[Ag^+]^2}[/tex]
Because the total reaction is:
[tex]Zn~+~2Ag^+~->~Zn^+^2~+~2Ag[/tex]
So, the value of "Q" is:
[tex]Q=\frac{[0.052 M]}{[0.0042]^2}=2947.84[/tex]
Now, we can plug all the values in the equation (n=2, because the amount of electrons transferred is 2). So:
[tex]E=1.56~V~+~\frac{0.059}{2}Log(2947.84)=1.66~V[/tex]
I hope it helps!
What happens in a neutralization reaction?
a
The hydrogen (H+) ions from the base and the hydroxide (OH-) ions from the acld come together to form water.
The hydrogen (H+) ions from the acid and the hydroxide (OH) ions from the base come together to form water.
A substance's pH is increased to a value greater than 7.
A solution of a known concentration and volume is added until the reaction is completed.
Answer:
Answer is letter B
Explanation:
The first one is wrong because acids release H+, not bases.
The third one is wrong because the pH is exactly 7, not greater.
The last one is wrong because it is vague and does not fit a neutralization reaction.
Barium is a very reactive metal in the presence of oxygen and water, thus its density cannot be measured by water displacement. Instead, mesitylene (C9H12, density = 0.86370 g/mL (at 20 o C)) is used. 77.240 g of Ba is placed into a flask, and mesitylene is added so that together the total volume is 100.00 mL. The mass of the mesitylene and Ba together is 148.792 g. What is the density (in g/mL) of the Ba at 20 o C?
Answer:
The correct answer is 4.502 g per ml.
Explanation:
Based on the given question, the sum of the mass of mesitylene and barium together is 148.792 grams. The mass of barium given is 77.240 grams. Therefore, the mass of mesitylene will be,
Mass of mesitylene = Total mass - Mass of barium
= 148.792 - 77.240
= 71.552 grams
The density of mesitylene is 0.86370 g per ml. To calculate the volume of mesitylene, the formula to be used is,
Volume = mass / density. Now, putting the values we get,
Volume = 71.552 / 0.86370 = 82.8436 ml.
As the total volume is 100 ml, therefore, the volume of Ba will be,
Volume of Ba = 100-82.8436 = 17.1564 ml
The density of Ba at 20 degree C can be calculated by using the formula,
Density = mass / volume. Now putting the values we get,
Density = 77.240 g / 17.1564 ml
= 4.502 g per ml
What can be known about the salt sample that Gerry is looking at?
Answer:
That its small pointed. Pink(Himalayan salt)or white(normal salt)
Explanation:
Summa dees questions are so stupid, deys makin me salty.
The standard free energy change, ΔG°', for this reaction is +6.7 kJ/mol. However, the observed free energy change (ΔG) for this reaction in pig heart mitochondria is +0.8 kJ/mol. What is the ratio of [isocitrate]/[citrate] in these mitochondria at 25.0 °C?
what type of bonds do compounds formed from non metal consist of?
Compounds formed from non-metals consist of molecules. The atoms in a molecule are joined together by covalent bonds. These bonds form when atoms share pairs of electrons.
Which of the following formulas represents an ionic compound?
1.HI 2.HCI 3.LiCI 4.SO2
Answer:
Numbers 4,3
Explanation:
Ionic bond is between nonmental and metals
An excess of sodium carbonate, Na2CO3, in solution is added to a solution containing 15.71 g CaCl2. After performing the experiment, 13.19 g of calcium carbonate, CaCO3, is produced. Calculate the percent yield of this reaction
Answer:
93.15 %
Explanation:
We have to start with the chemical reaction:
[tex]CaCl_2~+~Na_2CO_3~->~CaCO_3~+~NaCl[/tex]
Now, we can balance the reaction:
[tex]CaCl_2~+~Na_2CO_3~->~CaCO_3~+~2NaCl[/tex]
Our initial data are the 15.71 g of [tex]CaCl_2[/tex], so we have to do the following steps:
1) Convert from grams to moles of [tex]CaCl_2[/tex] using the molar mass (110.98 g/mol).
2) Convert from moles of [tex]CaCl_2[/tex] to moles of [tex]CaCO_3[/tex] using the molar ratio. ( 1 mol [tex]CaCl_2[/tex]= 1 mol of [tex]CaCO_3[/tex]).
3) Convert from moles of [tex]CaCO_3[/tex] to grams of [tex]CaCO_3[/tex] using the molar mass. (100 g/mol).
[tex]15.71~g~CaCl_2\frac{1~mol~CaCl_2}{110.98~g~CaCl_2}\frac{1~mol~CaCO_3}{1~mol~CaCl_2}\frac{100~g~CaCO_3}{1~mol~CaCO_3}=14.16~g~CaCO_3[/tex]
Finally, we can calculate the yield percent:
[tex]%~=~\frac{13.19~g~CaCO_3}{14.16~g~CaCO_3}*100=93.15~%[/tex]
I hope it helps!
The percentage yield obtained when excess sodium carbonate, Na₂CO₃, is added to a solution containing 15.71 g CaCl₂ is 93.2%
We'll begin by writing the balanced equation for the reaction. This is given below: [tex]Na_{2}CO_{3} + CaCl_{2} - > CaCO_{3} + 2NaCl[/tex]Molar mass of CaCl₂ = 40 + (35.5×2) = 111 g/mol
Mass of CaCl₂ from the balanced equation = 1 × 111 = 111 g
Molar mass of CaCO₃ = 40 + 12 + (16×3) = 100 g/mol
Mass of CaCO₃ from the balanced equation = 1 × 100 = 100 g
SUMMARY
From the balanced equation above,
111 g of CaCl₂ reacted to produce 100 g of CaCO₃
Next, we shall determine the theoretical yield of of CaCO₃. This can be obtained as follow:From the balanced equation above,
111 g of CaCl₂ reacted to produce 100 g of CaCO₃.
Therefore,
15.71 g of CaCl₂ will react to produce = [tex]\frac{15.71 * 100}{111} \\\\[/tex] = 14.15 g of CaCO₃.
Thus, the theoretical yield of of CaCO₃ is 14.15 g
Finally, we shall determine the percentage yield. This can be obtained as follow:Actual yield of CaCO₃ = 13.19 g
Theoretical yield of CaCO₃ = 14.15 g
Percentage yield =?[tex]Percentage yield = \frac{Actual}{Theoretical} * 100\\\\= \frac{13.19}{14.15} * 100\\\\[/tex]
= 93.2%Therefore, the percentage yield of the reaction is 93.2%
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Mass is:
measured in kilograms
measured using a scale
affected by gravity
all of the above
a gas obeys the equation of state p(v-b)=RT.for the gas b=0.0391L/mol.calculate the fugacity coefficient for the gas at 1000°c and 1000atm
Answer:
The fugacity coefficient is [tex][\frac{f}{p} ] = 1.45[/tex]
Explanation:
From the question we are told that
The gas obeys the equation [tex]p(v-b) = RT[/tex]
The value of b is [tex]b = b = 0.0391 \ L /mol[/tex]
The pressure is [tex]p = 1000 \ atm[/tex]
The temperature is [tex]T= 1000^oC = 1273 K[/tex]
generally
[tex]RT ln[\frac{f}{p} ] = \int\limits^{p}_{o} [ {v_{r} -v_{i}} ]\, dp[/tex]
Where [tex]\frac{f}{p}[/tex] is the fugacity coefficient
[tex]v_r[/tex] is the real volume which is mathematically evaluated from above equation as
[tex]v_r = \frac{RT}{p} + b[/tex]
[tex]v_r = \frac{RT}{p} + 0.0391[/tex]
and [tex]v_{i}[/tex] is the ideal volume which is evaluated from the ideal gas equation (pv = nRT , at n= 1) as
[tex]v_{i} = \frac{RT}{p}[/tex]
So
[tex]RT ln[\frac{f}{p} ] = \int\limits^{1000}_{o} [[ \frac{RT}{p} + 0.0391] - [\frac{RT}{p} ]} ]\, dp[/tex]
=> [tex]RT ln[\frac{f}{p} ] = \int\limits^{1000}_{o} [0.391 ]\, dp[/tex]
=> [tex]RT ln[\frac{f}{p} ] = [0.391p]\left | 1000} \atop {0}} \right.[/tex]
=> [tex]RT ln[\frac{f}{p} ] = 38.1[/tex]
So
[tex]ln[\frac{f}{p} ] = \frac{39.1}{RT}[/tex]
Where R is the gas constant with value [tex]R = 0.082057\ L \cdot atm \cdot mol^{-1}\cdot K^{-1}[/tex]
[tex][\frac{f}{p} ] = \frac{39.1}{ 2.303 *0.082057 * 1273}[/tex]
[tex][\frac{f}{p} ] = 1.45[/tex]
Calculate the percentage of the void space out of the total volume occupied by 1 mole of water molecules. The density of water is assumed to be 1.0 g/mL that is 1.0 g/cm3. The molar mass of water is 18.0 g/mol. The atomic radius of hydrogen is 37 pm and of oxygen is 73 pm. The formula for the volume of a sphere is 4/3(r3
Answer:
The percentage of the void space out of the total volume occupied is 93.11%
Explanation:
A mole of water contains 2 atoms of hydrogen and 1 atom of oxygen.
To calculate the volume of a mole of water, we calculate 2 times the volume of the hydrogen atom and 1 times the volume of the oxygen atom
Let's calculate this one after the other.
For the hydrogen, formula for the volume will be
[tex]V_{hydrogen[/tex] = 2 × 4/3 × π × [tex]r_{H}^{3}[/tex]
where [tex]r_{H}^{3}[/tex] = 37 pm which is read as 37 picometer (1 picometer = 10^-12 m) = 37 × [tex]10^{-12}[/tex] meters
Volume of the hydrogen = 8/3 × (37 × 10^-12)^3 = 4.05 * 10^-31
we multiply this by the avogadro's number = 6.02 * 10^23
= 4.05 * 10^-31 * 6.02 * 10^23 = 2.6 * 10^-8 m^3
We do same for thr oxygen, but this time we do not multiply the volume of the oxygen by 2 as we have only one atom of oxygen
Volume of oxygen = 4/3 * π * (73 * 10^-12) ^3 * avogadro's number = 9.81 * 10^-7 m^3
adding both volumes together, we have 1.24 * 10^-6 m^3 or simply 1.24 ml ( 0.01 m = 1 ml)
Dividing the molar mass of one mole of water by its density, we can get the volume of 1 mole of water
= (18g/mol)/(1 g/ml) = 18 ml/mol
Now we proceed to calculate the volume of void = Total volume - volume of molecule = 18 - 1.24 = 16.76 ml
Now, the percentage of void = volume of void/total volume * 100%
= 16.76/18 * 100% = 93.11%
Experiment predicted observation A student has two unopened cans containing carbonated water. Can A has been stored in the garage () and can B has been stored in the fridge (). The student opens one can at the time, both cans make a fizz.
A) The fizz will be the same for both cans
B) There is not enough information to predict which can will make the louder fizz
C) Can A will make a louder and stronger fizz than can B.
D) Can B will make a louder and stronger fizz than can A.
Answer:
Can A will make a louder and stronger fizz than can B.
Explanation:
Temperature has a direct effect on gas solubility. We know that carbonated water contains carbon dioxide dissolved in water. The extent of dissolution or solubility of this gas is dependent on the temperature of the system.
As the temperature of the system rises, the solubility of gas in solution decreases. It follows that can A, having been stored in a garage is definitely at a higher temperature than can B stored in the refrigerator.
Since solubility of gases decreases with increasing temperature, the carbon dioxide in can A will be less soluble than in can B. This will cause can A to make a louder and stronger fizz when opened than can B.
Given the density of iron (Fe) is 7.87 g/cm3, determine the mass of iron (in grams) in a rectangle block with the dimensions of 12.5 in long, 3.50 in wide, and 2.50 in high. (1in = 2.54 cm).
Answer:
[tex]m=14,105.71 g Fe[/tex]
Explanation:
Hello,
In this case, the first step is to compute the volume of the block considering the length, height and width:
[tex]V=L \times W \times H =12.5 in\times 3.50 in \times 2.50 in =109.375 in^3[/tex]
Then, we compute the volume in cubic centimetres:
[tex]V=109.375in^3\times \frac{16.3871 cm^3}{1in^3} =1792.34cm^3[/tex]
Finally, as the density is given by:
[tex]\rho =\frac{m}{V}[/tex]
We solve for the mass:
[tex]m=\rho \times V= 7.87\frac{g}{cm^3} \times 1792.34 cm^3\\\\m=14,105.71 g Fe[/tex]
Best regards.
Which statement BEST describes how a golf club does "work" on a golf ball?
(A) When the club hits the ball the club transfers all of its kinetic energy to the ball.
(B) All of the kinetic energy from the club is transferred to the ball as they both move through the air.
(C)
Some of the kinetic energy from the golf club is transferred to the ball and some transforms into sound
and heat, but the total energy remains the same.
(D) The golf club loses kinetic energy when it hits the ball and the ball gains kinetic energy from the air as it
travels
Answer:
C
Explanation:
It looks pretty reasonable to me
Which of the following provides evidence to support Thomson's hypothesis about electrons??
A.) Gold foil experiment
B.) Cathode ray experiments
C.)Spectrum of colors emitted by gas
D.) Radiation produced when beryllium is bombarded with alpha particles
Cathode ray experiments of the following provides evidence to support Thomson's hypothesis about electrons.
What is the hypothesis of Thomson's atomic model?Rutherford's gold leaf experiment demonstrated that the atom is essentially empty space with such a tiny, compact, positively-charged nucleus. Thomson had proposed the plum pie model of the atom, which featured negatively-charged electrons buried within a favorably "soup." Since most of the alpha particles flow through an atom directly without being deflected, contrary to what Thomson's model predicted, the majority of a space inside of an atom is empty. As a result, the Thomson model of a molecule was disproved.
Who disproved Thomson's theory?According to Thomson's model, every atom is made up of negative charges "plums" surrounded in positively charged "pudding," or electrons with a soup of positive ion to balance their negative charges. Hans Geiger and Arthur Marsden's 1909 gold foil test refuted the 1904 Thomson model.
To know more about Thomson hypothesis visit:
https://brainly.com/question/28824828
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An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.35×10-2 M CH2Cl2, 0.173 M CH4 and 0.173 M CCl4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.155 mol of CH4(g) is added to the flask?
Answer:
[CH₂Cl₂] = 7.07x10⁻² M
[CH₄] = 0.319 M
[CCl₄] = 0.164 M
Explanation:
The equilibrium reaction is the following:
2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)
The equilibrium constant of the above reaction is:
[tex] K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{0.173 M*0.173 M}{(5.35 \cdot 10^{-2} M)^{2}} = 10.5 [/tex]
When 0.155 mol of CH₄(g) is added to the flask we have the following concentration of CH₄:
[tex] C = \frac{\eta}{V} = \frac{0.155 mol}{1.00 L} = 0.155 M [/tex]
[tex]C_{CH_{4}} = 0.328 M[/tex]
Now, the concentrations at the equilibrium are:
2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)
5.35x10⁻² - 2x 0.328 + x 0.173 + x
[tex]K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{(0.328 + x)(0.173 + x)}{(5.35 \cdot 10^{-2} - 2x)^{2}}[/tex]
[tex]10.5*(5.35 \cdot 10^{-2} - 2x)^{2} - (0.328 + x)*(0.173 + x) = 0[/tex]
Solving the above equation for x:
x₁ = 0.076 and x₂ = -0.0086
Hence, the concentration of the three gases once equilibrium has been reestablished is:
[CH₂Cl₂] = 5.35x10⁻² - 2(-0.0086) = 7.07x10⁻² M
[CH₄] = 0.328 + (-0.0086) = 0.319 M
[CCl₄] = 0.173 + (-0.0086) = 0.164 M
We took x₂ value because the x₁ value gives a negative CH₂Cl₂ concentration.
I hope it helps you!
How many kg of gas fill a 11.6 gal gas tank
Answer:
43.964
Explanation:
i think i used a calculator so let me know if its wrong
Answer:
39.49 kg
Explanation:
:)
