Rock candy is formed through a chemical process known as crystallization which requires a supersaturated sugar solution. In the scenario, batch A prepared with hot water is more likely to form rock candy as it can dissolve more sugar creating a supersaturated solution. Batch B prepared at room temperature may not form as much rock candy due to lesser sugar dissolution.
Explanation:The question relates to the chemical process of crystallization, particularly in the formation of rock candy. When making rock candy, a supersaturated solution of sugar and water is required. This condition is achieved when as much sugar as possible is dissolved in hot water. Once cooled, the oversaturated solution starts to crystallize and forms rock candy. So in the given scenario, batch A is more likely to produce rock candy because it involves the preparation of a supersaturated solution through dissolving sugar in hot water. Batch B, prepared at room temperature, may not dissolve as much sugar as batch A, and thus, less or no rock candy might be formed.
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Batch A, which dissolved sugar in hot water that was then cooled, is more likely to form rock candy as the cooling process can lead to the crystallization of sugar from a supersaturated solution.
The process of making rock candy involves dissolving sugar in water to create a saturated solution from which sugar crystals can form upon cooling or evaporation of the solvent. The solubility of sugar increases with temperature, which means hot water can dissolve more sugar than room temperature water. Therefore, for batch A, dissolving sugar in hot water likely supersaturates the solution, and as it cools to room temperature, excess sugar will crystallize out.
Batch B, on the other hand, dissolves sugar at room temperature, potentially creating a saturated solution, but without the temperature change, it is less likely to form a supersaturated environment and thus may yield less crystallization compared to batch A. In summary, batch A is more likely to form rock candy due to the temperature-dependent solubility of sugar, and the cooling process allows crystals to form from the supersaturated solution.
What is the Kelvin temperature of the air in a tire if the pressure inside the tire is 188 kPa at 32°C. While driving under perilous road conditions on a hot road; the pressure in the tire has increased to 225 kPa.
Answer:
414KExplanation:
The question is not completely clear because some missing parts or grammar (syntax) errors.
Interpreting it, the question is what is the Kelvin temperatue of the air in a tire, when the pressure in the tire has increased to 225 kPa, if the initial conditions of the air inside the tire were 188kPa of pressure and a temperature of 32ºC.
To solve this, you can assume constant volume and use the law of Gay-Lussac for ideal gases:
[tex]\dfrac{P_1}{T_{1}}=\dfrac{P_2}{T_2}[/tex]
That equation works with absolute temperatures, i.e. Kelvin.
32ºC = 32 + 273.15 K = 305.15KThen solve for T₂, substitute and compute:
[tex]T_2=P_2\times \dfrac{T_1}{P_{1}}=225kPa\times \dfrac{305.15K}{188kPa}[/tex]
[tex]T_2=413.90K\approx 414K\longleftarrow answer[/tex]
Consider the three ligand field spectra corresponding to octahedral complexes A, B, and C, all formed from the same metal ion.
From the following list, Ti3 , Ni2 , Pt4 , Cu2 , to which metal ions could the spectra correspond and to which would it be very unlikely
Answer:
Ni^2+ is most likely
Ti^3+ is very unlikely
Explanation:
The Crystal Field Stabilization Energy almost always favors octahedral over tetrahedral in very many cases, but the degree of this favorability varies with the electronic configuration. In other words, for d1 there is only a small gap between the octahedral and tetrahedral lines, whereas at d3 and d8 is a very big gap. However, for d0, d5 high spin and d10, there is no crystal field stabilization energy difference between octahedral and tetrahedral. The ordering of favorability of octahedral over tetrahedral is:
d3, d8 > d4, d9> d2, d7 > d1, d6 > d0, d5, d10. This explains the answer choices above.
Ti^3+ being a d1 specie is least likely to exist in octahedral shape while Ni2+ a d8 specie is more likely to exist in octahedral shape.
Be sure to answer all parts. Use MO diagrams to place C2−, C2, C2+ in order of the following properties: (a) increasing bond energy C2− < C2 < C2+ C2− < C2+ < C2 C2 < C2− < C2+ C2 < C2+ < C2− C2+ < C2 < C2− C2+ < C2− < C2 (b) increasing bond length C2− < C2 < C2+ C2− < C2+ < C2 C2 < C2− < C2+ C2 < C2+ < C2− C2+ < C2 < C2− C2+ < C2− < C2
Answer:
Bond Order = ½ [# bonded e--# antibonded e-]
C2= ½ [6-2] = 2
C2+= ½ [5-2] = 1.5
C2- = ½ [7-2] = 2.5
Bond Length: C2+> C2> C2-
Bond Energy: C2-> C2> C2+
Explanation:
Bond order is a measurement of the number of electrons involved in bonds between two atoms in a molecule.
How to calculate bond order
1. Draw the Lewis structure.
2. Count the total number of bonds.
3. Count the number of bond groups between individual atoms.
4. Divide the number of bonds between atoms by the total number of bond groups in the molecule.
Bond energy (E) is defined as the amount of energy required to break apart a mole of molecules into its component atoms. It is a measure of the strength of a chemical bond.
The higher the bond order, the shorter the bond length and the greater the bond energy.
The molecular orbital configuration of a chemical specie shows the arrangement of electrons in such a specie into molecular orbitals in accordance with the Aufbau principle.
For C2;
σ1s2 σ*1s2 σ2s2 σ*1s2 π2py2 π2pz2
Bond order = 1/2(8 - 4) = 2
For C2^-;
σ1s2 σ*1s2 σ2s2 σ*1s2 π2py2 π2pz2 σ2px1
Bond order = 1/2(9 - 4) = 2.5
For C2^+;
σ1s2 σ*1s2 σ2s2 σ*1s2 π2py2 π2pz1
Bond order = 1/2(7 - 4) = 1.5
We must recall that the higher the bond order, the shorter the bond length and the greater the bond energy. Hence;
a) In order of increasing bond energy; C2^+ < C2 < C2^-
b) In order of increasing bond length; C2^- < C2 < C2^+
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Balance the redox reaction by inserting the appropriate coefficients.
H2O + Br- + Al3- = Al + BrO3- + H+
The balanced equation is given as,
2Al³⁺ + 3H₂O + Br⁻ → 2 Al + BrO³⁻ + 6H⁺
Explanation:
H₂O + Br⁻ + Al³⁺ → Al + BrO₃⁻ + H⁺
Here the half reactions are:
Al³⁺→ Al [reduction]
Br⁻ → BrO₃⁻ [oxidation]
Now we have to balance the half reactions as,
Al³⁺ + 3e⁻ → Al
To balance H atoms we have to add water and electrons.
3H₂O + Br⁻→ BrO³⁻ + 6H⁺ + 6e⁻
Now we have to balance the electrons as,
2Al³⁺ + 6e⁻ → 2 Al
3H₂O + Br⁻→ BrO³⁻ + 6H⁺ + 6e⁻
Now we have to add both the equations as,
2Al³⁺ + 6e⁻ + 3H₂O + Br⁻ → 2 Al + BrO³⁻ + 6H⁺ + 6e⁻
6 electrons on both sides of the equation gets cancelled and so the balanced reaction can be written as,
2Al³⁺ + 3H₂O + Br⁻ → 2 Al + BrO³⁻ + 6H⁺
In the following reaction HF(aq) + HPO42-(aq) = F-(aq) + H2PO4-(aq)
a. HPO42- is an acid and H2PO4- is its conjugate base.
b. H2PO4- is an acid and F- is its conjugate base.
c. HPO42- is an acid and HF is its conjugate base.
d. HF is an acid and F- is its conjugate base.
e. HF is an acid and HPO42- is its conjugate base.
d. HF is an acid and [tex]F^-[/tex] is its conjugate base.
What is acid/ base and its conjugate acid/base?
Whenever an acid donates a proton, the acid changes into a base, and whenever a base accepts a proton, an acid is formed. An acid and a base which differ only by the presence or absence of a proton are called a conjugate acid-base pair.For example: HCl is an acid and its conjugate base is [tex]Cl^-[/tex].Given chemical reaction:
[tex]HF(aq) + HPO_4^{2-}(aq)----> F^-(aq) + H_2PO_4^-(aq)[/tex]
In this reaction, HF is an acid and its conjugate base is [tex]F^-[/tex].
Thus out of all the options; the correct option is d.
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In the reaction HF(aq) + HPO4^2-(aq) = F^-(aq) + H2PO4^-(aq), HF is the acid and F^- is the conjugate base, making the correct answer (d).
Explanation:In the given reaction HF(aq) + HPO42-(aq) = F-(aq) + H2PO4-(aq), we need to identify the correct acid-base pairs. When looking at this reaction, HF donates a proton (H+) to form its conjugate base F-, which makes HF the acid. On the other hand, since HPO42- receives a proton (H+) to become H2PO4-, HPO42- clearly acts as the base in this reaction and H2PO4- is its conjugate acid, which shows that it is amphoteric.
Based on this explanation, the correct answer is (d) HF is an acid and F- is its conjugate base.
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A 3.00 L flask contains 2.33 g of argon gas at 312 mm Hg What is the temperature of the gas
Answer:
T= 257.36 k
Explanation:
using the ideal gas law
pv=nRt
first, convert pressure from mmhg to kpa
312 x (101.3\760)= 41.58 kpa
R is constant= 8.31
to get n(number of moles)
n=m\M (m is mass, M is molar mass)
molar mass of argon is 39.948
n= 2.33\39.948
n=0.0583
substitute;
41.58 x 3 = 0.0583 x 8.31 x T
T= (41.58 x 3)\ (0.0583 x 8.31)
T= 257.36 k
The temperature of the gas under ideal conditions is 257.36K
In order to get the temperature of the gas, we will use the ideal gas equation expressed according to the formula:
[tex]PV = nRT[/tex]
P is the pressure of the gas = 312mmHg = 41.58KPa
V is the volume of the gas = 3.00L
n is the moles of the gas = 0.1165moles
R is boltzmann constant = 8.31
T is the required temperature
Mole = mass/molar mass
Mole = 2.33/40
Mole of argon = 0.05825moles
Substitute the given parameters into the formula
[tex]T=\frac{PV}{nR}\\ T=\frac{41.58 \times 3}{0.05825\times8.31}\\T=257.36K[/tex]
Hence the temperature of the gas under ideal conditions is 257.36K
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Within each ______ in the periodic table, elements have similar properties because they have the same number of valence electrons.
Period
Row
Group
Metals
Plz this is really really really urgent
Answer:
group
Explanation:
elements in the same group have the same number of valence electrons
There are various kind of elements that are present in periodic table. Some elements are harmful, some are radioactive, some are noble gases. Therefore, the correct option is option C.
What is periodic table?Periodic table is a table in which we find elements with properties like metals, non metals, metalloids and radioactive element arranges in increasing atomic number.
Periodic table help a scientist to know what are the different types of elements are present in periodic table so that they can discover the new elements that are not being discovered yet.
Within each group in the periodic table, elements have similar properties because they have the same number of valence electrons.
Therefore, within each group in the periodic table, elements have similar properties because they have the same number of valence electrons. The correct option is option C.
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In a redox reaction, oxidation is defined by the:
1.gain of electrons, resulting in an increased oxidation number.
2.loss of electrons, resulting in a decreased oxidation number.
3.gain of electrons, resulting in a decreased oxidation number.
4.loss of electrons, resulting in an increased oxidation number.
Answer:
Option 4. loss of electrons, resulting in an increased oxidation number.
Explanation:
Oxidation is a process involving loss of electron(s). When this happens the oxidation number of the atom being oxidised increases. This can be seen when calcium (Ca) reacts with chlorine (Cl2) to form calcium chloride (CaCl2) according to the equation given below:
Ca + Cl2 —> CaCl2
The oxidation number of calcium increases from 0 to +2. This implies that calcium is being oxidised as it loses its electrons. The oxidation number of chlorine decreases from 0 to - 1 as it gains electron.
Now, we can see that the oxidation of calcium i.e lose of electrons increased its oxidation number from 0 to +2.
From the simple illustrations above, we can see clearly that oxidation involves loss of electrons, resulting in an increased oxidation number.
Consider this reaction occurring at 298 K: N2O(g) + NO2(g) 2 3 NO(g) a. Show that the reaction is not spontaneous under standard conditions by calculating A Grx b. If a reaction mixture contains only N20 and NO2 at partial pressures of 1.0 atm each, the reaction will be spontaneous until some NO forms in the mixture. What maximum partial pressure of NO builds up before the reaction ceases to be spontaneous? c. Can the reaction be made more spontaneous by an increase or decrease in temperature? If so, what temperature is required to make the reaction spontaneous under standard conditions?
Answer:
A.) ΔG = 107.8 kJ/mol
ΔG rxn is positive implies that the reaction is non spontaneous.
B) P (NO) = 5.0 x 10^-7 atm
C) T > 923.4 K, therefore, the temperature should be 924K in order to make the reaction spontaneous.
Explanation:
N2O(g) + NO2(g) - 3NO(g)
A) ΔG = ΔG products - ΔG reactants
ΔGf (N2O) = 103.7 kJ/mol
ΔGf (NO2) = 51.3 kJ/mol
ΔGf (NO) = 87.6 kJ/mol
ΔG rxn = 3*87.60 - (103.7 + 51.3)
= 107.8 kJ/mol
ΔG rxn is positive implies that the reaction is non spontaneous.
(B). P (N2O) = P (NO2) = 1 atm
ΔG rxn = ΔG°rxn + RTln(K)
ΔG rxn = ΔG°rxn + RT×ln (P NO)^3 / [P(N2O) × P(NO2)]
When the reaction ceases to be spontaneous, then ΔG rxn = 0.
0 = 107.8 × 10^3 + 8.314 × 298 × ln (P NO)^3 / (1 * 1)
107.8 × 10^3 = - 8.314 × 298 × ln (P NO)^3
ln (P NO)^3 = - 43.51
3 × ln (P NO) = - 43.51
ln (P NO) = -14.50
P (NO) = e^(-14.50)
P (NO) = 5.0 x 10^-7 atm
(C). For a reaction to be spontaneous, ΔG rxn should be negative.
It is known that:
ΔG= ΔH - TΔS
ΔG= ΔH - TΔS < 0
ΔH< TΔS
T < ΔH / ΔD
For the reaction:
ΔH = 3×NO - (N2O + NO2)
= 3×(91.3) - (81.6 + 33.2)
= 159.1 kJ/mol
ΔS = 3×210.8 - (220.0 + 240.1)
= 172.3 J/mol.K
= 0.1723 kJ/mol.K
From the above expression.
T > 159.1 / 0.1723
T > 923.4 K
Therefore, the temperature should be 924K in order to make the reaction spontaneous.
When electrodes are used to record the electrocardiogram, an electrolyte gel is usually put between them and the surface of the skin. This makes it possible for the metal of the electrode to form metallic ions that move into the electrolyte gel. Often, after prolonged use, this electrolyte gel begins to dry out and change the characteristic of the electrodes. Draw an equivalent circuit for the electrode while the electrolyte gel is fresh. Then discuss and illustrate the way you expect this equivalent circuit to change as the electrolyte gel dries out. In the extreme case where there is no electrolyte gel left, what does the equivalent circuit of the electrode look like
Answer:
The equivalent circuit for the electrode while the electrolyte gel is fresh
From the uploaded diagram the part A is the electrolyte, the part part B is the electrolyte gel when is fresh and the part C is the surface of the skin
Now as the electrolyte gel start to dry out the resistance [tex]R_s[/tex] of the gel begins to increase and this starts to limit the flow of current . Now when the gel is then completely dried out the resistance of the gel [tex]R_s[/tex] then increases to infinity and this in turn cut off flow of current.
The diagram illustrating this is shown on the second uploaded image
Explanation:
A gas is placed in a storage tank at a pressure of 30.0 atm at 20.3 C. As a safety device, a small metal plug in the tank is made of a metal alloy that melts at 130 C. If the tank is heated, what is the maximum pressure (in atm) that will be attained in the tank before the plug melts and releases gas?
Answer: 41.2 atm
Explanation
To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
where,
[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.
[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.
We are given:
[tex]P_1=30.0atm\\T_1=20.3^0C=(20.3+273)=293.3K\\P_2=?\\T_2=130^0C=(130+273)K=403K[/tex]
Putting values in above equation, we get:
[tex]\frac{30.0}{293.3K}=\frac{P_2}{403}\\\\P_2=41.2atm[/tex]
The maximum pressure (in atm) that will be attained in the tank before the plug melts and releases gas is 41.2
The maximum pressure (in atm) that will be attained in the tank before the plug melts and releases gas is 41.22 atm
From the question given above, the following data were obtained:
Initial pressure (P₁) = 30 atm
Initial temperature (T₁) = 20.3 °C = 20.3 + 273 = 293.3 K
Final temperature (T₂) = 130 °C = 130 + 273 = 403 K
Final pressure (P₂) =?The final pressure can be obtained as illustrated below:
[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\\\frac{30}{293.3} = \frac{P_{2}}{403}\\\\[/tex]
Cross multiply
293.3 × P₂ = 30 × 403
293.3 × P₂ = 12090
Divide both side by 293.3
[tex]P_{2} = \frac{12090}{293.3} \\\\[/tex]
P₂ = 41.22 atmTherefore, the maximum pressure (in atm) that will be attained in the tank before the plug melts and releases gas is 41.22 atm
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A σ bond arises from the straight-on overlap of two atomic orbitals. The electron density lies along the axis of the two bonded nuclei. Example: Sigma Bonding in methane, CH4 What atomic or hybrid orbital on the central I atom makes up the sigma bond between this I and an outer Br atom in iodine tribromide, IBr3 ?
In iodine tribromide, the I-Br sigma bond is formed through the straight-on overlap of sp3d hybrid atomic orbitals on iodine with p orbitals on bromine.
Explanation:In iodine tribromide (IBr3), the central iodine (I) atom is surrounded by three bromine (Br) atoms. The I-Br bond that forms is a sigma (σ) bond, which is the result of a straight-on overlap of atomic orbitals. Sigma bonding typically involves the hybridized orbitals. Specifically, in IBr3, the I atom undergoes hybridization and forms two lone pairs and three σ bonds with Br atoms using sp3d hybrid orbitals. So, the hybrid orbital on the iodine atom that makes up the σ bond with a bromine atom is the sp3d hybrid orbital.
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The sigma bond in IBr3 between the central I atom and an outer Br atom is formed by the overlap of a sp3d hybrid orbital from iodine with a p orbital from bromine to accommodate the T-shaped geometry of the molecule.
In iodine tribromide (IBr3), the sigma bond between the central iodine (I) atom and an outer bromine (Br) atom is typically formed by the overlap of a hybrid orbital from the iodine with a p orbital from the bromine. When we consider the valence shell electronic configuration of iodine (5s2 5p5), upon forming IBr3, three of the p orbitals would be used to form sigma bonds with the bromine atoms, and this would likely involve hybridization to accommodate the molecular geometry of IBr3. The exact nature of the hybrid orbitals would depend on the geometry of the molecule, which, due to the presence of only three bond pairs and two lone pairs, adopts a T-shaped geometry, pointing towards sp3d hybridization. However, the exact details of this hybridization can only be confirmed with molecular orbital calculations
How many liters would you need to make a 0.7 M solution if you have 3.34 mol of Potassium
Chloride?
The volume of Potassium chloride solution is 4.8 L.
Explanation:
Molarity is found by dividing the number of moles of the given substance by its volume in litres, and so the unit of molarity is mol/L.
Here number of moles is given as 3.34 mol
Molarity = 0.7 M
Volume = ? L
Molarity = [tex]$\frac{moles}{Volume (L) }[/tex]
To find the volume we have to rearrange the equation as,
Volume (L) = [tex]$\frac{moles}{molarity}[/tex]
Now, we have to plug in the values as,
Volume (L) = [tex]$\frac{3.34 mol}{0.7 mol/L}[/tex]
= 4.77 ≈ 4.8 L
So the volume of Potassium chloride solution is 4.8 L.
Two friends, Pamela and Elaine, compared snack items they planned to enjoy during a work break. Pamela's snack was a small bag of shortbread cookies composed of 2.6 grams fat, 21 grams carbohydrates, and 2.1 grams protein. Elaine had a bag of popcorn composed of 2.9 grams fat, 31 grams carbohydrates, and 3.5 grams protein.
Nutrient Fuel value (kJ/g):Carbohydrate 17, Fat 38, Carbs 17
(a) Using the fuel values provided, calculate the number of calories in each person's snack. Pamela's snack Cal Elaine's snack Cal
(b) Suppose the friends took a 15-minute walk after snacking. If Pamela burned 43 calories and Elaine expended 48 calories, who had the smaller net gain of calories after snacking and walking?
Answer:
a) Pamela's shortbread cookies has 117,471.3 calories = 117.5 kcal
Elaine's popcorn has 166,515.3 calories = 166.5 kcal
b) Pamela had the smaller net gain of calories after snacking and walking.
Explanation:
- Pamela's snack was a small bag of shortbread cookies composed of 2.6 grams fat, 21 grams carbohydrates, and 2.1 grams protein.
- Elaine had a bag of popcorn composed of 2.9 grams fat, 31 grams carbohydrates, and 3.5 grams protein.
Nutrient | Fuel value (kJ/g):
Carbohydrate | 17
Fat | 38
Protein | 17
Noting that instead of Carbs, Protein was intended to be written.
a) For Pamela
Carbohydrates = (21 g) × (17 KJ/g) = 357 KJ
Fats = (2.6 g) × (38 KJ/g) = 98.8 KJ
Protein = (2.1 g) × (17 KJ/g) = 35.7 KJ
Total fuel value = 357 + 98.8 + 35.7 = 491.5 KJ = 491,500 J
Note that 1 cal = 4.184 J
491,500 J = (491,500/4.184) = 117,471.3 calories = 117.5 kcal
Carbohydrates = (31 g) × (17 KJ/g) = 527 KJ
Fats = (2.9 g) × (38 KJ/g) = 110.2 KJ
Protein = (3.5 g) × (17 KJ/g) = 59.5 KJ
Total fuel value = 527 + 110.2 + 59.5 = 696.7 KJ
Note that 1 cal = 4.184 J
696,700 J = (696,700/4.184) = 166,515.3 calories = 166.5 kcal
b) Suppose the friends took a 15-minute walk after snacking. If Pamela burned 43 kcal and Elaine expended 48 kcal, who had the smaller net gain of calories after snacking and walking?
Pamela gains 117.5 kcal
And loses 43 kcal
Net gain = 117.5 - 43 = 74.5 kcal
Elaine gains 166.5 kcal
And loses 48 kcal
Net gain = 166.5 - 48 = 118.5 kcal
Pamela had the smaller net gain of calories after snacking and walking.
Hope this Helps!!!
Salad is a mixture because
the ingredients go
well together
the ingredients form
a pure substance
the ingredients are
not chemcically
combined
the ingredients are
chemically
combined
Answer:
are you asking if it's true or false?
Answer:
no
Explanation:
50.g of NaNO3 was dissolved in 1250 mL of water. what is the molality of the solution? [ Molar mass of NaNO3 = 85 g/mol
Answer:
Approximately [tex]0.47\; \rm mol \cdot L^{-1}[/tex] (note that [tex]1\; \rm M = 1 \; \rm mol \cdot L^{-1}[/tex].)
Explanation:
The molarity of a solution gives the number of moles of solute in each unit volume of the solution. In this [tex]\rm NaNO_3[/tex] solution in water,
Let [tex]n[/tex] be the number of moles of the solute in the whole solution. Let [tex]V[/tex] represent the volume of that solution. The formula for the molarity [tex]c[/tex] of that solution is:
[tex]\displaystyle c = \frac{n}{V}[/tex].
In this question, the volume of the solution is known to be [tex]1250\; \rm mL[/tex]. That's [tex]1.250\; \rm L[/tex] in standard units. What needs to be found is [tex]n[/tex], the number of moles of [tex]\rm NaNO_3[/tex] in that solution.
The molar mass (formula mass) of a compound gives the mass of each mole of units of this compound. For example, the molar mass of [tex]\rm NaNO_3[/tex] is [tex]85\; \rm g \cdot mol^{-1}[/tex] means that the mass of one mole of
[tex]\displaystyle n = \frac{m}{M}[/tex].
For this question,
[tex]\begin{aligned}&n\left(\mathrm{NaNO_3}\right) \\ &= \frac{m\left(\mathrm{NaNO_3}\right)}{M\left(\mathrm{NaNO_3}\right)}\\&= \frac{50\; \rm g}{85\; \rm g \cdot mol^{-1}} \\& \approx 0.588235\; \rm mol\end{aligned}[/tex].
Calculate the molarity of this solution:
[tex]\begin{aligned}c &= \frac{n}{V} \\&= \frac{0.588235\; \rm mol}{1.250\; \rm L} \\&\approx 0.47\;\rm mol \cdot L^{-1}\end{aligned}[/tex].
Note that [tex]1\; \rm mol \cdot L^{-1}[/tex] (one mole per liter solution) is the same as [tex]1\; \rm M[/tex].
Answer:
.47
Explanation:
just did it on CK-12
Microwave ovens work by irradiating food with microwaves, which are absorbed by the water molecules in the food and converted to heat. Assuming that microwave ovens emit radiation with a wavelength of 29.7 cm, calculate how many photons are emitted per second in a 600-Watt microwave oven. (1 W = 1 J/s)
Answer:
8.969×10²⁶ photons
Explanation:
From the question,
Using,
E = hc/λ........................... Equation 1
Where E = Energy of the photon, h = Planck's constant, c = speed of light, λ = wave length of the photon.
Given: h = 6.626×10⁻³⁴ J.s, c = 3×10⁸ m/s, λ = 0.297 m
Substitute into equation 1
E = 6.626×10⁻³⁴(3×10⁸)/0.297
E = 6.69×10⁻²⁵ J.
The energy of the photon in one seconds = 6.69×10⁻²⁵ J/s
If the power of the microwave oven = 600 J/s
Then,
Number of photons emitted per seconds = 600/(6.69×10⁻²⁵)
Number of photons emitted per seconds = 8.969×10²⁶ photons
2.088 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of dioxygen, producing 4.746 g CO 2 and 1.943 g H 2O. What is the empirical formula of the compound?
Answer:
The empirical formula is C3H6O
Explanation:
Step 1: Data given
Mass of the sample =2.088 grams
The mass contains carbon, hydrogen, and oxygen
Mass of CO2 produced = 4.746 grams
Mass of H2O produced = 1.943 grams
Molar mass of CO2 = 44.01 g/mol
Molar mass of H2O = 18.02 g/mol
Atomic mass of C = 12.01 g/mol
Atomic mass of H = 1.01 g/mol
Atomic mass of O = 16.0 g/mol
Step 2: Calculate moles CO2
Moles CO2 = mass CO2 / molar mass CO2
Moles CO2 = 4.746 grams/ 44.01 g/mol
Moles CO2 = 0.1078 moles
Step 3: Calculate moles C
For 1 mol CO2 we have 1 mol C
For 0.1078 moles CO2 we'll have 0.1078 moles C
Step 4: Calculate mass C
Mass C: moles C * atomic mass C
Mass C: 0.1078 moles * 12.01 g/mol
Mass C= 1.295 grams
Step 5: Calculate moles H2O
Moles H2O = 1.943 grams / 18.02 g/mol
Moles H2O = 0.1078 moles
Step 6: Calculate moles H
For 1 mol H2O we'll have 2 moles H
For 0.1023 moles H2O we'll have 2*0.1078 = 0.2156 moles H
Step 7: Calculate mass H
Mass H = 0.2046 moles * 1.01 g/mol
Mass H = 0.218 grams
Step 8: Calculate mass O
Mass O = 2.088 grams - 1.295 grams - 0.218 grams
Mass O = 0.575 grams
Step 9: Calculate moles O
Moles O = 0.575 grams / 16.0 g/mol
Moles O = 0.0359 moles
Step 10: Calculate the mol ratio
We divide by the smallest amount of moles
C: 0.1078 moles / 0.0359 moles = 3
H: 0.2156 moles / 0.0359 moles = 6
O: 0.0359 moles / 0.0359 moles =1
The empirical formula is C3H6O
The empirical formula of the compound is C₃H₆O
We'll begin by calculating the mass of C, H and O in the compound. This can be obtained as follow:
For C:Mass of CO₂ = 4.746 g
Molar mass of CO₂ = 44 g/mol
Molar mass of C = 12 g/mol
Mass of C =?Mass of C = 12/44 × 4.746
Mass of C = 1.294 gFor H:Mass of H₂O = 1.943 g
Molar mass of H₂O = 18 g/mol
Molar mass of H₂ = 1 × 2 = 2 g/mol
Mass of H =?Mass of H = 2/18 × 1.943
Mass of H = 0.216 gFor O:Mass of C = 1.294 g
Mass of H = 0.216 g
Mass of compound = 2.088 g
Mass of O =?Mass of O = (Mass of compound ) – (mass of C + mass of H)
Mass of O = 2.088 – (1.294 + 0.216)
Mass of O = 0.578 gFinally, we shall determine the empirical formula of the compound. This can be obtained as follow:C = 1.294 g
H = 0.216 g
O = 0.578 g
Empirical formula =?Divide by their molar mass
C = 1.294 / 12 = 0.108
H = 0.216 / 1 = 0.216
O = 0.578 / 16 = 0.036
Divide by the smallest
C = 0.108 / 0.036 = 3
H = 0.216 / 0.036 = 6
O = 0.036 / 0.036 = 1
Therefore, the empirical formula of the compound is C₃H₆O
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Which of the following solutions is a good buffer system?Which of the following solutions is a good buffer system?A solution that is 0.10 M Li OH and 0.10 MHNO3A solution that is 0.10 MHCN and 0.10 MLiCNA solution that is 0.10 MNaCl and 0.10 MHClA solution that is 0.10 M Li and 0.10 MHNO3A solution that is 0.10 MHCN and 0.10 M Li Cl
Answer:
A solution that is 0.10 M HCN and 0.10 M LiCN.
Explanation:
A buffer solution is a solution of a weak acid and its conjugate base or a solution of a weak base and its conjugate acid, so the pH of the solution changes in a very little range when an acid or base is added to the solution.
Let's evaluate each statement:
(a) A solution that is 0.10 M LiOH and 0.10 M HNO₃
This is not a buffer solution since the HNO₃ is not the acid conjugate of the LiOH, and also, the LiOH is a strong base and the HNO₃ is a strong acid.
(b) A solution that is 0.10 M HCN and 0.10 M LiCN
This is a buffer solution, with the following reaction:
HCN + H₂O ⇄ CN⁻Li⁺ + H₃O⁺
The acid HCN is a weak acid and the LiCN is its conjugate base. The fact that the concentrations are equal for both is appropriate for the buffer solution.
(c) A solution that is 0.10 M NaCl and 0.10 M HCl
This is not a buffer solution since the HCl is a strong acid. It dissociates in water to form Cl⁻ and H⁺.
(d) A solution that is 0.10 M Li and 0.10 M HNO₃
This is not a buffer solution since the HNO₃ is not the conjugate base of Li, the HNO₃ is a strong acid that dissociates in water to form H⁺ and NO₃⁻.
(e) A solution that is 0.10 M HCN and 0.10 M LiCl
This is not a buffer solution since the LiCl is not the conjugate base of the acid HCN.
Therefore, the correct option is: A solution that is 0.10 M HCN and 0.10 M LiCN.
I hope it helps you!
The Mannich reaction is one of the few three-component reactions in organic chemistry. In this reaction, a ketone, an aldehyde and an amine react together under acid catalyzed conditions to form the final product. The mechanism involves the following steps: 1. Following initial protonation of the carbonyl oxygen, nucleophilic attack by the amine forms a protonated carbinolamine 1; 2. Proton transfer and elimination of water forms iminium ion 2; 3. The enol form of the ketone attacks the iminium ion to form adduct 3; 4. Deprotonation of adduct 3 leads to the final product. Write out the mechanism on a separate sheet of paper and then draw the structure of iminium ion 2.
Answer:
Step 1, formation of the iminium ion
2) enol reaction with iminium ion to form beta-amino enone
Explanation:
Please find attached the detailed reaction mechanism and the structure of iminium ion separately
The Mannich reaction is a three-component reaction that forms a ß-amino ketone or aldehyde. It involves protonation, nucleophilic attack, formation of an iminium ion, enol attack, and deprotonation.
The Mannich reaction is a three-component organic reaction involving a ketone, an aldehyde, and an amine under acid-catalyzed conditions to form a ß-amino ketone or ß-amino aldehyde. The mechanism of the reaction can be broken down into several steps:
Protonation of the carbonyl oxygen: Initially, the carbonyl oxygen of the ketone or aldehyde is protonated by an acid catalyst.Nucleophilic attack by the amine: The nucleophilic amine attacks the protonated carbonyl, forming a protonated carbinolamine.Formation of iminium ion: Following a proton transfer, the carbinolamine dehydrates to form an iminium ion.Enol attack: The enol form of the ketone then attacks the iminium ion, leading to the formation of adduct 3.Deprotonation: Finally, adduct 3 undergoes deprotonation to yield the final ß-amino product.Given these steps, the structure of iminium ion 2 involves a nitrogen atom double-bonded to a carbon atom derived from the original carbonyl.
A salt bridge: Question 3 options: provide a pathway through which the electrons travel from the cathode to the anode. provide a pathway through which the electrons travel from the anode to the cathode. provide a source of counterions to prevent the build-up of charge at both the cathode to the anode. provide a physical connection between the two cells, allowing the solutions in both cells to slowly mix.
Answer:
The correct option is: provide a source of counterions to prevent the build-up of charge at both the cathode to the anode.
Explanation:
A salt bridge is a U-shaped glass tube that is used in a voltaic cell or galvanic cell to connect the oxidation and reduction half-cells and complete the electric circuit.
It allows the ions to pass through it, thus preventing the accumulation of charge on the anode and cathode as the chemical reaction proceeds.
Therefore, the correct option is: provide a source of counterions to prevent the build-up of charge at both the cathode to the anode.
The reaction: 2 ClO2 (aq) + 2OH- (aq)→ ClO3- (aq) + ClO2- + H2O (l) was studied with the following results: Experiment [ClO2] (M) [OH-] (M) Initial Rate (M/s) 1 0.060 0.030 0.0248 2 0.020 0.030 0.00276 3 0.020 0.090 0.00828 a. Determine the rate law for the reaction. b. Calculate the value of the rate constant with the proper units. c. Calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M.
Answer:
Explanation:
question solved.
The rate law is defined as the rate of a chemical reaction, which is dependent on the concentration of the reactants.
It can be calculated by the formula:
[tex]\text v_o &= \text k [\text A]^x [\text B]^y[/tex]
where,
Vo = ratek = rate constantA = concentration of species Ax = order of reaction with respect to AB = concentration of species By = order of reaction with respect to BIn the given reaction:
[tex]\begin{aligned}2\;\text{ClO}_2\left(aq \right )+2\text{OH}^-\left(aq \right )\rightarrow\text{ClO}_3^-\left(aq \right )+\text{ClO}_2^-+\text{H}_2\text{O} \end{aligned}[/tex]
Using the rate law formula, we get:
Rate = K [Cl]ⁿ [OH⁻]ⁿ
0.0248 = K [0.060]ⁿ [0.030 ]ⁿ ........(1)
0.00276 = K [0.020]ⁿ [0.030 ]ⁿ .........(2)
0.00828 = K [0.020]ⁿ [0.090]ⁿ ..........(3)
Dividing (1) equation by (2), we get:
[tex]\begin {aligned} \dfrac {0.0248}{0.00276}&= \dfrac{\text K (0.06)^n (0.03)^n} {\text K (0.02)^n (0.03)^n}\\\\9 &= \dfrac{0.06}{0.02}^n\\\\3 ^2 &= 3 ^ m\\\text m &=2\\\end {aligned}[/tex]
Dividing the equation (2) by (3), we get:
[tex]\begin {aligned} \dfrac {0.00276}{0.00828 }&= \dfrac{\text K (0.02)^n (0.03)^n} {\text K (0.02)^n (0.09)^n}\\\\\dfrac{1}{3} &= \dfrac{0.03}{0.09}^n\\\\ \dfrac{1}{3} ^1 &= \dfrac{1}{3} ^n\\\\\text n &=2\\\end {aligned}[/tex]
Based on the rate law:
a. Rate = K [ClO₂]²⁻ [OH⁻]¹
b. 0.0248 = K [0.06]² [0.03]¹
K = [tex]\dfrac {0.0248}{(0.06)^2 \times 0.03}[/tex] = 229.63 m⁻²s⁻¹
c. Rate = K [ClO]₂ [OH⁻]ⁿ
Rate = 229.62 x (0.2)² (0.05)¹
Therefore, the rate of the reaction is 0.45926 m/sec.
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Isocyanates are good electrophiles that have been used for protein modification. However, they have limited stability in water. As a result isothiocyanates have been developed at less hydrolytically sensitive variants. Molecules like fluorescein isothiocyanate (FITC) have been used extensively to fluorescently modify proteins. Draw the product of a lysine side chain with FITC. (for the sake of simplicity we are just using a model for lysine's side chain butyl amine)
Find figure in the attachment
Answer:
Explanation:
solution to the question is found below
Ethylenediamine is a bidentate ligand. The oxalate group, used in this experiment was also a didentate ligand. The structure of ethylenediamine is NH2CH2CH2NH2. The oxalate group formed a coordinated compound using the negative charges on the oxygen. Explain how the ethylenediamine compound will bond.
Answer:
Ethylene diamine will bond to the Central metal via a lone pairs of electrons on nitrogen
Explanation:
Complexes are formed by coordinate bond formation. Before a coordinate bond is formed, one of the species must have a lone lair of electrons available for donation into empty orbitals on the central metal.
Ethylene diammine contains nitrogen which has a lone pair of electrons. The two lone pairs on the two nitrogen atoms can bond with the central metal. This makes ethylene diammine a bidentate ligand (two bonding atoms).
Ethylenediamine will bond to a metal ion through both of its nitrogen atoms, forming a chelate complex.
Ethylenediamine (en) is a bidentate ligand, which means it has two donor atoms that can simultaneously bond to a central metal ion. In the case of ethylenediamine, the donor atoms are the nitrogen atoms, each of which has a lone pair of electrons available for donation. When ethylenediamine bonds to a metal ion, it does so through both nitrogen atoms, creating a five-membered ring structure known as a chelate. The term chelate comes from the Greek word chele, which means claw. This chelation results in a more stable complex because the metal ion is bound to two sites on the ligand, reducing the likelihood of the ligand dissociating from the metal center.The formation of a chelate complex with ethylenediamine can be represented as follows: [tex]\[ \text{M}^{n+} + \text{en} \rightarrow \text{M(en)}^{n+} \][/tex] Here, M represents the metal ion, n+ is its oxidation state, and en is the ethylenediamine ligand. The resulting complex M(en)n+ has the metal ion bound within the chelate ring formed by the ethylenediamine ligand. In contrast, the oxalate group is also a bidentate ligand but with two oxygen atoms as the donor atoms. The oxalate group typically bonds to a metal ion through its negatively charged oxygen atoms, forming a similar chelate complex but with a different donor atom and potentially different coordination geometry. The stability of chelate complexes is often discussed in terms of the chelate effect, which states that metal complexes with chelating ligands are more stable than those with similar non-chelating (monodentate) ligands. This increased stability is due to the entropy effect, where the formation of a chelate ring results in the release of fewer solvent molecules compared to the formation of separate bonds with monodentate ligands.
18.67 Consider the reaction Given that DG8 for the reaction at 258C is 173.4 kJ/mol, (a) calculate the standard free energy of formation of NO, and (b) calculate KP of the reaction. (c) One of the starting substances in smog formation is NO. Assuming that the temperature in a running automobile engine is 11008C, estimate KP for the above reaction. (d) As farmers know, lightning helps to produce a better crop. Why
Answer:
a. 86.7KJ/mol
b. 2.5 * 10^30
c. 1.42 * 10^55
d. Lightning produces more amount of NO from nitrogen and oxygen thus giving a better crop as NO is essential for crop growth
Explanation:
Complete question is as follows;
Consider the reaction N2(g)+ O2(g) —-> 2NO(g)
Given that ΔG0 for the reaction at 25 degrees celsius is 173.4 kJ/mol a) calculate the standard freeenergy of formation of NO and B) calculate KP ofthe reaction. C) One of the starting substances in smog formationis NO. Assuming that the temperature in a running automobile engine is 1100 degreesC, estimate KP of the above reaction.D) As farmers know, lighting helps to produce a better crop.Why?
solution
Please check attachment for complete solution and step by step explanation
A chemistry student weighs out 0.120 g of acetic acid into a 250. mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700 M NaOH solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Be sure your answer has the correct number of significant digits.
Answer:
28.6 mL
Explanation:
Step 1: Write the balanced neutralization reaction between acetic acid and sodium hydroxide
CH₃COOH(aq) + NaOH(aq) = CH₃COONa(aq) + H₂O(l)
Step 2: Calculate the moles of acetic acid
The molar mass of acetic acid is 60.05 g/mol. The moles corresponding to 0.120 g are:
[tex]0.120g \times \frac{1 mol}{60.05g} =2.00 \times 10^{-3} mol[/tex]
Step 3: Calculate the moles of sodium hydroxide
The molar ratio of CH₃COOH to NaOH is 1:1. The reacting moles of NaOH are 2.00 × 10⁻³ mol.
Step 4: Calculate the volume of the 0.0700 M NaOH solution
[tex]2.00 \times 10^{-3} mol \times \frac{1L}{0.0700mol} =0.0286 L = 28.6 mL[/tex]
Answer:
We have to add 286 mL of NaOH
Explanation:
Step 1: Data given
Mass of acetic acid (CH3COOH)= 0.120 grams
Volume of acetic acid = 250 mL = 0.250 L
Molarity of NaOH = 0.0700 M
Step 2: The balanced equation
CH3COOH + NaOH → CH3COONa + H2O
Step 3: Calculate moles acetic acid
Moles acetic acid = mass / molar mass
Moles acetic acid = 0.120 grams / 60.05 g/mol
Moles acetic acid = 0.00200 moles
Step 4: Calculate molarity acetic acid
Molarity acetic acid = moles / volume
Molarity acetic acid = 0.00200 moles /0.250 L
Molarity acetic acid = 0.008 M
Step 5: Calculate the volume of solution the student will need to add to reach the equivalence point
C1*V1 = C2*V2
⇒with C1 = the molarity of acetic acid = 0.008 M
⇒with V1 = the volume of acetic acid = 0.250 L
⇒with C2 = the molarity of NaOH = 0.0700 M
⇒with V2 = the volume of NaOH neede = TO BE DETERMINED
0.008 M * 0.250 L = 0.0700 M * V2
V2 = (0.008M * 0.250 L) / 0.0700 M
V2 = 0.286 L = 286 mL
We have to add 286 mL of NaOH
Esters can be synthesized by an acid-catalyzed nucleophilic acyl substitution between an alcohol and a carboxylic acid; this process is called the Fischer esterification reaction. Because the alcohol oxygen is a poor nucleophile, the carbonyl carbon is made a better electrophile by protonation of the carbonyl oxygen. The steps of the synthesis are all reversible. The reaction is generally driven to completion by using an excess of the liquid alcohol as a solvent, or by distilling off the product as it forms. Draw curved arrows to show the movement of electrons in this step of the mechanism.
Answer:
See attachment for mechanism.
Explanation:
The Fischer esterification reaction is a nucleophilic substitution in the acyl group of a carboxylic acid, catalyzed by an acid.
1) The protonation of the oxygen of the carbonyl group activates the carboxylic acid...
2) ... against a nucleophilic attack by the alcohol, and produces a tetrahedral intermediate.
3) The transference of a proton from an oxygen atom to another produces a second tetrahedral intermediate and converts the -OH into a good leaving group.
4) The loss of a proton and the expulsion of H₂O regenerates the acid catalyzer and gives an ester as a product.
The Fischer esterification reaction involves reacting a carboxylic acid with an alcohol to form an ester. The reaction is driven to completion by using an excess of alcohol or by distilling off the product.
Explanation:
In the
Fischer esterification reaction
, an alcohol (for example, ethanol) reacts with a carboxylic acid (like acetic acid) to form an ester. The carbonyl carbon of the acid is protonated to make it a better electrophile that can react with the alcohol's oxygen. Here's a simplified version of what happens:
R-C-0-H (a carboxylic acid) + :O: (the oxygen of an alcohol) -> R-C(=O)-O-R' (an ester)
, where R and R' represent any organic groups. The reaction is balanced by distilling off the ester product or using an excess of the alcohol solvent. In a more detailed mechanism, curved arrows would be used to show the movement of electrons, but those can't be represented in text form.
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Calcium hydroxide, Ca(OH)2, is used as a calcium nutritional supplement in some foods and beverages, such as orange juice. What is the pH of a solution of 0.0012M calcium hydroxide at 25.0∘C?
Answer:
pH = 11.38
Explanation:
The pH of a solution is given as:
pH = pKw − pOH = 14.00 + log[OH-]
Since the concentration of Ca(OH)2 is 0.0012 M, the [OH−] is twice that, or 0.0024 M, resulting in pOH = 2.62 and pH = 11.38.
Final answer:
The pH of a solution of 0.0012M calcium hydroxide at 25.0°C is 11.38.
Explanation:
The pH of a solution of 0.0012M calcium hydroxide at 25.0°C can be calculated using the concentration of hydroxide ions ([OH-]).
Calcium hydroxide is a strong base that dissociates completely in water to form two hydroxide ions for every formula unit dissolved.
The concentration of the solute is 0.0012M, but because Ca(OH)2 is a strong base, the actual concentration of hydroxide ions ([OH-]) is two times this, or 2 × 0.0012M = 0.0024M.
Since [OH-] = 0.0024M, the pOH can be calculated as pOH = -log([OH-]) = -log(0.0024) = 2.62. The pH can then be calculated using the expression pH = 14 - pOH = 14 - 2.62 = 11.38.
A typical refrigerator is kept at 4˚C, and a soda can has a pressure of
1.18 atm. The inside of a car can reach up to 60˚C (140˚F) when it is left in
direct sunlight on a hot day. If you left the soda can in the car, what would
be the new pressure if it reached 60˚C?
Answer: The new pressure will be 1.42 atm
Explanation:
To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
where,
[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.
[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.
We are given:
[tex]P_1=1.18atm\\T_1=4^0C=(4+273)K=277K\\P_2=?\\T_2=60^0C=(60+273)K=333K[/tex]
Putting values in above equation, we get:
[tex]\frac{1.18}{277}=\frac{P_2}{333}\\\\P_2=1.42[/tex]
Hence, the new pressure will be 1.42 atm
Ampicillin is a bacteriostatic antibiotic. (Instead of killing bacteria, it inhibits their growth.) Ramon plated bacteria transformed with a plasmid that confers ampicillin resistance. He plated the bacteria on Thursday and left them in a 37 oC incubator overnight. On Friday, he observed moderately sized bacterial colonies. He decide to leave them in the incubator a little longer before picking the colonies that he wanted to work with further. Unfortunately, he forgot about the plate and left it in the incubator over the weekend. On Monday, his plate had large colonies that were each surrounded by very small colonies
A) Interpret and explain Ramon’s observations.
B) Predict what he will find when he picks some large and some small colonies and follows the plasmid isolation protocol for each of the colonies.
Answer:
A) That resistance in bacteria is produced due to inactivation of ampicillin by the beta lactamase enzyme. This enzyme is expressed by the bla gene found in the plasmid. This enzyme is secreted into the culture medium, thereby inactivating ampicillin. Thanks to this inactivation, the bacteria colonies will be able to develop. After a day of incubation, only those bacteria that took the plasmid that gives them resistance to ampicillin will grow after transformation. After prolonged incubation, two types of colonies can be observed in the culture medium. One large colony with ampicillin resistance, and another small colony, both of which are sensitive to ampicillin.
B) Large colonies are characterized by being resistant to ampicillin. When Ramón isolates the plasmid, he will have the gene that provides resistance to antibiotics. Said plasmid can be used again on those bacteria that are sensitive to ampicillin.
On the other hand, satellite colonies are sensitive to ampicillin. These types of colonies do not have the plasmid that contains the gene that gives ampicillin resistance. It is not possible to isolate any plasmids from these satellite colonies. These satellite bacteria will not be able to grow if they are transferred to a plate containing fresh ampicillin, while large colonies, which possess the plasmid that gives them resistance to ampicillin, will be able to grow on that plate.
Explanation: