The rate law for the overall reaction derived from a two-step reaction mechanism is determined by the slowest (rate-determining) step. In this case, the rate law would be rate = k * K_eq * [A]^2[B], where k is the rate constant for the slow step, K_eq is the equilibrium constant for the fast step, and [A] and [B] represent the concentrations of A and B, respectively.
Explanation:In the context of chemistry, one step in a multistep reaction mechanism is often significantly slower than the others. This slow step is known as the rate-determining step or the rate-limiting step. The reaction cannot proceed faster than this slowest step.
In your specific example, the given mechanism consists of two steps. The rate law for each step is generally expressed in terms of the concentration of the reactants involved in that step.
For Step 1 (A + B ⇄ C), assuming that the reaction reaches an equilibrium, the concentrations of A, B, and C would remain constant over time, and won't affect the overall rate. Therefore, we would ignore this step when deriving the rate law for the overall reaction.
On the other hand, Step 2 (C + A ⟶ D) is the slow step, and thus determines the rate of the overall reaction. The rate law for this step would be rate = k * [C][A]. But since C is also a product of Step 1, we need to express C in terms of A and B. From equilibrium of Step 1, we know [C] = K_eq*[A][B] (where K_eq is equilibrium constant).
Substituting this in rate law of step 2, we get rate = k * K_eq * [A]^2[B] for the overall reaction.
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The rate law for the overall reaction 2A + B ⟶ D, considering Step 2 is the rate-determining step, is calculated by using the equilibrium from Step 1 to express the concentration of intermediate C in terms of A and B. After substitution and simplification, the rate law for the overall reaction is rate = k [A]²[B], indicating second-order dependence on A and first-order dependence on B.
Explanation:To determine the rate law for the overall reaction, we must look at the mechanism provided. Since Step 2, which involves the conversion of C and A to D, is the rate-determining step, the rate law for the overall reaction will reflect this slowest step. However, since C is an intermediate that we cannot measure directly, we must use the equilibrium established in Step 1 to express the concentration of C in terms of the concentrations of A and B.
Assuming Step 1 is at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, and thus:
Rate of forward reaction (k₁ [A][B]) = Rate of reverse reaction (k₁₁ [C])
We can rearrange this to solve for [C]:
[C] = k₁/k₁₁ [A][B]
Now, since the rate-determining step is Step 2, we write the rate law based on this step:
rate = k₂ [C][A]
Substituting in the expression for [C] gives us:
rate = k₂ (k₁/k₁₁ [A][B])[A]
rate = (k₂ * k₁/k₁₁) [A]²[B]
Thus, after simplifying and combining the rate constants into a single overall rate constant (k), we have:
rate = k [A]²[B]
This shows that the reaction is second order with respect to A and first order with respect to B.
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a metal sample has a mass of 7.56 g. the sample is placed into a graduated cylinder previously filled with 20.00 mL of water. If the final volume in the cylinder is 21.68 mL, what is the density of the substance?
Answer:
4.5g/ml
Explanation:
metal sample has a mass = 7.56 g
cylinder previously filled with water = 20.00 mL
final volume in the cylinder = 21.68 mL
[tex]density = \frac{mass}{V_f - V_i}[/tex]
[tex]density = \frac{7.56}{(21.68 - 20.00)}[/tex]
density = 4.5g/ml
The aggregation of nonpolar molecules or groups in water is thermodynamically due to the ________.
A. increased entropy of the water molecules.
B. decreased enthalpy of the system.
C. very strong van der Waals forces among the nonpolar molecules or groups.
D. increased entropy of the nonpolar molecules when they associate.
Answer:
A. Increased entropy of the water molecules.
Explanation:
Entropy is the quantitative measure of disorder or randomness in a system or element. In thermodynamics, entropy or hydrophobic effect is the free energy change of water enclosing a solute. Hence the existing negative free charges enhances the effect of hydrophilicty hence the aggregation of non-polar molecules.
The aggregation of nonpolar molecules in water is primarily due to the increased entropy of the water molecules, as they rearrange themselves to accommodate nonpolar molecules.
Explanation:The aggregation of nonpolar molecules or groups in water is thermodynamically due to the increased entropy of the water molecules. When nonpolar molecules or groups are added to water, water molecules arrange themselves in order to minimize the disruption of hydrogen bonds, which increases the entropy of water. In other words, the introduction of nonpolar substances leads to a more disordered and random arrangement of water molecules, increasing the system's overall entropy.
Options B, C, and D although relevant, are not the primary reason. While enthalpy may decrease due to aggregation and Van der Waals forces might influence nonpolar association, the primary factor contributing to the phenomenon is the change in water's entropy.
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Calculate the volume (in mL) of 6.25 x 10-4 M ferroin solution that needs to be added to a 10.0 mL volumetric flask and diluted with deionized (DI) water in order to prepare a calibration standard solution with a concentration of 2.50 x 10-5 M ferroin. As part of your preparation for performing this experiment, repeat this calculation for each of the calibration standards you will need to prepare and record the information in your lab notebook so that you have it ready during the lab session. Group of answer choices 0.200 mL 0.400 mL 0.600 mL 0.800 mL none of the above
The volume V2 is 0.400 ml
Explanation:
The dilution equation is the product of initial values of molarity and volume which is equal to the product of final values of molarity and volume.
The dilution equation is given by
M1 [tex]\times[/tex] V1 = M2 [tex]\times[/tex] V2
where,
M represents the molarity of the solution
V represents the volume of the solution
M1 and V1 are the initial values of the solution
M2 and V2 are the final values of the solution
M1 [tex]\times[/tex] V1 = M2 [tex]\times[/tex] V2
(2.5 [tex]\times[/tex] [tex]10^{-5}[/tex]) [tex]\times[/tex] 10 = (6.25 [tex]\times 10^{-4}[/tex]) [tex]\times[/tex] V2
V2 = 2.5 [tex]\times 10^{-4}[/tex] / (6.25 [tex]\times 10^{-4}[/tex])
V2 = 0.4
The volume V2 is 0.400 ml
The volume of solution required is 0.400 mL.
In this case, we have to use the dilution formula;
C1V1 = C2 V2
In this case;
C1 = 6.25 x 10-4 M
V1 = ?
C2 = 2.50 x 10-5 M
V2 = 10.0 mL
Making V1 the subject of the formula;
V1 = C2V2/C1
V1 = 2.50 x 10-5 M x 10.0 mL/6.25 x 10-4 M
V1 = 0.400 mL
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The spectator ions in the reaction between aqueous perchloric acid and aqueous barium hydroxide are ________. OH- and ClO4- H , OH- , ClO4-, and Ba2 H and OH- H and Ba2 ClO4- and Ba2
Answer:
ClO4-, and Ba2+.
Explanation:
A spectator ion is an ion that exists as a reactant and a product in a chemical equation.
Equation of the reaction.
2HClO4 + Ba(OH)2 --> Ba(ClO4)2 + 2H2O
ClO4- + OH- --> 2ClO4- + H2O
The spectator ions in the reaction between aqueous perchloric acid and aqueous barium hydroxide are ClO₄⁻ and Ba⁺². These ions do not participate in the reaction and remain unchanged in the solution after the reaction.
The spectator ions in the reaction between aqueous perchloric acid and aqueous barium hydroxide are ions that do not participate in the chemical reaction and do not change during the course of the reaction. Aqueous solutions of acids and bases typically dissociate into their constituent ions in water. For perchloric acid (HClO₄), it dissociates into H⁺ and ClO⁴⁻ ions, while barium hydroxide (Ba(OH)₂) dissociates into Ba²⁺ and OH⁻ ions.
In a neutralization reaction, the H+ ions from the acid combine with the OH⁻ ions from the base to form water (H₂O), meaning they are not spectator ions. Therefore, the spectator ions would be the ions that remain unchanged, which are ClO⁴⁻ and Ba²⁺. The complete ionic equation for this reaction would show all the ions separated, but the net ionic equation would exclude these spectators, focusing only on the ions that participate directly in forming the product, water.
Mercury and oxygen react to form mercury(II) oxide, like this: 2 Hg(l)+02(g)--HgO(s) At a certain temperature, a chemist finds that a 6.9 L reaction vessel containing a mixture of mercury, oxygen, and mercury(II) oxide at equilibrium has the following composition: compound amount Hg 16.9 g O 10.9 g HgO 23.8 g Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.
Answer:
Kc = 20
Explanation:
We have the equilibrium:
2 Hg (l) + O₂ ( g) ⇄ HgO (s)
Kc = 1/ [O₂]
The key here is to remember that pure solids and liquids do not enter into the calculation for the equilibrium expression, and Hg is a pure liquid and HgO is a solid
So what we need to do to solve this question is to calculate the concentration of oxygen at equilibrium. We are given its mass, and the volume so we are equipped to calculate the concentration of oxygen as follows:
[O₂] = # moles O₂ / V
# moles O₂ = mass / molar mass = 10.9 g / 32g/mol = 0.34 mol
[O₂] = 0.34 mol / 6.9 L = 0.049 M
⇒ Kc = 1 / 0.049 = 20 ( rounded to 2 significant figures )
Consider the second-order reaction:
2HI(g)→H2(g)+I2(g)
Use the simulation to find the initial concentration [HI]0 and the rate constant k for the reaction. What will be the concentration of HI after t = 4.53×1010 s ([HI]t) for a reaction starting under the condition in the simulation?
Given from simulation:
Rate Law: k[HI]^2
k= 6.4 x 10^-9 l/(mol x s) at 500K
Initial Rate= 1.6 x 10^-7 mol/(l x s)
Answer:
Initial concentration of HI is 5 mol/L.
The concentration of HI after [tex]4.53\times 10^{10} s[/tex] is 0.00345 mol/L.
Explanation:
[tex]2HI(g)\rightarrow H_2(g)+I_2(g) [/tex]
Rate Law: [tex]k[HI]^2 [/tex]
Rate constant of the reaction = k = [tex]6.4\times 10^{-9} L/mol s[/tex]
Order of the reaction = 2
Initial rate of reaction = [tex]R=1.6\times 10^{-7} Mol/L s[/tex]
Initial concentration of HI =[tex][A_o][/tex]
[tex]1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2[/tex]
[tex][A_o]=5 mol/L[/tex]
Final concentration of HI after t = [A]
t = [tex]4.53\times 10^{10} s[/tex]
Integrated rate law for second order kinetics is given by:
[tex]\frac{1}{[A]}=kt+\frac{1}{[A_o]}[/tex]
[tex]\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}[/tex]
[tex][A]=0.00345 mol/L[/tex]
The concentration of HI after [tex]4.53\times 10^{10} s[/tex] is 0.00345 mol/L.
Consider the reaction: N2(g) + 2O2(g)2NO2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.90 moles of N2(g) react at standard conditions. S°system = J/K Submit Answer
Explanation:
It is known that standard entropies for [tex]N_{2}(g)[/tex] is 191.6 J/mol K, [tex]O_{2}(g)[/tex] = 205 J/mol K, and [tex]NO_{2}(g)[/tex] is 239.7 J/mol K at 298 K.
Therefore, we will calculate the value of [tex]\Delta S^{o}_{reaction}[/tex] from standard absolute entropies as follows.
[tex]\Delta S^{o}_{reaction} = \sum \Delta S^{o}_{products} - \sum \Delta S^{o}_{reactants}[/tex]
= 2 mole of [tex]NO_{2}(g)[/tex] - 1 mole of [tex]N_{2}(g)[/tex] + 2 mole of [tex]O_{2}(g)[/tex]
= [tex]2 \times 239.7 J/mol K - 1 \times 191.6 J/mol K + 2 \times 205 J/mol K[/tex]
= -122.2 J/K
The entropy change for 1.90 moles of [tex]N_{2}(g)[/tex] reacting is as follows.
[tex]\Delta S^{o}_{system}[/tex] = 1.90 moles of [tex]N_{2}(g) \times 122.2 J/K/ 1 \text{mol of} N_{2}(g)[/tex]
= 232.18 J/K
Thus, we can conclude that the entropy change for the given system is 232.18 J/K.
Compare the mass of the Mg ribbon with the mass of the magnesium oxide. Notice that the the mass of the magnesium oxide is greater than the mass of the Mg. How do you account for this apparent increase in mass?
The mass of magnesium oxide is greater than that of the Magnesium ribbon because the Magnesium ribbon reacts with oxygen in the air to form magnesium oxide. The extra mass is from the oxygen atoms.
Explanation:When comparing the mass of the Mg ribbon and the magnesium oxide, you may notice that the mass of the magnesium oxide seems to be greater. This apparent increase in mass can be attributed to a chemical reaction. In this case, when the Mg ribbon is exposed to air, it reacts with the oxygen present in the air to create magnesium oxide (MgO).
The increase in mass is due to the additional weight of the oxygen atoms that are now part of the compound. Moreover, smaller pieces of magnesium metal will react more rapidly than larger pieces because there is more reactive surface available. The increase in mass is thus directly related to the magnesium oxide formation with the interaction of magnesium and oxygen.
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The increase in mass of Magnesium Oxide compared to Magnesium ribbon is due to the incorporation of Oxygen during the oxidation process which forms Magnesium Oxide.
Explanation:
In comparing the mass of a Magnesium (Mg) ribbon and that of Magnesium Oxide (MgO), it's observed that the mass of the MgO is greater. This apparent increase in mass can be attributed to the chemical reaction that takes place when Magnesium reacts with Oxygen in the air to form Magnesium Oxide.
Magnesium + Oxygen -> Magnesium Oxide
Magnesium (Mg) atoms combine with Oxygen (O) molecules from the air. As a result, the atoms bond to form Magnesium Oxide (MgO) which results in an increase in mass since the mass of the Oxygen is now being taken into consideration.
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Write the overall molecular equation for the reaction of hydroiodic acid ( HI ) and potassium hydroxide. Include physical states. Enter the formula for water as H 2 O .
The molecular equation for the reaction between hydroiodic acid (HI) and potassium hydroxide (KOH) is:
HI(aq) + KOH(aq) ⇒ KI(aq) + H₂O(l).
The molecular equation represents the chemical reaction between hydroiodic acid (HI) and potassium hydroxide (KOH). In the aqueous phase (aq), hydroiodic acid dissociates into hydrogen ions (H+) and iodide ions (I-), while potassium hydroxide dissociates into potassium ions (K+) and hydroxide ions (OH-).
During the reaction, the hydrogen ions from hydroiodic acid react with the hydroxide ions from potassium hydroxide, forming water (H₂O) in liquid (l) state. Additionally, the remaining potassium ions from potassium hydroxide and iodide ions from hydroiodic acid combine to form potassium iodide (KI) in aqueous state. This balanced equation illustrates the rearrangement of ions and atoms during the chemical reaction,
HI(aq) + KOH(aq) ⇒ KI(aq) + H₂O(l).
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The reaction between hydroiodic acid and potassium hydroxide forms potassium iodide and water. The balanced molecular equation is HI(aq) + KOH(s) → KI(aq) + H₂O(l). All the physical states are indicated for each compound.
In this reaction, hydroiodic acid (HI) is added to solid potassium hydroxide (KOH).
This is an acid-base reaction, which typically results in the formation of a salt and water. The balanced molecular equation for the reaction is given below:HI(aq) + KOH(s) → KI(aq) + H₂O(l)In the equation, HI is in aqueous state (aq), KOH is in solid state (s), KI is in aqueous state (aq), and water is in liquid state (l).The reaction can be summarized as follows:
Reactants: Hydroiodic acid (HI) and Potassium hydroxide (KOH)Products: Potassium iodide (KI) and Water (H₂O)Correct question is: Write the overall molecular equation for the reaction of hydroiodic acid ( HI ) and potassium hydroxide. Include physical states. Enter the formula for water as H₂O .
PCL was mixed with gelatin to make a blend for elctrospun fibrous scaffold encapsulating growth factor that was admixed in the polymer solution and then filled in the syringe for electrospinning. Three scaffolds were made with (A: only PCL), (B: PCL:gelatin=3:1), (C: PCL:gelatin = 1:1). The scaffolds were immersed in PBS at 37 °C! and its degradation rate was noted: Scaffold A, B and C degraded 1%, 25% and 50% respectively, in 21 days. Assuming the growth factor is trapped and cannot freely diffuse out of the scaffold. Assuming, growth factor release is dependent on scaffold degradation A. Which scaffold can be used for delivering growth factor (encapsulated in the scaffold) needed to be delivered in first 7 days of incubation in an in-vivo experiment and why? (5pts) B. The scaffold A and C were prepared and freeze dried. Their weights were Dry weight of A: 10 mg Dry weight of C: 10 mg Both (A and C) the scaffolds were kept in PBS for 7 days at 37 °C. The scaffolds were freeze dried and their weight was recorded. Weight of A after degradation: 9.967 mg Weight of C after degradation: 8.33 mg Calculate the percentage weight remaining for A and C scaffold and comment on why one scaffold degraded faster than the other.
Answer:
Explanation:
First PCL to form a scaffold, combined with gelatin.
They are made by first three forms A) made by just PCL.B) made by gelatin ratio PCL is 3:1 and last is C) made by gelatin PCL.
The decoration rate is 1%, 25% and 50% respectively.
A) Growth factor is stuck in 21 days, and can not spread. In this case in vivo experiment for 7 days used highly degradable scaffold use the ability to break down due to decomposition in vivo degradation rate depends on the scaffold's acid byproduct impact.
B) the amount of scaffolded degradation.
First, with scaffold A.
10 mg scaffold weight A= 1 per cent degradation.
Following degradation wt is 9.967=? Degradation per cent.
So, degradation (9.967* 1)/10= 0.9967 per cent.
Likewise for C) scaffold c 10 mg wt. Loss of 50 per cent.
After wt 8.33=? Degradation per cent.
Degradation (8.33* 50)/10=41.65 per cent.
Scaffold c degrades significantly, since the loss of wt is even greater.
Consider the following mechanism for the decomposition of NO2Cl to NO2 and Cl2: (1) NO2Cl ⇌ NO2 + Cl (2) NO2Cl + Cl → NO2 + Cl2 (a) Use steady-state approximation to express the rate of Cl2 production. Select the single best answer
Final answer:
The rate of Cl2 production can be determined using the steady-state approximation, which states that the rate of the forward reaction in the first step is equal to the rate of the reverse reaction. Therefore, the rate of Cl2 production is given by the rate of the forward reaction of step 1. This can be expressed as k1 [NO2Cl] [Cl].
Explanation:
The rate of Cl2 production can be determined using the steady-state approximation. According to the given mechanism, the first step is in equilibrium, so the forward and reverse reaction rates are equal. This allows us to express the rate of the forward reaction of step 1 as:
rate of forward reaction of step 1 = k1 [NO2Cl] [Cl]
Since the reverse reaction of step 1 is negligible, the overall rate of Cl2 production is equal to the rate of the forward reaction of step 1. Therefore, the rate of Cl2 production is given by:
rate of Cl2 production = k1 [NO2Cl] [Cl]
Final answer:
To express the rate of Cl2 production using the steady-state approximation, one must balance the rate of chlorine atom creation with its consumption and solve for its concentration in terms of the reactants and rate constants. This value is then used in the rate equation for Cl2 production.
Explanation:
The question pertains to the mechanism of the decomposition of nitryl chloride (NO2Cl) to nitrogen dioxide (NO2) and chlorine gas (Cl2). Since the steady-state approximation is used, we consider the intermediate species such as chlorine atoms (Cl) to be in a steady state, meaning their concentrations do not change over time. The provided steps do not correspond directly to the decomposition of NO2Cl, but they outline a similar mechanism which can be analyzed in the same manner using steady-state approximation.
If we follow a similar approach for NO2Cl decomposition, we would set the rate of creation of Cl equal to the rate of its consumption. This would lead to a system of equations that can then be solved to express the rate of Cl2 production in terms of the concentrations of the reactants and the rate constants of the elementary steps.
The rate of reaction (2) can be determined using the rate law, which is determined experimentally. However, since we are not given any experimental data or rate law, we cannot provide a specific answer.
To summarize, to find the rate of Cl2 production in the decomposition of NO2Cl, we need the rate law for reaction (2). Without the rate law, we cannot determine the specific rate of Cl2 production.
e sure to answer all parts. Calculate the pH of the following aqueous solutions at 25°C: (a) 9.5 × 10−8 M NaOH (b) 6.3 × 10−2 M LiOH (c) 6.3 × 10−2 M Ba(OH)2
Final answer:
To calculate the pH of the given aqueous solutions at 25°C, we can use the formula pH = -log[H3O+]. For NaOH, LiOH, and Ba(OH)2, the concentration of hydroxide ions is equal to the concentration of the base, which allows us to find the pH.
Explanation:
To calculate the pH of an aqueous solution, we need to determine the concentration of the hydronium ions (H3O+). Since NaOH, LiOH, and Ba(OH)2 are strong bases that ionize completely, we can assume that the concentration of the hydroxide ions (OH-) is equal to the concentration of the base. To find the pH, we can use the formula:
pH = -log[H3O+]
For example:
For 9.5 × 10^(-8) M NaOH, the concentration of hydroxide ions is 9.5 × 10^(-8) M. Taking the negative logarithm, the pH is approximately 7.02.
For 6.3 × 10^(-2) M LiOH, the concentration of hydroxide ions is 6.3 × 10^(-2) M. Taking the negative logarithm, the pH is approximately 11.20.
For 6.3 × 10^(-2) M Ba(OH)2, the concentration of hydroxide ions is twice the concentration of the base, so it is 2 * 6.3 × 10^(-2) M = 1.26 × 10^(-1) M. Taking the negative logarithm, the pH is approximately 1.90.
Which of the following is a false statement about the isotopes carbon-12 (12C) and carbon-13 (13C)? Please choose the correct answer from the following choices, and then select the submit answer button. 12C and 13C have the same atomic weight. 12C and 13C have different numbers of neutrons. 13C has one more electron than 12C. 13C has more subatomic particles in its nucleus than 12C. 13C has the same atomic number as 12C.
The given isotopes of carbon, i.e., 12C and 13C, have the same number of protons but the number of neutrons differs. So, the statement 12C and 13C have different numbers of neutrons, is correct.
Isotopes are members of the same element's family but have variable numbers of neutrons despite having the same number of protons.
The element carbon can have different isotopes, based on the number of neutrons.
The isotopes of carbon include, C10, C11, C12, C13, C14, C15, and C16.
The given isotope, 12C and 13C, possess similar chemical properties and the same atomic number.
But the mass of the 13C isotope of carbon is more, because it possesses, one extra neutron, as compared to the 12C isotope of carbon.
Thus, the statement about the isotopes, “12C and 13C have different numbers of neutrons”, is correct.
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The statement that 13C has one more electron than 12C is incorrect. Both carbon-12 (12C) and carbon-13 (13C) isotopes have the same number of electrons due to both having the same atomic number. They only differ in the number of neutrons they possess.
Explanation:The statement '13C has one more electron than 12C' is false. Both isotopes, carbon-12 (12C) and carbon-13 (13C) have the same atomic number '6', which represents the number of protons and, under stable conditions, also views it as the same number of electrons.
So, while 12C and 13C indeed differ in the number of neutrons (12C has 6 neutrons and 13C has 7), they possess the same number of protons and electrons. Furthermore, their atomic weights are different as a result of the difference in the number of neutrons. The atomic weight of 12C is closer to 12 and that of 13C is closer to 13.
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Suppose a current of 60. A flows through a copper wire for 22.0 minutes. Calculate how many moles of electrons travel through the wire. Be sure your answer has the correct unit symbol and the correct number of significant digits.
Answer: 0.821 moles
Explanation:
Moles of electron = 1 mole
According to mole concept:
1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of particles.
We know that:
Charge on 1 electron = [tex]1.6\times 10^{-19}C[/tex]
Charge on 1 mole of electrons = [tex]1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C[/tex]
To calculate the charge passed, we use the equation:
[tex]I=\frac{q}{t}[/tex]
where,
I = current passed = 60 A
q = total charge = ?
t = time required = 22.0 minutes =[tex]22.0\times 60=1320sec[/tex]
Putting values in above equation, we get:
[tex]60A=\frac{q}{1320}\\\\q={60A}\times {1320s}=79200C[/tex]
As 96500 C contains = 1 mole of electrons
79200 C contains = [tex]\frac{1}{96500}\times 79200 =0.821[/tex] mole of electrons
Thus 0.821 moles of electrons travel through the wire.
Write the dissolution reaction for iron(III) nitrate in water. Use the pull-down menus to specify the state of each reactant and product. + Is iron(III) nitrate considered soluble or not soluble ? ... ... A. Soluble ... B. Not soluble Based upon this, the equilibrium constant for this reaction will be: ... ... A. Greater than 1 ... B. Less than 1
Explanation:
As Iron (III) nitrate is an ionic compound so when it is dissolved in water then it will dissociate to give nitrate ions and ferric ions. Also, we know that like dissolves like and here iron (III) nitrate being an ionic compound will readily dissolve in water (polar solvent).
The chemical equation for this reaction is as follows.
[tex]Fe(NO_{3})_{3}(s) + H_{2}O(l) \rightarrow Fe^{3+}(aq) + 3NO^{-}_{3}(aq)[/tex]
Hence, iron(III) nitrate considered soluble in water.
Now, equilibrium constant expression for this reaction is as follows.
[tex]K_{eq} = [Fe^{3+}][(NO_{3})_{3}]^{3}[/tex]
Therefore, the equilibrium constant for this reaction will be greater than 1.
The solubility product constant for MX2 is 7.2 x 10-8. How many grams of MX2 (108.75 g/mol) will dissolve in 276 ml of water at 25°C. M is the metal and X is the anion. Enter as a number to 4 decimal places.
Answer: The mass of [tex]MX_2[/tex] that will dissolve is 0.0786 grams
Explanation:
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The chemical equation for the ionization of [tex]MX_2[/tex] follows:
[tex]MX_2(aq.)\rightleftharpoons M^{2+}(aq.)+2X^-(aq.)[/tex]
s 2s
The expression of [tex]K_{sp}[/tex] for above equation follows:
[tex]K_{sp}=s\times (2s)^2[/tex]
We are given:
[tex]K_{sp}=7.2\times 10^{-8}[/tex]
Putting values in above expression, we get:
[tex]7.2\times 10^{-8}=s\times (2s)^2\\\\s=2.62\times 10^{-3}M[/tex]
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
Molar mass of [tex]MX_2[/tex] = 108.75 g/mol
Molarity of solution = [tex]2.62\times 10^{-3}mol/L[/tex]
Volume of solution = 276 mL
Putting values in above equation, we get:
[tex]2.62\times 10^{-3}mol/L=\frac{\text{Mass of }MX_2\times 1000}{108.75/mol\times 276}\\\\\text{Mass of }MX_2=\frac{2.62\times 10^{-3}\times 108.75\times 276}{1000}=0.0786g[/tex]
Hence, the mass of [tex]MX_2[/tex] that will dissolve is 0.0786 grams
The mass of MX₂ that will dissolve is 0.0786 grams.
What is Solubility Product?Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The chemical equation for the ionization of follows:
[tex]MX_2--- > M^{2+}+2X^-[/tex]
s 2s
Calculation for "s" :
Given: [tex]K_{sp}= 7.2*10^{-8}[/tex]
[tex]K_{sp}=s*(2s)^2\\\\7.2*10^{-8}=s*(2s)^2\\\\s=2.62*10^{-3}M[/tex]
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:
M = n/ V
Molar mass of = 108.75 g/mol
Molarity of solution = [tex]2.62*10^{-3}mol/L[/tex]
Volume of solution = 276 mL
On substituting the values:
[tex]2.62*10^{-3} mol/L=\frac{\text{ Mass of } MX_2 * 1000}{108.75g/mol*276} \\\\\text{ Mass of } MX_2=\frac{2.62*10^{-3} mol/L*108.75g/mol*276}{1000} \\\\\text{ Mass of } MX_2=0.0786 g[/tex]
Hence, the mass of MX₂ that will dissolve is 0.0786 grams.
Find more information about Solubility product here:
brainly.com/question/9807304
Now use the density formula to calculate the density of each object. Round your answers to the nearest one-tenth of a gram (one decimal place). a. Bowling Ball: 8.5 g/in3 Feathers: 0.1 g/in3 b. Bowling Ball: 8.5 g/cm3 Feathers: 0.1 g/cm3 c. Bowling Ball: 0.1 lb/in3 Feathers: 8.5 lb/in3 d. Bowling Ball: 0.1 g/cm3 Feathers: 8.5 g/cm3
Question:
Imagine a six-pound bowling ball. That's generally the lightest you'll find in a bowling alley. Also imagine six pounds of feathers. Your average feathered bedroom pillow weighs about 2 pounds, so imagine three pillows in a stack. You now have six pounds of bowling ball and six pounds of feathers. If you were to take each and throw them into a swimming pool, you would expect the bowling ball to sink and the feathers to float, but why?First, let's figure out their volume.A bowling ball has a radius of 4.25 in, which converts to 10.8 cm. Using the formula to find the volume of a sphere:
Note; The result is 322 cm³(= error by calculating in 4.25 cm instead).
Next, the standard size for a bedroom pillow is 26 in x 20 in x 4 in. When we convert to centimeters, our stacked pillows are 66 cm x 51 cm x 10 cm. The volume is then 33,660 cm³. We already know the mass of each object (six pounds), but remember that mass in this formula is measured in grams. You'll first need to convert six pounds to grams.
One pound converts to 453.6 g. Round your calculation to the nearest full gram and write it down.Now use the density formula to calculate the density of each object. Round your answers to the nearest one-tenth of a gram (one decimal place).
Bowling Ball: 0.1 g/cm³Feathers: 8.5 g/cm³ Correct!
Bowling Ball: 8.5 g/cm³ Feathers: 0.1 g/cm³
Bowling Ball: 0.1 lb/in³ Feathers: 8.5 lb/in³
Bowling Ball: 8.5 g/in³ Feathers: 0.1 g/in³
Answer:
The correct answer to the question is
b. Bowling Ball: 8.5 g/cm³ Feathers: 0.1 g/cm³.
Explanation:
To solve the question, we note that both the sphere and the Bowling Ball have a mass of 6 lb which is equal to 2721.554 g
The radius of a bowling ball is 4.25 cm = 1.67 in which gives its volume as
322 cm³
The mass of the bowling ball 6 lb while
The volume of a feathered pillow = 26 in × 20 in × 4 in = 2080 in³ = 58899040.911 cm³
The density of the Bowling Ball = mass/volume = 2721.554 g/322 cm³
= 8.452 g/cm³
The density of the feathers = mass/volume
= 2721.554 g/34085 cm³ = 0.0798 g/cm³ which to one decimal place
= 0.1 g/cm³
Therefore the density of the Bowling Ball to the nearest one-tenth
= 8.5 g/cm³ and
The density of the feathers to the nearest one-tenth = = 0.1 g/cm³
13. In animals, nitrogenous wastes are produced mostly from the catabolism of
A. phospholipids and glycolipids
B. proteins and nucleic acids.
C. starch and cellulose
D. triglycerides and steroids
E. fatty acids and glycerol
Answer:
The answer is B
Explanation:
Determine the change in entropy for 2.7 moles of an ideal gas originally placed in a container with a volume of 4.0 L when the container was expanded to a final volume of 6.0 L at constant temperature.
Answer:
The value of entropy change for the process [tex]dS = 0.009 \frac{KJ}{K}[/tex]
Explanation:
Mass of the ideal gas = 0.0027 kilo mol
Initial volume [tex]V_{1}[/tex] = 4 L
Final volume [tex]V_{2}[/tex] = 6 L
Gas constant for this ideal gas ( R ) = [tex]R_{u} M[/tex]
Where [tex]R_{u}[/tex] = Universal gas constant = 8.314 [tex]\frac{KJ}{Kmol K}[/tex]
⇒ Gas constant R = 8.314 × 0.0027 = 0.0224 [tex]\frac{KJ}{K}[/tex]
Entropy change at constant temperature is given by,
[tex]dS = R log _{e} \frac{V_{2}}{V_{1}}[/tex]
Put all the values in above formula we get,
[tex]dS = 0.0224 log _{e} [\frac{6}{4}][/tex]
[tex]dS = 0.009 \frac{KJ}{K}[/tex]
This is the value of entropy change for the process.
With respect to NAD+ and NADP+, which electron carrier is generally preferred for anabolic reactions? Group of answer choices Just NAD+ Just NADP+ Either of these Neither of these
Answer:
Just NADP+.
Explanation:
Two main type of biochemical reaction are anabolic reaction and catabolic reaction. Anabolic reactions joins the small molecules to form the large products.
The electron carrier may be defined as the molecule that has the ability to transport the electron from one molecule to the other molecule. NADP+ acts as electron carrier in the synthesis of molecule especially in the cholesterol metabolism. NAD+ acts as electron carrier in the catabolic reactions.
Thus, the correct answer is option (2).
When individuals are looking for jobs but are unable to find work, they are said to be
.
Answer:
Unemployed
Explanation:
(of a person) without a paid job but available to work.
"I was unemployed for three years" - ----- Example
Similar:
jobless
out of work
out of a job
not working
between jobs .
The rate constant for this second‑order reaction is 0.380 M − 1 ⋅ s − 1 0.380 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶products How long, in seconds, would it take for the concentration of A A to decrease from 0.860 M 0.860 M to 0.230 M?
Answer: 8.38 seconds
Explanation:
Integrated rate law for second order kinetics is given by:
[tex]\frac{1}{a}=kt+\frac{1}{a_0}[/tex]
[tex]a_0[/tex] = initial concentartion = 0.860 M
a= concentration left after time t = 0.230 M
k = rate constant =[tex]0.380M^{-1}s^{-1}[/tex]
[tex]\frac{1}{0.860}=0.380\times t+\frac{1}{0.230 }[/tex]
[tex]t=8.38s[/tex]
Thus it will take 8.38 seconds for the concentration of A to decrease from 0.860 M to 0.230 M .
Final answer:
To find out how long it would take for the concentration of A to decrease from 0.860 M to 0.230 M in a second-order reaction, we can use the integrated rate law.
Explanation:
The reaction in question is second order and the rate constant is 0.380 M⁻¹⋅s⁻¹ at 300 ∘C. To find out how long it would take for the concentration of A to decrease from 0.860 M to 0.230 M, we can use the integrated rate law for a second-order reaction:
t = 1 / (k * [A])
Substituting the given values:
t = 1 / (0.380 M⁻¹⋅s⁻¹ * 0.860 M)
t ≈ 2.80 s
How many mL of 0.05 M sodium acetate should be added to 100 mL of 0.05M acetic acid to make a buffer of pH 5.1? What is the molarity of the resulting buffer with respect to acetate (Acetate + Acetic Acid)? pKa acetic = 4.76
Explanation:
Let us assume that volume of acetic acid added is V ml.
So, [tex][CH_{3}COOH] = \frac{0.05 \times 100}{100 + V}[/tex]
and, [tex][CH_{3}COONa] = \frac{0.05 \times V}{100 + V}[/tex]
Expression for the buffer solution is as follows.
pH = [tex]pK_{a} + log \frac{[CH_{3}COONa]}{[CH_{3}COOH]}[/tex]
5.1 = [tex]4.76 + log \frac{0.05 \times V}{0.05 \times 100}[/tex]
0.34 = log V - 2
log V = 2.34
or, V = 218.77 ml
Now, we will calculate the molarity of the buffer with respect to acetate as follows.
= [tex][CH_{3}COO^{-}] + [CH_{3}COOH][/tex]
= [tex]\frac{0.05 \times 218.77}{318.77} + \frac{0.05 \times 100}{318.77}[/tex]
= 0.0499 M
or, = 0.05 M (approx)
Thus, we can conclude that molarity of the resulting buffer with respect to acetate is 0.05 M.
Final answer:
To make a buffer of pH 5.1 using 0.05 M acetic acid, the Henderson-Hasselbalch equation is used to calculate the amount of 0.05 M sodium acetate needed. The result is based on achieving the correct ratio of acetate ion to acetic acid. The molarity of the resulting buffer depends on the total moles of acetate and acid in the final solution volume.
Explanation:
To determine how many mL of 0.05 M sodium acetate should be added to 100 mL of 0.05M acetic acid to make a buffer of pH 5.1, we can use the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the acetate ion and [HA] is the concentration of acetic acid. Given that the pKa of acetic acid is 4.76, plugging in the pH 5.1 gives us the equation 5.1 = 4.76 + log([A-]/[0.05]).
Solving for [A-], we find that the ratio of [A-] to [HA] needed is approximately 2.2. Since the volume of the acetic acid solution is 100 mL and its concentration is 0.05M, to achieve the desired ratio, the amount of sodium acetate needed can be calculated based on the molarity and the final volume of the solution. The resulting molarity of the buffer with respect to acetate (acetate + acetic acid) will be determined by the total moles of acetate ions and acetic acid divided by the total volume of the solution after the addition of sodium acetate.
Net Ionic Equations for mixing Strong Acids with Strong Bases Consider a reaction between hydrochloric acid and potassium hydroxide.
a) In an aqueous solution of hydrochloric acid, check all of the major species found in solution (ignore the trace hydronium ions and hydroxide ions that would come from the autoionization of water). HCI (1) CIo (aq) CI (aq) OH (ag)H20 () H3o (aq)
b) In an aqueous solution of potassium hydroxide, check all of the major species found in solution (ignore the trace hydronium ons and hydroxide ions that would come from the autoionization of water). H20 () K+ (aq) OH (aq) Hyo (aq) KOH (5)
c) When the acid and base react together, they will neutralize each other to form water and a salt. Give the chemical formula for the salt formed. chemPad Help Greek acid and potassium hydroxide. Write it out in this order. Remember that, by convention, a net ionic equation has a single reaction arrow
d) When you cancel out the spectator ions, what is the net ionic equation that remains for the reaction between hydrochloric Help chemPad Oreek ▼
Answer:
sorry but I am just answering the questions because I need points
Explanation:
thank you
A chemist adds of a sodium carbonate solution to a reaction flask. Calculate the mass in grams of sodium carbonate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.
The question is incomplete, here is the complete question:
A chemist adds 180.0 mL of a 1.42 M sodium carbonate solution to a reaction flask. Calculate the mass in grams of sodium carbonate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.
Answer: The mass of sodium carbonate that must be added are 40.9 grams
Explanation:
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
Molar mass of sodium carbonate = 106 g/mol
Molarity of solution = 1.42 M
Volume of solution = 180.0 mL
Putting values in above equation, we get:
[tex]1.42mol/L=\frac{\text{Mass of sodium carbonate}\times 1000}{160g/mol\times 180}\\\\\text{Mass of sodium carbonate}=\frac{160\times 1.42\times 180}{1000}=40.9g[/tex]
Hence, the mass of sodium carbonate that must be added are 40.9 grams
Write the balanced chemical equation for the combustion of ethane, , and answer these questions. (Use the lowest possible coefficients. Omit states of matter.) How many molecules of oxygen would combine with 16 molecules of ethane in this reaction
Answer:
1. 2C2H6 + 7O2 —> 4CO2 + 6H2O
2. 56moles of O2
Explanation:
Ethane undergo Combustion to produce CO2 and H20 according the equation below:
C2H6 + O2 —> CO2 + H2O
Let us balance the equation. There are 6 atoms of H on the left side and 2 atoms on the right side. It can be balanced by putting 3 in front of H2O as shown below:
C2H6 + O2 —> CO2 + 3H2O
There are 2 atoms of C on left and 1atom on the right. It can be balanced by putting 2 in front of CO2 as shown below:
C2H6 + O2 —> 2CO2 + 3H2O
There are a total of 7 atoms of O on the right and 2 atoms on the left. It can be balanced by putting 7/2 in front of O2 as shown below:
C2H6 + 7/2O2 —> 2CO2 + 3H2O
Now we multiply through by 2 to remove the fraction as shown below
2C2H6 + 7O2 —> 4CO2 + 6H2O
Now the equation is balanced
2. 2C2H6 + 7O2 —> 4CO2 + 6H2O
From the equation above,
2 moles of ethane(C2H6) combined with 7moles of O2.
Therefore, 16moles of ethane(C2H6) will combine with = (16x7)/2 = 56moles of O2
Place the following substances in order of increasing boiling point. Ne Cl2 O2
a. O2 < Cl2 < Ne
b. Cl2 < Ne < O2
c. Cl2 < O2 < Ne
d. Ne < O2 < Cl2
e. Ne < Cl2 < O2
Answer: = D
Explanation:
The atomic mass increases from Ne to O2 to Cl2 hence the boiling point also increases, therefore
Ne < O2 < Cl2
A student has 500.0 mL of a 0.1133 M aqueous solution of BaI2 to use in an experiment. She accidentally leaves the container uncovered and comes back the next week to find only a solid residue. The mass of the residue is 28.28 g. Determine the chemical formula of this residue.
Answer:
[tex]BaI_2\ ^.6H_2O[/tex]
Explanation:
Hello,
In this case, by knowing the volume and the molarity of the barium iodide, is it possible to compute the residue's mass as shown below:
[tex]m_{BaI_2}=0.500L*0.1133\frac{molBaI_2}{L}*\frac{391.136gBaI_2}{1molBaI_2} =22.16gBaI_2[/tex]
Nevertheless, the obtained value is lower than the obtained by 6.133 g which means that mass corresponds to water forming a hydrate. In such a way, one could know how many waters in form of hydrate remain with the residue by a trial-error procedure as shown below:
[tex]m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+18)gBaI_2\ ^.H_2O}{1molBaI_2} =23.17g\rightarrow No\\m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+2*18)gBaI_2\ ^.2H_2O}{1molBaI_2} =24.20g\rightarrow No\\m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+3*18)gBaI_2\ ^.3H_2O}{1molBaI_2} =25.22g\rightarrow No\\m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+4*18)gBaI_2\ ^.4H_2O}{1molBaI_2} =26.24g\rightarrow No\\[/tex]
[tex]m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+5*18)gBaI_2\ ^.5H_2O}{1molBaI_2} =27.26g\rightarrow No\\m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+6*18)gBaI_2\ ^.6H_2O}{1molBaI_2} =28.28g\rightarrow Yes\\[/tex]
Therefore, the formula is barium iodide hexahydrate:
[tex]BaI_2\ ^.6H_2O[/tex]
Best regards.
One way to represent this equilibrium is: N2O4(g)2NO2(g) Indicate whether each of the following statements is true, T, or false, F. AT EQUILIBRIUM we can say that: 1. The concentration of NO2 is equal to the concentration of N2O4. 2. The rate of the dissociation of N2O4 is equal to the rate of formation of N2O4. 3. The rate constant for the forward reaction is equal to the rate constant of the reverse reaction. 4. The concentration of NO2 divided by the concentration of N2O4 is equal to a constant.
Answer:
1. The concentration of [tex]NO_2[/tex] is equal to the concentration of [tex]N_2O_4[/tex] : False
2. The rate of the dissociation of [tex]N_2O_4[/tex] is equal to the rate of formation of [tex]N_2O_4[/tex]: True
3. The rate constant for the forward reaction is equal to the rate constant of the reverse reaction: False
4. The concentration of [tex]NO_2[/tex] divided by the concentration of [tex]N_2O_4[/tex] is equal to a constant : False
Explanation:
[tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]
The concentrations of reactant and product is constant and not equal.
The rate of forward and backward reactions are equal. Thus rate of the dissociation of [tex]N_2O_4[/tex] is equal to the rate of formation of [tex]N_2O_4[/tex]
For a reversible reaction, the equilibrium constant for the forward reaction is inverse of the equilibrium constant for the backward reaction and not equal.
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.
[tex]K_c=\frac{[NO_2]^2}{[N_2O_4]}[/tex]
Final answer:
Statements 1 and 3 are False, and 2 and 4 are True. At equilibrium, the rate of formation and dissociation of N₂O₄ are equal but their concentrations are not necessarily the same. Rate constants for forward and reverse reactions are different yet define the equilibrium constant. The concentration ratio of NO₂ to N₂O₄ is constant at a given temperature.
Explanation:
When dinitrogen tetroxide (N₂O₄) is in equilibrium with nitrogen dioxide (NO₂), the following statements can be made:
1. The concentration of NO₂ is equal to the concentration of N₂O₄. False - The stoichiometry of the reaction is 1:2, so for every molecule of N₂O₄ that dissociates, two molecules of NO₂ are formed.2. The rate of the dissociation of N₂O₄ is equal to the rate of formation of N₂O₄. True - At equilibrium, the rate of the forward reaction (dissociation of N₂O₄) is equal to the rate of the reverse reaction (formation of N₂O₄).3. The rate constant for the forward reaction is equal to the rate constant of the reverse reaction. False - The rate constants for the forward and reverse reactions are generally different, but their ratio is constant at a given temperature, defining the equilibrium constant.4. The concentration of NO₂ divided by the concentration of N₂O₄ is equal to a constant. True - At equilibrium, this ratio is equal to the equilibrium constant (Keq) for the reaction at a given temperature.Check the box next to each molecule on the right that has the shape of the model molecule on the left: model molecules (check all that apply X 5 ? | O CH20 CNH, You can drag the slider to rotate the model molecule. + 1 O Brf 4 CH2Cl2 Note for advanced students: the length of bonds and size of atoms in the model is not necessarily realistic. The model is only meant to show you the general geometry and 3D shape of the molecule.
Answer:
[tex]NH^{+} _{4}[/tex] ammonium ion.
Explanation:
Check the box next to each molecule on the right that has the shape of the model molecule on the left: model molecules (check all that apply X 5 ? | O CH20 CNH, You can drag the slider to rotate the model molecule. + 1 O Brf 4 CH2Cl2 Note for advanced students: the length of bonds and size of atoms in the model is not necessarily realistic. The model is only meant to show you the general geometry and 3D shape of the molecule.
when you look at the diagram from the source page
one can conclude that the diagram is tetrahedral and the angle between the molecules is 109.5 deg
There is one central atom bonded to four atoms in a tetrahedral molecule .it has no lone electron pairs.m
NH4+ is the answer
Other molecules that are tetrahedral in shape are methane ion and phosphate ion.
Answer:
CH3O- , BrF4- and NH4+ have tetrahedral geometry on the basis of their electron domain geometry..
Explanation:
The object on the picture as shown on the fig below describes a compound with a total of 4 pair electron domain with it's electron domain typically described as tetrahedral.
The task is to sort out which of those in the options fort into the category.
Although NH3 and NH4+ ion both have the SP3 hybridization their electron pair geometry differs. In the NH3 molecule one lone pair and three bond pairs are present. While Distortion is caused by repulsion between lone pair and bond pair the geometry of NH3 causes it to become pyramidal in NH4+ irrespective of possessing the sp3 hybridization.
Its resulting trigonal pyramidal geometry is thus described as tetrahedral. It consequently has 3 bonding domains and 1 nonbinding domain.
CH3O- is also tetrahedral with an idealized bond angle of 109.5°.
BrF4- It has 4 bond pair present hence the tetrahedral geometry. The presence of two lone pair makes it square planar described sometimes as AE2X4. It has 6 electron regions.
C2Cl4 has a linear geometry it has one triple bond and two single bonds this giving hints that its coordinate and steric number is 2 and its bond angle is 180°.
So, CH3O- , BrF4- and NH4+ have tetrahedral geometry