Answer:
b. the heating of 10 g of liquid water from 0°C to 100°C.
Explanation:
Hello,
In this case, we must notice a., c. and d. processes are not actually absorbing heat but releasing it since cooling, freezing and condensation are processes with negative heat sign since matter changes from a state of more energy to a state of less energy. We can prove this by realizing that freezing enthalpy of water is -6.00 kJ/mol, condensation enthalpy of eater is -40.8 kJ/mol and a change of temperature from 100 °C to 0 °C is negative.
In such a way, the only process absorbing heat is b. the heating of 10 g of liquid water from 0°C to 100°C since energy must be added to the system, or absorbed by it in order to attain the heating.
Regards.
The process having the greatest amount of heat is:
b. the heating of 10 g of liquid water from 0°C to 100°C.
Looking at all the options:The options a., c. and d. processes are not actually absorbing heat but releasing it since cooling, freezing and condensation are processes with negative heat sign since matter changes from a state of more energy to a state of less energy.
The freezing enthalpy of water is -6.00 kJ/mol, condensation enthalpy of eater is -40.8 kJ/mol and a change of temperature from 100 °C to 0 °C is negative.
So out of all the options, only process at b is a heating process thus it will absorb greatest amount of heat.
Find more information about Heat here:
brainly.com/question/13439286
A student has an unknown sample of solution X. This solution is placed in a 1.00 cm wide cuvet and inserted into the spectrometer, producing an absorbance reading of 0.275 at a wavelength of 525.0 nm. What is the concentration of solution X in the unknown sample
Answer:
The concentration of the sample is 3.564x10⁻³M.
Explanation:
Using Lambert-Beer law, absorbance of a sample is directely proportional to its concentration.
The general graph of the absorbance of the standards with different concentrations is:
Y = 75.9X + 0.0045
R² = 0.9946
Where Y is the absorbance of the sample and X its concentration in mole/L.
If a solution has an absorbance of 0.275:
0.275 = 75.9X + 0.0045
0.2705 = 75.9X
3.564x10⁻³M = X → The concentration of the sample.
Which compound has the lowest melting point? KCl CaCl2 Na2O C6H12O6
A certain substance X condenses at a temperature of 120.7 degree C. But if a 500, g sample of X is prepared with 55.4 g of urea (NH_2)_2 CO) dissolved in it, the sample is found to have a condensation point of 125.2 degree C instead. Calculate the molal boiling point elevation constant K_b of X. Round your answer to 2 significant digits.
Answer: The molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]
Explanation:
Formula used for Elevation in boiling point :
[tex]\Delta T_b=k_b\times m[/tex]
or,
[tex]T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]
where,
[tex]T_b-T^o_b =(125.2-120.7)^0C=4.5^0C[/tex]
[tex]k_b[/tex] = boiling point constant = ?
m = molality
[tex]w_2[/tex] = mass of solute (urea) = 55.4 g
[tex]w_1[/tex] = mass of solvent X = 500 g
[tex]M_2[/tex] = molar mass of solute (urea) = 60 g/mol
Now put all the given values in the above formula, we get:
[tex]4.5^oC=k_b\times \frac{55.4g\times 1000}{60\times 500g}[/tex]
[tex]k_b=2.4^0C/m[/tex]
Thus the molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]
At 298K, the equilibrium constant for the following reaction is 4.20×10-7: H2CO3(aq) + H2O H3O+(aq) + HCO3-(aq) The equilibrium constant for a second reaction is 4.80×10-11: HCO3-(aq) + H2O H3O+(aq) + CO32-(aq) Use this information to determine the equilibrium constant for the reaction: H2CO3(aq) + 2H2O 2H3O+(aq) + CO32-(aq)
Answer:
The correct answer is 2.016 x 10⁻¹⁷
Explanation:
We have the following chemical reactions and their equilibrium constants (K):
(1) H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + HCO₃⁻(aq) K₁= 4.20×10⁻⁷
(2) HCO₃⁻(aq) + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq) K₂= 4.80×10⁻¹¹
And we have to obtain K for the following reaction:
H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq)
If we add equations (1) and (2) we obtain the the desired equation. Remember that when we add chemical equations, the global equilibrium constant is the product of the constants.
H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + HCO₃⁻(aq) K₁= 4.20×10⁻⁷
+
HCO₃⁻(aq) + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq) K₂= 4.80×10⁻¹¹
-------------------------------------------------------------
H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq) K= K₁ x K₂
K = K₁ x K₂ = (4.20×10⁻⁷) x (4.80×10⁻¹¹) = 2.016 x 10⁻¹⁷
need help and quick answer as fast as possible
Which metal can replace another metal in a reaction
Answer:
The products of the reaction are aqueous magnesium nitrate and solid copper metal. This subcategory of single-replacement reactions is called a metal replacement reaction because it is a metal that is being replaced (zinc)
Explanation:
The products of the reaction are aqueous magnesium nitrate and solid copper metal. This subcategory of single-replacement reactions is called a metal replacement reaction because it is a metal that is being replaced (zinc)
Covalent bonds can be best described as
Answer:
neutral atoms coming together to share electrons
Answer:
a
Explanation:
neutral atoms coming together to share electrons
Three different students determined the density of a metal object. Here are their results: 15.12 g/mL, 15.09 g/mL, and 15.12 g/mL. The actual density of the object was 14.41 g/mL. Calculate the percent error. Make sure to include units with your answer, units are %.
Answer:
The correct answers are 4.93 %, 4.72 % and 4.93 %.
Explanation:
Based on the given question, 14.41 g per ml is the actual density of the object. However, the density determined by three different students of the object is 15.12 g per ml, 15.09 g per ml, and 15.12 g per ml. The percent error can be calculated by using the formula,
% error = (actual value - calculated value) / actual value * 100
By 1st student, the calculated value is 15.12 g per ml, the percent error will be,
% error = (14.41 - 15.12) / 14.41 * 100
= 0.71/14.41 * 100
= 4.93 %
By 2nd student, the calculated value is 15.09 g per ml, the percent error will be,
% error = (14.41-15.09)/14.41 * 100
= 0.68/14.41 * 100
= 4.72 %
By 3rd student, the calculated value is 15.12 g per ml, the percent error will be,
% error = (14.41-15.12)/14.41 * 100
= 0.71/14.41 * 100
= 4.93 %
12.39 g sample of phosphorus (30.97 g/mol) reacts with 52.54 g of chlorine gas, Cl2
(70.91 g/mol) to form only phosphorus trichloride, PC13 (137.33 g/mol). Which is the
limiting reactant?
Answer:
P is the limiting reagent
Explanation:
P = phosphorus = 30.97g/mol
Cl2 = Chlorine = 70.91g/mol
PCl3 = Phosphorus Trichloride = 137.33g/mol
P + Cl2 = PCl3
Left Side
P = 1
Cl = 2
Right Side
P = 1
Cl = 3
So equation needs to be balanced first
2P + 3Cl = 2PCl3
Left Side
P = 2
Cl = 6
Right Side
P = 2
Cl = 6
That's better.
Ok so we have 12.39g of P so we have 0.4 moles of it
We then have 52.54g of Cl so we have 0.74 moles of it
For every P we need 1.5 Cl so we have an excess of Cl
What is the number of valence electrons in a nitrogen atom in the ground state
Answer: 5
Explanation:
It just is
Answer:
5
Explanation:
Bc valence electron means last # in the electron configuration
Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 3.4 g of octane is mixed with 15.6 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.
Answer:
10 g of CO2
Explanation:
Equation of the reaction:
CH3(CH2)6CH3 + 17O2 ----> 18H2O + 8CO2
Fom the above balanced equation,
1 mole of Octane gas reacts with 17 moles of oxygen gas to produce 8 moles of CO2
Molar mass of Octane = 114 g/mol
Molar mass of oxygen gas = 32 g/mol
Molar mass of CO2 = 44 g/mol
Therefore, 114 g of Octane reacts completely with 17 * 32g (= 544 g) of oxygen to produce 8 * 44 g(=352g) of CO2.
From the given mass of reactants;
3.4 g of Octane will react with (544 * 3.4)/114 g of oxygen = 16.22g of oxygen.
Therefore oxygen is the limiting reactant.
15.6 g of oxygen will react with (114 * 15.6)/544 g of CO2 = 3.27 g of octane.
Mass of CO2 produced will be
(352 * 15.6)/544 = 10 g of CO2
If a jet’s cruising altitude is 32,200ft(to three significant figures),the distance in km is :(1 mile=1.61km;1 mile=5280 ft)
Answer:
9.82 km.
Explanation:
Hello,
In this case, given the conversion factors from miles to kilometres and from miles to feet, we can directly compute the jet’s cruising altitude in kilometres as shown below:
[tex]32,200ft\times \frac{1mile}{5280ft}\times \frac{1.61km}{1mile} \\\\=9.82km[/tex]
Best regards.
Show work plzzz
Unknown Metal Bar #8
Mass of Unknown Metal bar 11.3g
Length of bar 13.90cm
Width of bar 2.9cm
Thickness of bar 0.081cm
1. Calculate the volume of the bar:
2. Calculate the (experimental) density of the bar:
3. Based on the provided list of (true) densities, what is the possible identity of the Unknown metal?
4. What is the percent difference between the true density of your metal and the calculated density?
= | − | ∗ 100%
Answer:
1= Volume
= Length x breath x height
= 13.90 x 2.9 x 0.081
=3.26511
2= Density = Mass ÷ volume
= 11.3 ÷ 3.26511
= 3.461 (3d.p)
idk the rest because you haven't shown a picture of the rest
Answer:
1. 3.3 cm³; 2. 3.5 g/cm³; 3. barium; 4. 4%
Explanation:
Experimental data:
Mass = 11.3 g
Length = 13.90 cm
Width = 2.9 cm
Thickness = 0.081 cm
Calculations:
1. Volume of bar
V = lwh = 13.90 cm × 2.9 cm × 0.081 cm = 3.3 cm³
2. Experimental density
[tex]\text{Density} = \dfrac{m}{V} = \dfrac{\text{11.3 g}}{\text{3.27 cm}^{3}} = \textbf{3.5 cm}^{\mathbf{3}}[/tex]
3. Identity of metal
The three most likely metals are scandium (3.00 g/cm³), barium (3.59 g/cm³), and yttrium (4.47 g/cm³)
The metal is probably barium.
4. Percent difference
[tex]\begin{array}{rcl}\text{Percent difference}&= &\dfrac{\lvert \text{ True - Calculated}\lvert}{ \text{True}} \times 100 \,\%\\\\& = & \dfrac{\lvert 3.59 - 3.5\lvert}{3.59} \times 100 \, \% \\\\& = & \dfrac{\lvert 0.1\lvert}{3.59} \times 100 \, \%\\ \\& = & 0.04 \times 100 \, \%\\& = & \mathbf{4 \, \%}\\\end{array}\\\text{The percent difference is $\large \boxed{\mathbf{4 \, \%} }$}[/tex]
Take a series of observations to determine if process is spontaneous. Based upon those observations, you will create an activity series, listing the metals in order of their reactivity. Second, you will construct a series of virtual galvanic cells and use those to power a stopwatch. Third, you will determine the standard reduction potential of an unknown metal; comparing its reduction potential to a standard list, you will identify the unknown. Finally, you will create a situation in which the cells are not in the standard condition and measure the cell potential; using the Nernst equation, you will determine the concentration of an unknown solution
Answer the below questions for the portion of the activity in which Sn(s) is placed in AgNO3(aq)
1. Is there a reaction? (circle the correct response) Yes / No
2. How many electrons are transferred 4 electrons
3. Write the balanced redox reaction for the combination of AgNO3(aq) and Sn(s)Sn(s) + Ag+(aq) Sn2+(aq) + Ag(s)
Answer:
Explanation:
2AgNO₃ + Sn ⇄ Sn( NO₃)₂ + 2Ag
Ag⁺/Ag = .80 V
Sn⁺²/Sn = - .14 V
Hence Ag will be reduced and Sn will be oxidised . Hence the reaction will take place . YES .
2 ) 2 electrons are transferred .
3 )
2Ag⁺ + 2e = 2Ag
Sn = Sn⁺² + 2e
---------------------------
2Ag⁺ + Sn = Sn⁺² + 2Ag .
One method of experimentally determining whether a species is paramagnetic is to weigh it in an instrument called a magnetic susceptibility balance. This is a balance with a strong electromagnet placed next to the sample holder. If the species is paramagnetic, the mass reading of the balance will increase when the field is switched on.Classify these species as paramagnetic or diamagneticWhich species will have the strongest mass shift on a magnetic susceptibility balance?
Answer:
Diamagnetic have paired electrons while paramagnetic have at least one unpaired electron.
Explanation:
F2, C2 and N2 are diamagnetic while O2 and B2 are paramagnetic. Diamagnetic are those atoms which have paired electrons while paramagnetic are those atoms which contain at least one unpaired electron so we can say that F2, C2 and N2 have paired electrons while O2 and B2 have unpaired electrons. When diamagnetic materials are allowed to contact with external magnetic field so they will be repelled while paramagnetic materials are attracted by magnetic field due to the presence of unpaired electrons.
Iron oxide (FeO) is the strongest paramagnetic material having the value of 720.
When a sample of Mg(s) reacts completely with O2(g), the Mg(s) loses 5.0 moles of electrons. How many moles of electrons are gained by the O2(g)? *
Answer:
if magnesium looses five moles of electrons, oxygen will also gain five moles of electrons.!
Explanation:
Oxidation refers to the loss of electrons. Any specie that looses electrons in a redox reaction is said to be the reducing agent. Hence the reducing agent participates in the oxidation half equation. In this case, magnesium is the reducing agent.
Reduction has to do with the gain of electrons. The oxidizing agent participates in the reduction half equation. Hence the oxidizing agent is reduced in the redid reaction. The reducing agent in this case is the oxygen molecule.
Oxidation half equation;
Mg(s)-----> Mg^2+(aq) + 2e
Reduction half equation;
O2(g) + 2e ------> 2O^2-(aq)
From the balanced reaction equation, two moles of electrons is transferred.
Hence if magnesium looses five moles of electrons, oxygen will also gain five moles of electrons.
If a gas occupies 12.60 liters at a pressure of 1.50 atm, what will its pressure at a volume of 2.50 liters?
Answer:
7.56 atm
Explanation:
Boyle's law states that the pressure and volume of a gas are proportional to each other
The formular for Boyle's law is
P1V1=P2V2
According to the question above, the values given are
P1=1.50 atm
P2= ?
V1=12.60 litres
V2= 2.50 litres
Let us make P2 the subject of formular
P2= P1V1/V2
P2= 1.50×12.60/2.50
P2= 18.9/2.50
P2= 7.56 atm
Hence when the volume of a gas is 2.50 litres then it's pressure is 7.56 atm
Discuss any give ways by which
the falling moral standards of Ghanaian
youth can be minimised.
Answer:
The falling standards of Ghanaian youths can be minimized by proper upbringing of the children by their parents. The youths should be taught about what is wrong or right and there should be a corresponding reward for those who do good and exceptional in order to encourage others in towing that line and punishment should also be meted out to those who break the law. Mediocrity shouldn’t be celebrated and the elders should lead by example.
These will make the falling standards of Ghanaian youth get reduced.
5. Rosalind Franklin was a key figure in the discovery of the structure of DNA, yet she
was not included in the Nobel Prize which was awarded to Watson and Crick. Carry out
some research to find out how she contributed to this work and use the space below
to write up your findings
Answer:
Explanation:
Search for "Rosalind Franklin: DNA's unsung hero - Cláudio L. Guerra" which basically summarizes what Rosalind did and how we was snubbed from receiving the noble prize even though she had vast and critical evidence to highlight the structure of DNA. You can look for more sources but I can tell you a quick recap:
Rosalind Franklin was born in an era where women scientists or workers were very uncommon and they were even discriminated and looked down upon. After her phD., she was working to find the structure of DNA and soon she was able to form an x-ray image of it. However, her lab colleague took the picture and showed it to other scientists (Watson and Crick) without the knowledge or permission of Rosalind. Here Rosalind was working on analyzing her data and on other part of world Watson and Crick were doing the same. Based on Watson and Crick's analysis, they came up with the correct structure of DNA and soon Rosalind got done as well. Both submitted their paper to journal, however, the journal placed Watson and Crick paper before Rosalind (making it look like Rosalind just confirmed what Watson and Crick proposed). This made it look like Watson and Crick were geniuses behind DNA structure whereas, in reality, it was Rosalind. She would have received Nobel Prize but she died of Cancer and Nobel prizes are not awarded to dead people.
Click on the Delta H changes sign whan a process is reversed button within the activity and analyze the relationship between the two reactions that are displayed. The reaction that was on the screen when you started and its derivative demonstrate that the reaction enthalpy, ΔH, changes sign when a process is reversed. Consider the reaction H2O(l)→H2O(g), ΔH =44.0kJ What will ΔH be for the reaction if it is reversed?
Answer:
ΔH = - 44.0kJ
Explanation:
H2O(l)→H2O(g), ΔH =44.0kJ
In the reaction above, liquid water changes to gaseous water. This occurs through a process known as boiling. This process requires heat, hence the ΔH is positive.
If he reaction is reversed, we have;
H2O(g)→H2O(l)
In this reaction, gaseous water changes to liquid water. This process is known as condensation. The water vapor loses heat in this reaction. Hence ΔH would be negative but still have the same value.
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.
The breaking buffer that we use this week contains 10mM Tris, pH 8.0, 150mM NaCl. The elution buffer is breaking buffer that also contains 300mM imidazole. Describe how the instructor made the 0.25L elution buffer for all the students this week given 500ml of 1M of Tris (121.1 g/mole) (pH8.0), 750ml of 5M NaCl (MW
Answer:
Explanation:
From the given information ;the objective is to determine how the instructor made the 0.25L elution buffer
0.25 L elution buffer = 250 mL elution butter
The breaking buffer that we use this week contains
10mM Tris = 0.01 M
150mM NaCl = 0.15 M
300mM imidazole. = 0.3 M
The stock concentration of Tris in 1M
Therefore ; by using the formula: [tex]M_1V_1 = M_2 V_2[/tex]; we can determine the volume in the preparation; so;
[tex]1*V_1 = 0.0 1 \ M * 250 \ mL[/tex]
[tex]V_1 = \dfrac{0.0 1 \ M * 250 \ mL}{1 }[/tex]
[tex]V_1 = 2.5 \ mL[/tex]
In NaCl, The amount of stock concentration is 5 M
so; using the same formula; we have:
[tex]5*V_1 = 0.15 \ M * 250 \ mL[/tex]
[tex]V_1 = \dfrac{0. 15 \ M * 250 \ mL}{5 }[/tex]
[tex]V_1 = 7.5 \ mL[/tex]
From Imidazole ; the amount of stock concentration is
[tex]1*V_1 = 0.3 \ M * 250 \ mL[/tex]
[tex]V_1 = \dfrac{0. 3 \ M * 250 \ mL}{1 }[/tex]
[tex]V_1 = 75 \ mL[/tex]
Thus; we can have a table as shown as :
Stock concentration volume to be added Final concentration
1 M of Tris 2.5 mL 10 mM
5 M of NaCl 7.5 mL 150 mM
1 M of Imidazole 75 mL 300 mM
In conclusion. the addition of all the volume make up the 250 mL elution buffer that is equivalent to 0.25 L.
3 Pieces of Key Information elements, compounds, mixtures also state a real world example of elements, compounds, mixtures and lastly why is elements, compounds, mixtures are important
Answer:
Element and compounds are the pure substances but mixture is not a pure substance.
Explanation:
Element and compounds are the pure substances in which element comprise of only one atom while compound is formed by the chemical combination of more than one element in a fixed ratio by mass while mixture is also made up of more that one substances but they are combine physically not chemically. Elements, compounds, mixtures are very important because all the materials we used in our daily life are formed from elements, compounds and mixtures.
What is the limiting reactant if 8 g of Ba reacts with 2.8 g of Al2(SO4)3?
Answer:
Al2(SO4)3 is the limiting reactant
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
3Ba + Al2(SO4)3 → 2Al + 3BaSO4
Next, we shall determine the mass of Ba and the mass of Al2(SO4)3 that reacted from the balanced equation. This is illustrated below:
Molar mass of Ba = 137g/mol
Mass of Ba from the balanced equation = 3 x 137 = 411g
Molar mass of Al2(SO4)3 = 2x27 + 3[32 + (16x4)]
= 54 + 3[32 + 64]
= 54 + 3[96]
= 54 + 288 = 342g/mol
Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g
Summary:
From the balanced equation above,
411g of Ba reacted with 342g of Al2(SO4)3.
Finally, we shall determine the limiting reactant as follow:
From the balanced equation above,
411g of Ba reacted with 342g of Al2(SO4)3.
Therefore, 8g of Ba will react with
= (8 x 342/411 = 6.66g of Al2(SO4)3.
From the calculations made above, we can see that it will take a higher mass of Al2(SO4)3 i.e 6.66g than what was given i.e 2.8g to react completely with 8g of Ba.
Therefore, Al2(SO4)3 is the limiting reactant and Ba is the excess reactant.
Compound H is optically active and has the molecular formula C6H10 and has a five carbon ring. On catalytic hydrogenation, H is converted to I (C6H12) and I is optically inactive. Propose structures for H and I. (Draw a three-dimensional formula for each using dashes and wedges around chiral centers.)
Answer:
Explanation:
Given that ;
Compound H is optically active and have a molecular formula of C6H10 and therefore undergo catalytic hydrogenation. Catalytic hydrogenation involves the use Platinum/Nickel to produce C6H12
i.e
[tex]C_6H_{10} +H_2 \to ^{Pt/Ni} \ \ \ C_6H_{12}[/tex]
The proposed H and I structures are shown in the diagrams attached below .
compound H represents 3- methyl cyclopentene
compound I represents methyl cyclopentane
However; 3- methyl cyclopentene posses just only one chiral carbon which is optically active at the third position and it R and S enantiomers are shown in the second diagram below.
The starting material is 3-methylcyclopentene while the optically inactive product is 1-methyl cyclopentane.
Hydrogenation refers to the addition of hydrogen across the double bond of an unsaturated compound. Hydrogenation results in the formation of a saturated compound having two more hydrogen atoms than the starting material.
The starting material is optically active 3-methylcyclopentene. The R and S enantiomers of the starting material is shown in image (1) attached. The optically inactive product is, 1-methyl cyclopentane is shown in image (2) attached.
Learn more: https://brainly.com/question/6249935
In E. coli, the enzyme hexokinase catalyzes the reaction: Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102. In the living E. coli cells, [ATP] = 7.9 mM; [ADP] = 1.04 mM, [glucose] = 2 mM, [glucose 6-phosphate] = 1 mM. Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.
Answer:
Explanation:
Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.
In the living E. coli cells,
[ATP] = 7.9 mM;
[ADP] = 1.04 mM,
[glucose] = 2 mM,
[glucose 6-phosphate] = 1 mM.
Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.
The reaction is given as
Glucose + ATP → glucose 6-phosphate + ADP
Now reaction quotient for given equation above is
[tex]q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}[/tex]
[tex]q=\frac{(1mm)\times (1.04 mm)}{(7.9mm)\times (2mm)} \\\\=6.582\times 10^{-2}[/tex]
so,
[tex]q<<K_e_q[/tex] ⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable until q = Keq
Give the IUPAC name for the following structure
Answer:
6-metyl-2-heptyne
Explanation:
C-C-C-C-C-C-C hept
2
C-C≡C-C-C-C-C 2-heptyne
C
| 6
C-C≡C-C-C-C-C
6-metyl-2-heptyne
The IUPAC name for the above structure is 6 methyl, hept-2-yne.
What is IUPAC?IUPAC stands for international Union of pure and applied chemistry. It is the body in charge of naming organic chemical compounds.
The naming is is based on a molecule's longest chain of carbons connected by single/double/triple bonds, whether in a continuous chain or in a ring etc.
According to this question, a structure is given. The following applies;
The compound has a triple bond located on the second carbon, hence, belongs to alkyne group. It has seven carbon atoms, hence, is heptyne. The methyl group is on the sixth carbon.Learn more about IUPAC at: https://brainly.com/question/33646537
#SPJ6
Classify the following unbalanced chemical reaction Na3PO4(aq) + FecCl2(aq) = Fe3(PO4)2(s) + NaCl(aq)
1. Acid-Base Reaction
2. Precipitation Reaction
3. Oxidation-Reduction Reaction
4. Combustion Reaction
Answer:
2. Precipitation Reaction
Explanation:
Na3PO4(aq) + FecCl2(aq) = Fe3(PO4)2(s) + NaCl(aq)
Fe3(PO4)2(s) - solid, it means it will precipitate.
It is a precipitation reaction.
What is precipitate give example?A precipitate is a solid that forms out of a solution. A common example is that of the mixing of two clear solutions: (1) silver nitrate (AgNO3) and (2) sodium chloride (NaCl): The reaction is. The precipitate forms because the solid (AgCl) is insoluble in water.
What is precipitate formation?A precipitate is a solid formed in a chemical reaction that is different from either of the reactants. This can occur when solutions containing ionic compounds are mixed and an insoluble product is formed. The identity of the precipitate can often be determined by examining solubility rules.
Learn more about precipitation reaction here https://brainly.com/question/12144162
#SPJ2
10. How many grams of NH, are present in 6 moles
of NH,?
Answer:
90.08784 grams
Explanation:
idk
Classify the following unbalanced chemical reaction Fe(s) + Cl2(aq) = Fe2+(aq) + Cl-(aq)
1. Acid-Base Reaction
2. Precipitation Reaction
3. Oxidation-Reduction Reaction
4. Combustion Reaction
Answer:
3. Oxidation-Reduction Reaction
Explanation:
Fe(s) + Cl2(aq) = Fe2+(aq) + Cl-(aq)
Fe(s) -2e- ----> Fe2+(aq) oxidation
Cl2(aq) + 2e- -----> 2Cl-(aq) reduction
The given unbalanced chemical reaction is the oxidation-reduction reaction. Therefore, option (3) is correct.
What is an oxidation-reduction reaction?Redox reactions can be defined as oxidation-reduction chemical reactions in which the reactants of the reaction undergo a change in their oxidation states. All the redox reactions are further broken down into two different processes: a reduction process and an oxidation process.
The oxidation and reduction reactions take place simultaneously in an Oxidation-Reduction reaction. The substance that is getting reduced in a reaction is known as the oxidizing agent, while a substance that is getting oxidized is the reducing agent.
The given chemical reaction is:
[tex]Fe(s) + Cl_2(aq) \longrightarrow Fe^{2+}(aq) + Cl^-(aq)[/tex]
The oxidation reaction for this reaction is: Fe (s) → Fe²⁺ (aq) + 2e⁻
The reduction reaction: Cl₂ (g) + 2e⁻ → 2Cl⁻ (aq)
Therefore, the given reaction between the iron and chlorine gas is the oxidation-reduction reaction or redox reaction.
Learn more about oxidation-reduction reactions, here:
https://brainly.com/question/16867778
#SPJ2
