Answer:
I should use a volumetric flask.
Explanation:
If the accuracy of the concentration is important, we need to use a volumetric flask.
Answer:
The glassware appropriate for dilution preparation is a (250 mL) volumetric flask
Explanation:
We know that molarity or concentration of a solution is the number of moles per litres of solution, the molarity and volume of the solution in question can be used to find out how much of the stock solution to be diluted using the formula;
[tex]c = \frac{n}{v}[/tex]
Once the variables are confirmed, a 250 mL solution of 1.50M HCl can be prepared by measuring determined stock solute into a 250mL volumetric flask and the making it upto the 250mL mark. (a measuring cylinder can be used to check accuracy of volume added).
use the weak acid HCHO2 to defend or contradict the statement "A weak acid has a stronger conjugate base". Let me know if you are using HCHO2 to defend the statement or contradict it. Then use equations/formulas and no more than 2 sentences to do so.
Answer:
Formic acid can react to water, to give protons to medium:
HCOOH + H₂O → H₃O⁺ + HCOO⁻
Weak acid Base Acid Strong conjugate base
HCOO⁻ + H₂O ⇄ HCOOH + OH⁻ Kb
Strong base Acid Weak acid Base
So the formate can take a proton from water to become formic acid again and that's why it is a conjugate strong base.
Explanation:
HCOOH → Formic acid
To determine the conjugate pair and to know if they are weak or strong, we should know, if they can react to water. This is called hydrolisis.
For example: formic acid is a weak acid, so the formed formate will be its conjugate base and it will be strong because the formate can react to water, to make formic again.
Weak acid → Strong conjugate base
Strong acid → Weak conjugate base
Weak base → strong conjugate acid
Strong base → weak conjugate acid
For example HCl is a strong acid. When it is in aqueous solution, we have protons and chlorides.
HCl + H₂O → H₃O⁺ + Cl⁻
Chloride will be the weak conjugate base, because it can't react to water.
We can not make HCl again, according to this equation:
Cl⁻ + H₂O ← HCl + OH⁻ This is impossible.
Formic acid can react to water, to give protons to medium:
HCOOH + H₂O → H₃O⁺ + HCOO⁻
Weak acid Base Acid Strong conjugate base
So the formate can take a proton from water to become formic acid again and that's why it is a conjugate strong base.
HCOO⁻ + H₂O ⇄ HCOOH + OH⁻ Kb
Strong base Acid Weak acid Base
HCHO2 can be used to defend the statement 'A weak acid has a stronger conjugate base' as its conjugate base CHO2- readily accepts a proton thus demonstrating its strength.
I'm using HCHO2 to defend the statement 'A weak acid has a stronger conjugate base'. The weak acid formula for HCHO2 is HCHO2 → H+ + CHO2-.
The conjugate base here is CHO2-, which gains a proton (H+) to become HCHO2, demonstrating it's stronger since it readily accepts a proton.
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Why have some chemicals been banned from usage in agriculture/food production
Answer:
okay so to tell the story short:
Explanation:
over the years farmers in agriculture businesses have decide to use safer chemicals. because since people have been getting food poisoning and been getting sick. but now farmer businesses now use some chemicals that have very low toxicity. When pesticides were first introduced, farmers were using chemicals that were very toxic. but then that's when they realised that the they had to be removed from the application and today they have been replaced with better and healthier ones just like (glyphosate.)
Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 x 10^-5), with 0.1000 M NaOH solution after the following additions of titrant.
(a) 10.00 mL: pH =
(b) 20.10 mL: pH =
(c) 25.00 mL: pH =
Answer:
pH after the addition of 10 ml NaOH = 4.81
pH after the addition of 20.1 ml NaOH = 8.76
pH after the addition of 25 ml NaOH = 8.78
Explanation:
(1)
Moles of butanoic acid initially present = 0.1 x 20 = 2 m moles = 2 x 10⁻³ moles,
Moles of NaOH added = 10 x 0.1 = 1 x 10⁻³ moles
CH₃CH₂CH₂COOH + NaOH ⇄ CH₃CH₂CH₂COONa + H₂O
Initial conc. 2 x 10⁻³ 1 x 10⁻³ 0
Equilibrium 1 x 10⁻³ 0 1 x 10⁻³
Final volume = 20 + 10 = 30 ml = 0.03 lit
So final concentration of Acid = [tex]\frac{0.001}{0.03} = 0.03mol/lit[/tex]
Final concentration of conjugate base [CH₃CH₂CH₂COONa][tex]=\frac{0.001}{0.03} = 0.03 mol/lit[/tex]
Since a buffer solution is formed which contains the weak butanoic acid and conjugate base of that acid .
Using Henderson Hasselbalch equation to find the pH
[tex]pH=pK_{a}+log\frac{[conjugate base]}{[acid]} \\\\=-log(1.54X10^{-5} )+log\frac{0.03}{0.03} \\\\=4.81[/tex]
During titration of butanoic acid with NaOH, we can calculate the pH at various points using the Henderson-Hasselbalch equation for buffer scenarios. After 10.00mL of NaOH, the pH will be 4.74. After 20.10 mL, the pH will be 8.27, and after 25.00 mL, the pH will be 12.30.
Explanation:This involves calculating the pH at various stages during a titration procedure. Here the titration involves a weak acid, butanoic acid, with a strong base, NaOH. We can simplify the reaction as follows: CH₃CH₂CH₂COOH + OH- --> CH₃CH₂CH₂COO- + H₂O.
(a) After 10.00 mL of NaOH is added, the system isn't at equivalence. Here, the reaction hasn't fully completed and a buffer solution is present. Using the Henderson-Hasselbalch equation, we can find the pH: pH = pKa + log([base]/[acid]). After calculating, we can find pH = 4.74.
(b) After 20.10 mL NaOH is added, the system reaches past equivalence. The pH can be determined by finding pOH using the remaining OH- concentration and then subtracting from 14. After the calculation, the pH = 8.27.
(c) For 25.00 mL of NaOH, the system is beyond equivalence and extra OH- ions increase pH. The pH calculation is like previous step and the result will be pH = 12.30.
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why does temperture stay constant during a phase change
Answer:
Explanation:
During a phase change, the temperature remains constant because the energy supplied via heat is used differently. While not changing states, the heat (energy) goes into changing the kinetic energy of every particle in the body that receives it.
The following reaction is exothermic. C6H12O6(s)+6O2(g)⇌6CO2(g)+6H2O(g)C6H12O6(s)+6O2(g)⇌6CO2(g)+6H2O(g) Predict the effect (shift right, shift left, or no effect) of increasing and decreasing the reaction temperature.
Answer:
According to Le Chatelier's principle, increasing the reaction temperature of an exothermic reaction causes a shift to the left and decreasing the reaction temperature causes a shift to the right.
Explanation:
C6H12O6(s) + 6O2(g) ⇌6CO2(g) + 6H2O(g)
We are told that the forward reaction is exothermic, meaning heat is removed from the reacting substance to the surroundings.
According to Le Chatelier's principle,
1. for an exothermic reaction, on increasing the temperature, there is a shift in equilibrium to the left and formation of the product is favoured.
2. if the temperature of the system is decreased, the equilibrium shifts to right and the formation of the reactants is favoured.
3. if the reaction temperature is kept constant, the system is at equilibrium and there is no shift to the right nor to the left.
Answer:
Increasing temperature = balance will shift to the left
Decreasing temperature = balance will shift to the right
Explanation:
Step 1: Data given
The increase or decrease in temperature can have an influence on the position of the equilibrium.
If the temperature is increased, the system will ensure that less heat is released. So the balance will shift to the left.
When the temperature drops, however, the system will produce more heat: the balance will shift to the right.
Step 2: The balanced equation
C6H12O6(s) + 6O2(g) ⇌ 6CO2(g) + 6H2O(g)
This is an endothermic reaction
Step 3: Increasing the temperature
If the temperature were increased, the heat content of the system would increase.
In exothermic reactions, increase in temperature decreases the K value. This means less products will be formed. The balance will shift to the left.
Step 4: Decreasing the temperature
If the temperature were decreased, the heat content of the system would increase.
In exothermic reactions, decrease in temperature increases the K value. This means more products will be formed, less reactants. The balance will shift to the right.
Salicylic acid will reach with methanol (CH3OH), and form methyl salicylate. This is another esterification reaction, only this reaction occurs with the carboxylic acid group (-COOH), not the alcohol (-OH) group on salicylic acid. Draw the reaction, circle the ester group in the product. Make sure the reaction is balanced (there is a second product.)
Answer:
C7H603 + CH3OH --> C8H803 + H2O
Balanced equation with corresponding stoichiometric numbers.
Explanation:
the salicylic acid, also called aspirin, when reacting with methanol produced an irreversible reaction giving methyl salicylate and water
For the decomposition of hydrogen iodide on a gold surface at 150 °C 2 HI(g)H2(g) + I2(g) the average rate of disappearance of HI over the time period from t = 0 s to t = 3646 s is found to be 1.20×10-4 M s-1.\
Answer:
The average rate of formation of I₂(g) is 6.00x10⁻⁵Ms⁻¹
Explanation:
The question is:
The average rate of formation of I2 over the same time period is______M s-1.
Based in the reaction:
2 HI(g) ⇄ H₂(g) + I₂(g)
If 2 moles of HI(g) disappears in a rate of 1.20x10⁻⁴Ms⁻¹, 1 mole of I₂(g) will appears at:
1 mole I₂(g) × (1.20x10⁻⁴Ms⁻¹ / 2 mol) = 6.00x10⁻⁵Ms⁻¹ I₂(g)
Calculate the [H+] in a solution that is 0.803 M in NaX and 0.677 M in HX given that the Ka of HX is 8.64 ⋅ 10 − 7 8.64⋅10-7. Report your answer in scientific notation to 3 sig figs.
The [H+] in the solution is 6.94752 * 10^−7 M.
Explanation:To calculate the [H+] in the given solution, we need to first calculate the concentration of HX and then use the equilibrium constant (Ka) to find the concentration of [H+]. The equation for the dissociation of HX is: HX ⇌ H+ + X-. Since the dissociated amount is small compared to 0.803 M, we can assume that the concentration of HX is approximately 0.803 M. Now, using the equation for Ka and the concentration of HX, we can find the concentration of [H+]:
Ka = [H+][X-] / [HX]
[H+] = Ka * [HX] = (8.64 * 10^−7) * 0.803 = 6.94752 * 10^−7 M
Therefore, the [H+] in the solution is 6.94752 * 10^−7 M.
How many joules of energy are needed to change 37.5g of ice at 0.00 oC to water at 45.0oC?
(Heat of fusion = 335J/g,
Specific heat of liquid water= 4.184J/g.oC,
Heat of vaporization = 2259 J/g
Answer:
19,623 J or 19.6 kJ of heat is needed to change ice at 0°C to water at 45°C
Explanation:
To calculate the energy needed to change 37.5g of ice at 0°C to water at 45.0°C, we obtain the individual values of energy needed to convert the ice from 0° to water at 0 °C and the value of energy needed to convert the water from 0 °C to water at 45°C and then add the values together.
Heat (q) = mΔHf + mCpΔT
So;
1. heat needed to change from solid to liquid = m ΔHf
q = 37.5 * 335
q = 12,562.5 Joules
2. heat needed to convert the water at 0C to water at 45 C
q = mcΔT
q= 37.5 * 4.184 * ( 45-0)
q = 37.5 * 4.184 * 45
q = 7,060.5 J
The heat needed to change the ice to water at 45 C = 12, 562.5 + 7.060.5 = 19,623 J or 19.6 kJ of heat.
Answer:
We need 19620 joules
Explanation:
Step 1: Data given
Mass of ice = 37.5 grams
Temperature of ice = 0.00 °C
Final temperature of water = 45.0°C
(Heat of fusion = 335J/g,
Specific heat of liquid water= 4.184J/g°C
Heat of vaporization = 2259 J/g
Step 2: Calculate the energy needed to melt ice to water at 0°C
Q = m*ΔHfus
Q = 37.5 grams * 335J/g
Q = 12562.5 J = 12.56 kJ
Step 3: Calculate energy needed to heat water from 0 to 45 °C
Q = m*c*ΔT
⇒with Q = the enegy needed to heat water from 0 to 45 °C
⇒with m =the mass of water = 37.5 grams
⇒with ΔT = the change of temperature = 45 °C
⇒with c = the specific heat of water = 4.184 J/g°C
Q = 37.5g * 4.184 J/g°C * 45 °C
Q = 7060.5 J = 7.06 kJ
Step 4: Calculate the total heat needed
Total heat = 12.56 kJ + 7.06 kJ
Total heat = 19.62 kJ = 19620 J
We need 19620 joules
What is the percent composition by mass of ammonium chloride in a saturated solution of ammonium chloride at 45°C
Explanation:
To alive this you need the solubility product constant of NH4Cl at 45°C
The percent composition by mass of ammonium chloride in a saturated solution at 45°C is approximately 27.17%.
To find the percent composition, we use the formula:
[tex]\[ \text{Percent composition by mass} = \left( \frac{\text{mass of solute}}{\text{mass of solution}} \right) \times 100 \][/tex]
Given that the mass of ammonium chloride (the solute) is 37.2 grams and the mass of water (the solvent) is 100 grams, the total mass of the solution is the sum of the mass of the solute and the mass of the solvent:
[tex]\[ \text{Mass of solution} = \text{Mass of solute} + \text{Mass of solvent} \][/tex]
[tex]\[ \text{Mass of solution} = 37.2 \text{ g} + 100 \text{ g} \][/tex]
[tex]\[ \text{Mass of solution} = 137.2 \text{ g} \][/tex]
Now, we can calculate the percent composition:
[tex]\[ \text{Percent composition by mass} = \left( \frac{37.2 \text{ g}}{137.2 \text{ g}} \right) \times 100 \][/tex]
[tex]\[ \text{Percent composition by mass} = \left( \frac{1}{3.68} \right) \times 100 \][/tex]
[tex]\[ \text{Percent composition by mass} \approx 27.17 \% \][/tex]
For each pair below, select the sample that contains the largest number of moles.
Pair A
2.50 g O2
2.50 g N2
Answer:2.50 gN2
Explanation:
Pair A is:
2.50 g N2
Pair B is:
21.5 g N2
Pair C is:
0.081 CO2
Ammonium perchlorate is the solid rocket fuel used by the U.S. Space Shuttle. It reacts with itself to produce nitrogen gas , chlorine gas , oxygen gas , water , and a great deal of energy. What mass of oxygen gas is produced by the reaction of 4.1 g of ammonium perchlorate?
Answer:
1.12 g of oxygen.
Explanation:
First step is to determine the decomposition reaction:
Ammonium perchlorate → NH₄ClO₄
2NH₄ClO₄ (s) → N₂(g) + Cl₂(g) + 2O₂(g) + 4H₂O(g)
We define the moles of the reactant:
4.1 g . /117.45 g/mol = 0.0349 moles
As ratio is 2:2, If I use 0.0349 moles of perchlorate I would produce the same amount of moles of oxygen.
0.0349 mol of O₂ are produced by the reaction
We convert to mass: 0.0349 mol . 32 g /1mol = 1.12 g
The acetyl group _______ some of the electron density _____ the nitrogen, making it much _____ basic; the nitrogen of this amide is ________ under the reaction conditions. The N retains enough electron density to share with the benzene ring, so the NHCOCH3 group is still an activating ________-director, though weaker than NH2.
Answer:
The gaps are filled using the following;
1. removes
2. from
3. less
4. Protonated
5. Ortho, para
Explanation:
Electron density is removed from nitrogen in acetyl group which is why it is less basic
The complete sentences is now written as;
The acetyl group removes some of the electron density from the nitrogen, making it much less basic; the nitrogen of this amide is protonated under the reaction conditions. The N retains enough electron density to share with the benzene ring, so the NHCOCH3 group is still an activating ortho, para director, though weaker than NH2.
0.20 mol solute 0.300 kg solvent
calculate the molality of the solution
The molality of the solution is 0.67 m, which is calculated using the moles of solute and mass of the solvent. The molar mass of the solute would depend on additional data such as osmotic pressure and temperature, and requires using the van't Hoff equation.
Explanation:Calculating Molality and Molar Mass
To calculate the molality of a solution, which is the number of moles of solute per kilogram of solvent, you can use the formula:
Molality (m) = Moles of solute \/ Mass of solvent in kilograms
According to the given information, we have 0.20 moles of solute dissolved in 0.300 kg of solvent. Therefore, using the formula, we get:
Molality = 0.20 mol \/ 0.300 kg = 0.67 m (molal)
To determine the molar mass of a solute B from the given osmotic pressure, temperature, and the amount of solute (0.200 g) dissolved in a known molar volume (0.0180 L/mol) of solvent (1.00 mol), use the van't Hoff factor (i = 1 for non-electrolytes) and the equation:
y = i * (M/R * T)
Where,
n = number of moles of solute
R = gas constant (0.0821 L atm mol^-1 K^-1)
T = temperature in Kelvin
y = osmotic pressure of the solution
From the osmotic pressure formula, we can rearrange to solve for n, and then use the mass of the solute to find the molar mass:
n = y * V/(R * T)
and
Molar mass = mass of solute \/ n
The molar mass calculation will depend on more specific data such as the osmotic pressure and temperature data provided in your hypothetical variation of this exercise. To complete the calculation, one would need to work through the osmotic pressure formula with the given measurements (0.640 atm at 298 K).
The rate constant for a first order reaction is 0.060s-1. when the temperature is increased from 298K to 331 K , the rate constant increases to 0.18s-1. Calculate the activation energy for this reaction. Use kJ/mol for the units. The activation energy in kJ/mol equals (include the units in your answer):
Answer:
27.3 kJ/mol
Explanation:
You would use the Arrhenius Equation to solve this problem.
[tex]ln(\frac{k_{2} }{k_{1} } )=\frac{E_{a} }{R} (\frac{1}{T_{1} } -\frac{1}{T_{2} } )[/tex]
Plug the rate constants into k1 and k2. Plug the temperatures into T1 and T2. R is a constant and is equal to 8.314. Solve for Ea.
Your answer should be 27,303.03 J/mol or 27.3 kJ/mol.
Hope this helps! <3
The activation energy will be equal to "12.40 kJ/mol". To understand the calculation, check below.
Activation energyAccording to the question,
For 1st order, rate constant = 0.060 s⁻¹
Temperature, T₁ = 298 K
T₂ = 331 K
Increased rate constant = 0.18s⁻¹
We know the formula,
→ ln k = ln A - [tex]\frac{E_a}{RT}[/tex]
By substituting the rate constant value,
ln 0.06 = ln A - [tex]\frac{E_a}{298 \ R}[/tex] ...(equation 1)
ln 0.18 = ln A - [tex]\frac{E_a}{318 \ R}[/tex] ...(equation 2)
From "equation 1 and 2", we get
Activation energy, 1.09 = [tex]\frac{83 \ E_a}{113538 \ R}[/tex]
[tex]E_a[/tex] = 12.40 kJ/mol
Thus the above answer is correct.
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Calculate the concentration of the lactate ion in a solution that is 0.100 Min lactic acid (CH3CH(OH)COOH, pKa = 3.86) and 0.080 Min HCl.
Answer:
Explanation:
pKa = 3.86
-log Ka = pKa
- log Ka = 3.86
Ka = 1 / ( 10^(3.86)) = 1.38 × 10⁻⁴
lactic → (H⁺) + lactate
but Ka ( equilibrium constant ) = ( H⁺) (lactate) / ( lactic acid)
when HCl dissociate, it forms
HCl → H ⁺ + Cl⁻
0.08 M of HCl will yield 0.08 M H⁺, lactate ion = 0.1 M
1.38 × 10⁻⁴ = (0.08) (lactate) / 0.1
(1.38 × 10⁻⁴ × 0.1) / 0.08 = lactate
lactate = 1.73 × 10⁻⁴
Ammonia gas combines with hydrogen chloride gas, forming solid ammonium chloride.a.Write a balanced chemical equation for this reaction.b.In a reaction mixture of 3.0 g ammonia and 5.0 g hydrogen chloride, which of the two is the limiting reagent?c.How many grams of ammonium chloride could form from the reaction mixture in part b?d.How much of the reactant is left over in the reaction mixture o
Answer:
a. NH₃(g) + HCl(g) → NH₄Cl(s)
b. HCl is the limiting reagent
c. 7.35 g of NH₄Cl
d. 0.039 moles of NH₃ remains after the reaction is complete (0.663 grams of ammonia)
Explanation:
First of all, we define the reaction where the reactants are the ammonia and the hydrogen chloride. The product is the ammonium chloride.
NH₃(g) + HCl(g) → NH₄Cl(s)
First of all, we need to determine the limiting reactant. As the ratio is 1:1, the compound that has the lowest mol, will be the limiting.
3 g / 17g/mol = 0.176 moles NH₃
5 g / 36.45 g/mol = 0.137 moles HCl (limiting reactant)
For 0.176 moles of ammonia, we need the same of HCl, but we have 0.137 moles. We convert the moles to mass, in order to know how many grams of NH₄Cl are produced → 0.137 mol . 53.45 g /1mol = 7.32 g
As ratio is 1:1, and the limiting reagent is the HCl, after the reaction goes complete, some ammonia remains. This is because, the ammonia is the excess reagent.
0.176 mol - 0.137 moles = 0.039 moles of NH₃ remains after the reaction is complete; we convert the moles to mass → 0.039 mol . 17 g /mol = 0.663 g are left over in the reaction mixture.
Final answer:
The balanced chemical equation for the reaction between ammonia and hydrogen chloride to form ammonium chloride is NH3(g) + HCl(g) → NH4Cl(s). Ammonia is the limiting reagent in the given mixture, and the theoretical yield of ammonium chloride is 9.42 g. No excess reactant will be left over after the reaction.
Explanation:
Chemical Reaction Between Ammonia and Hydrogen Chloride
When ammonia gas (NH3) combines with hydrogen chloride gas (HCl), solid ammonium chloride (NH4Cl) is formed. The balanced chemical equation for this reaction is:
NH3(g) + HCl(g) → NH4Cl(s).
Limiting Reagent and Yield
In a reaction mixture with 3.0 g of ammonia and 5.0 g of hydrogen chloride, to find the limiting reagent we must compare the mole ratio of the reactants to their stoichiometric ratio in the balanced equation. Using molar masses (NH3: 17.03 g/mol, HCl: 36.46 g/mol), we find that:
NH3: 3.0 g × (1 mol / 17.03 g) ≈ 0.176 moles
HCl: 5.0 g × (1 mol / 36.46 g) ≈ 0.137 moles
The stoichiometry of the reaction is 1:1, so ammonia is the
limiting reagent
because we have fewer moles of it relative to HCl.
The mass of ammonium chloride formed is based on the limiting reagent. Since all of the ammonia will react, the theoretical yield of NH4Cl is the moles of NH3 times the molar mass of NH4Cl (53.49 g/mol):
0.176 moles × 53.49 g/mol ≈ 9.42 g of NH4Cl.
Excess Reactant Left Over
To determine the amount of excess reactant left over, we subtract the number of moles of HCl that reacted (equivalent to the moles of NH3) from the initial moles of HCl:
0.137 moles HCl - 0.176 moles NH3 = -0.039 moles HCl (not possible, indicating no excess HCl remains).
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:
2 S(s) + 3 O2(g) → 2 SO3(g) ΔH°rxn = ?
Given:
SO2(g) → S(s) + O2(g) ΔH°rxn = +296.8 kJ
2 SO2(g) + O2(g) → 2 SO3(g) ΔH°rxn = -197.8 kJ
a. -494.6 kJ
b. -692.4 kJ
c. -791.4 kJ
d. 1583 kJ
Answer:
The answer is -791.5 kJ (option c)
Explanation:
You know:
SO₂ (g) → S (s) + O₂ (g) ΔH°rxn = +296.8 kJ
2 SO₂ (g) + O₂ (g) → 2 SO₃ (g) ΔH°rxn = -197.8 kJ
You must add them to obtain the desired equation:
2 S(s) + 3 O₂(g) → 2 SO₃(g) ΔH°rxn = ?
You want to calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction. The calculation is made using Hess's law. This law states: when the reactants are converted to products, the enthalpy change is the same, regardless of whether the reaction is carried out in one step or in a series of steps.
In Hess's law he explains that the enthalpy changes are additive, ΔHneta = ΣΔHr and contains three rules:
If the chemical equation is inverted, the ΔH symbol is inverted as well. If the coefficients are multiplied, multiply ΔH by the same factor. If the coefficients are divided, divide ΔH by the same divisor.The sum of the fitted equations should give the problem equation. In this case:
2*( S(s) + O₂(g) → SO₂ ) To obtain the desired reaction, this equation must be inverted, so the enthalpy value is also inverted. It must also be multiplied by 2, then the whole equation is multiplied, both reactants and products and the value of the enthalpy. So ΔH°rxn = (-296.8 kJ)*2= -593.6 kJ
2 SO₂(g) + O₂(g) → 2 SO₃ (g) ΔH°rxn = -197.8 kJ
____________________________________________________
2 S(s) + 3 O₂(g) → 2 SO₃(g) ΔH°rxn = -791.4 kJ ( Enthalpies are added algebraically)
The answer is -791.5 kJ (option c)
The value of ΔH°rxn for the 2S(s) + 3O₂(g) → 2SO₃(g) is -791.4 kJ.
What is enthalpy?Enthalpy of the reaction tells about the amount of heat released or absorbed during any chemical reaction.
Given that,
2SO₂(g) + O₂(g) → 2SO₃(g) ΔH°rxn = -197.8 kJ
SO₂(g) → S(s) + O₂(g) ΔH°rxn = 296.8 kJ
To obtain the required reaction first we have to invert the second equation and multiply that by 2 and then add with the left hand side as well as the right hand side of the reaction, we get:
2S(s) + 3O₂(g) → 2SO₃(g)
ΔH°rxn of the reaction will be calculated as-
ΔH°rxn = -2(296.8 kJ) + (-197.8 kJ)
ΔH°rxn = -791.4 kJ
Hence, option (C) is correct i.e. -791.4 kJ.
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Exactly 2.00 g of an ester A containing only C, H, and O was saponified with 15.00 mL of a 1.00 M NaOH solution. Following the saponification, the solution required 5.30 mL of 1.00 M HCl to titrate the unused NaOH. Ester A, as well as its acid and alcohol saponification products B and C, respectively, were all optically active. Compound A was not oxidized by K2Cr2O7, nor did compound A decolorize Br2 in CH2Cl2. Alcohol C was oxidized to acetophenone by K2Cr2O7. When acetophenone was reduced with NaBH4, a compound D was formed that reacted with the acid chloride derived from B to give two optically active compounds: A (identical to the starting ester) and E. Propose a neutral structure for each compound that is consistent with the data. Note that the absolute stereochemical configurations of chiral substances cannot be determined from the data. Arbitrarily draw a configuration for an enantiomer (and be consistent with the derivatives); omit wedge/dash bonds for a racemate.
Answer:
Explanation:
check the attachment for the propose neutral structure for each compound that is consistent with the data.
The structures are:
A: 1-phenylethyl lactate
B:lactic acid
C: 1-phenylethanol
D: 1-phenylethanol
E: Ester formed from 1-phenylethanol lactic acid chloride
To solve this problem, we will follow the information given in the question step by step to deduce the structures of compounds A, B, C, D, and E.
1. Saponification Reaction: The ester A reacts with NaOH to produce an alcohol and a carboxylate salt. The fact that 5.30 mL of 1.00 M HCl is required to titrate the unused NaOH tells us that the rest of the NaOH was used up in the saponification reaction. We can use the moles of NaOH reacted to find the moles of ester A that reacted.
First, calculate the moles of NaOH used in the saponification:
Since the reaction is 1:1, the moles of ester A are also 9.70 mmol.
2. Molecular Formula of A: Since we have 2.00 g of A a[tex]\[ \text{Moles of NaOH used} = (15.00 \text{ mL} - 5.30 \text{ mL}) \times \frac{1.00 \text{ mmol}}{\text{mL}} \] \[ \text{Moles of NaOH used} = 9.70 \text{ mmol} \][/tex]nd we know the moles of A, we can calculate the molar mass of A:
[tex]\[ \text{Molar mass of A} = \frac{\text{mass of A}}{\text{moles of A}} \] \[ \text{Molar mass of A} = \frac{2.00 \text{ g}}{9.70 \text{ mmol}} \] \[ \text{Molar mass of A} \approx 206.19 \text{ g/mol} \][/tex]
Given that A contains only C, H, and O, we can use the molar mass to propose a molecular formula. A common approach is to start with the simplest ratio and increase it until the calculated molar mass is close to the experimental value. For example, if we start with C₃H₆O₂ (molar mass 88.09 g/mol), we can see that it's not enough. We can keep increasing the number of carbons until we reach a molar mass close to 206.19 g/mol. Let's assume the formula is C₁₂H₂₄O₄ (molar mass 248.34 g/mol), which is a common ester formula that would give us a molar mass close to the experimental value.
3. Structure of A: Since A is not oxidized by K₂Cr₂O₇ and does not decolorize Br₂ in CH₂Cl₂, it suggests that A does not have any carbon-carbon double bonds or carbon-carbon triple bonds. Therefore, A is a saturated ester. Given that A is optically active, it must contain a chiral center. A simple saturated ester with a chiral center could be butanoic acid ethyl ester with a methyl group replaced by a hydroxyl group on the alpha carbon, making it chiral.
4. Structure of B and C: B is the acid product, and C is the alcohol product from the saponification of A. Since C is oxidized to acetophenone by K2Cr2O7, C must be a secondary alcohol with a benzene ring attached to the carbon adjacent to the hydroxyl group. Therefore, C is 1-phenylethanol. B, being the acid counterpart to C, must be benzoic acid.
5. Structure of D: D is formed by the reduction of acetophenone with NaBH4. This reaction would convert the ketone to a secondary alcohol. Therefore, D is 1-phenylethyl alcohol, which is the same as C.
6. Structure of E: E is formed when D reacts with the acid chloride derived from B. Since A is reformed, E must be the product of the reaction between benzoyl chloride (from B) and 1-phenylethyl alcohol (D). E is therefore a benzoate ester with a phenylethyl group.
Based on the above analysis, the proposed structures are:
A: 1-phenylethyl lactate
B:lactic acid
C: 1-phenylethanol
D: 1-phenylethanol
E: Ester formed from 1-phenylethanol lactic acid chloride
This proposal is consistent with the given data and the reactivity of the compounds involved.
Ammonia can be determined spectrophotometrically by reaction with phenol in the presence of hypochlorite (OCl−):
A 4.37-mg sample of protein was chemically digested to convert its nitrogen into ammonia and then diluted to 100.0 mL. Then 10.0 mL of the solution were placed in a 50-mL volumetric flask and treated with 5 mL of phenol solution plus 2 mL of sodium hypochlorite solution. The sample was diluted to 50.0 mL, and the absorbance at 625 nm was measured in a 1.00-cm cuvet after 30 min. For reference, a standard solution was prepared from 0.010 0 g of NH4Cl (FM 53.49) dissolved in 1.00 L of water. Then 10.0 mL of this standard were placed in a 50-mL volumetric flask and analyzed in the same manner as the unknown. A reagent blank was prepared by using distilled water in place of unknown.
Sample Absorbance at 625 nm
Blank 0.140
Reference 0.308
Unknown 0.592
(a) Calculate the molar absorptivity of the blue product.
(b) Calculate the weight percent of nitrogen in the protein.
Answer:
See explaination
Explanation:
Molar absorptivity, also known as the molar extinction coefficient, is a measure of how well a chemical species absorbs a given wavelength of light.
It is commonly used in chemistry and should not be confused with the extinction coefficient, which is used more often in physics.
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Assuming that the distances between the two ions are the same in all cases, which of the following ion pairs has the greatest electrostatic potential energy (i.e., largest in magnitude)? Please explain your answer.a.) Na+ - Cl- b.) Na+ - O-2. c.) Al+3 - O-2. d.) Mg+2-O-2 e.) Na- -Mg+2
Answer:
Correct option: C
Explanation:
As given in the question that distance between two ions are same in all cases hence r is same for all.
potential energy:
[tex]P.E =\frac{k\times q_{1} \times q_{2}}{r}[/tex]
therefore potential energy depend on the two charge muliplication
so higher the charge multiplication higer will be the potential energy.
magnitude of charge multiplication follow as:
a. 1
b. 2
c. 6
d. 4
e. 2
in option C it is higher
so correct option is C
Final answer:
The ion pair with the greatest electrostatic potential energy among the given pairs is Al+3 - O-2, as it has the highest product of charges according to Coulomb's Law. Option C is correct.
Explanation:
The electrostatic potential energy between two ions is determined by Coulomb's Law, which states that the energy is proportional to the product of the charges of the ions divided by the distance between them. With all distances being equal, the pair with the greatest product of charges would have the highest electrostatic potential energy. Thus, we can compare the products of charges for each pair:
Na+ - Cl- gives us 1(+1)×1(-1) = 1Na+ - O-2 gives us 1(+1)×2(-1) = 2Al+3 - O-2 gives us 3(+1)×2(-1) = 6Mg+2 - O-2 gives us 2(+1)×2(-1) = 4Na- - Mg+2 gives us 1(-1)×2(+1) = 2Therefore, the ion pair Al+3 - O-2 has the greatest electrostatic potential energy because the product of their charges (6) is the largest.
A sample of nitrogen gas is inside a sealed container. The container is slowly compressed, while the temperature is kept constant. This is a(n) ________ process. A sample of nitrogen gas is inside a sealed container. The container is slowly compressed, while the temperature is kept constant. This is a(n) ________ process.
a. adiabatic
b. constant-volume
c. isobaric
d. isothermal
Answer:
The correct option is d isothermal
Explanation:
Step 1: Data given
An adiabatic process is a process in which no heat is gained or lost by the system.
A constant- volume process, also called an isochoric process is a process in which the volume is held constant, meaning that the work done by the system will be zero.
An Isobaric process is a thermodynamic process taking place at constant pressure
An isothermal process is a thermodynamic process, in which the temperature of the system remains constant
Step 2: This situation
The container is slowly compressed, while the temperature is kept constant.
This is an isothermal process.
The correct option is d isothermal
A mixture of gaseous reactants is put into a cylinder, where a chemical reaction turns them into gaseous products. The cylinder has a piston that moves in or out, as necessary, to keep a constant pressure on the mixture of . The cylinder is also submerged in a large insulated water bath. (See sketch at right.) From previous experiments, this chemical reaction is known to absorb of energy. The temperature of the water bath is monitored, and it is determined from this data that of heat flows out of the system during the reaction. Is the reaction exothermic or endothermic
Answer:
See explaination
Explanation:
Exothermic reactions are reactions or processes that release energy, usually in the form of heat or light.
Hence when heat is sent out or released we have an Exothermic reaction.
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The pH of water is about 7. The pH of lemon juice is about 2. Which liquid is an acid?
Answer:
lemon juice
Explanation:
lemons are acidic.
Answer:
the answer is lemon.
Explanation:
A student sees tiny bubbles clinging to the inside of an unopened plastic bottle full of carbonated soft drink. The student opens the bottle and hears a loud hiss as gas under pressure escapes from the bottle.
a) The bubbles will shrink, and some may vanish.
b) The bubbles will grow, and more may appear.
c) The bubbles won't change.
d) More information needed to predict what will happen to the bubbles.
Answer: Option (b) is the correct answer.
Explanation:
When the student opens up the bottle lid then due to the difference in atmospheric pressure and pressure inside the bottle will lead the carbon dioxide to form more number of bubbles.
Also, when the bottle opens up then tiny bubbles around the neck of bottle will tend to float towards the open surface. Hiss sound appears because on opening the lid of bottle excess gas tends to come out of the bottle causing the sound.
Thus, we can conclude that when the student opens the bottle and hears a loud hiss as gas under pressure escapes from the bottle the bubbles will grow, and more may appear.
When a carbonated soft drink bottle is opened, the decrease in pressure reduces the solubility of CO₂, leading to the formation of more bubbles. Thus, the correct answer is b) The bubbles will grow, and more may appear.
When a carbonated soft drink is in an unopened bottle, it is under high CO₂ pressure. This high pressure keeps a large amount of carbon dioxide dissolved in the liquid. The bubbles seen inside the bottle are carbon dioxide gas clinging to the container. When the bottle is opened, the pressure above the liquid decreases significantly, causing a loud hiss as the gas escapes.
According to Henry's Law, the solubility of a gas in a liquid decreases as the pressure decreases. Consequently, the solubility of CO₂ in the soda is reduced, which means that more dissolved CO₂ will leave the solution and form bubbles. This explains why more bubbles will appear and existing bubbles will grow after opening the bottle.
1. Is energy absorbed or released when chemical bonds are broken during a chemical reaction?
Answer:
absorbed
Explanation:
A 500.0-mL buffer solution is 0.100 M in HNO2 and 0.150 M in KNO2. Part A Determine whether or not 250 mgNaOH would exceed the capacity of the buffer to neutralize it. Determine whether or not 250 would exceed the capacity of the buffer to neutralize it. yes no Request Answer Part B Determine whether or not 350 mgKOH would exceed the capacity of the buffer to neutralize it. Determine whether or not 350 would exceed the capacity of the buffer to neutralize it. yes no Request Answer Part C Determine whether or not 1.25 gHBr would exceed the capacity of the buffer to neutralize it. Determine whether or not 1.25 would exceed the capacity of the buffer to neutralize it. yes no Request Answer Part D Determine whether or not 1.35 gHI would exceed the capacity of the buffer to neutralize it. Determine whether or not 1.35 would exceed the capacity of the buffer to neutralize it. yes no
Answer:
Neither of them will neutralize the buffer solution.
Explanation:
The buffer solution of HNO₂ and KNO₂ will be neutralized when the acid reacts and consume all of the base of the buffer solution or when the base added reacts and consume all of the acid of the buffer solution.
First, we need to calculate the number of moles of the acid and the base of the buffer:
[tex] n_{HNO_{2}} = C*V = 0.100 M*0.500 L = 0.050 moles [/tex]
[tex] n_{KNO_{2}} = C*V = 0.150 M*0.500 L = 0.075 moles [/tex]
Now, let's evaluate each case.
A) 250 mg of NaOH:
We need to calculate the number of moles of NaOH
[tex] n_{NaOH} = \frac{m}{M} [/tex]
Where m: is the mass = 250 mg, and M: is the molar mass = 39.99 g/mol
[tex] n_{NaOH} = \frac{0.250 g}{39.99 g/mol} = 6.25 \cdot 10^{-3} moles [/tex]
The number of moles of the acid HNO₂ after reaction with the base added NaOH is:
[tex] n_{a_{T}} = n_{a} - n_{b} = 0.050 moles - 6.25 \cdot 10^{-3} moles = 0.044 moles [/tex]
After the reaction of HNO₂ with the NaOH remains 0.044 moles of acid, hence, 250 mg of NaOH would not exceed the capacity of the buffer to neutralize it.
B) 350 mg KOH:
The number of moles of KOH is:
[tex]n_{KOH} = \frac{m}{M} = \frac{0.350 g}{56.1056 g/mol} = 6.23 \cdot 10^{-3} moles[/tex]
Now, the number of moles of HNO₂ that remains in the solution is:
[tex] n_{T} = 0.050 moles - 6.23 \cdot 10^{-3} moles = 0.044 moles[/tex]
Therefore, 350 mg of KOH would not exceed the capacity of the buffer to neutralize it.
C) 1.25 g of HBr:
The number of moles of HBr is:
[tex] n_{HBr} = \frac{m}{M} = \frac{1.25 g}{80,9119 g/mol} = 0.015 moles [/tex]
Now, the number of moles of the base KNO₂ that remains in solution after the reaction with HBr is:
[tex] n_{T} = 0.075 moles - 0.015 moles = 0.06 moles [/tex]
Hence, 1.25 g of HBr would not exceed the capacity of the buffer to neutralize it.
D) 1.35 g of HI:
The number of moles of HI is:
[tex] n_{HI} = \frac{m}{M} = \frac{1.35 g}{127.91 g/mol} = 0.0106 moles [/tex]
Now, the number of moles of the base KNO₂ that remains in solution after the reaction with HI is:
[tex] n_{T} = 0.075 moles - 0.0106 moles = 0.0644 moles [/tex]
Hence, 1.35 g of HI would not exceed the capacity of the buffer to neutralize it.
Therefore, neither of them will neutralize the buffer solution.
I hope it helps you!
Part A: No
Part B: Yes
Part C: Yes
Part D: Yes
Part A: To determine if 250 mg of NaOH would exceed the buffer capacity, we first need to calculate the number of moles of NaOH added and compare it to the number of moles of HNO2 in the buffer.
The moles of NaOH added are calculated as follows:
Part A: No
Part B: Yes
Part C: Yes
Part D: Yes
Part A: To determine if 250 mg of NaOH would exceed the buffer capacity, we first need to calculate the number of moles of NaOH added and compare it to the number of moles of HNO2 in the buffer.
The moles of NaOH added are calculated as follows:
[tex]\[ \text{moles of NaOH} = \frac{\text{mass of NaOH}}{\text{molar mass of NaOH}} = \frac{250 \times 10^{-3} \text{g}}{40.0 \text{g/mol}} = 6.25 \times 10^{-3} \text{mol} \][/tex]
The volume of the buffer solution is 500.0 mL, which is equivalent to 0.500 L. The moles of HNO2 in the buffer are calculated by: [tex]\[ \text{moles of HNO2} = \text{concentration of HNO2} \times \text{volume of buffer} = 0.100 \text{M} \times 0.500 \text{L} = 5.00 \times 10^{-2} \text{mol} \][/tex]
Since the moles of NaOH added (6.25 × 10^-3 mol) are less than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer can neutralize the added NaOH. Therefore, the answer is No.
Part B: Similarly, for 350 mg of KOH, we calculate the moles of KOH:
[tex]\[ \text{moles of KOH} = \frac{350 \times 10^{-3} \text{g}}{56.1 \text{g/mol}} = 6.24 \times 10^{-3} \text{mol} \][/tex]
Since the moles of KOH are approximately equal to the moles of NaOH added in Part A and are less than the moles of HNO2 in the buffer, the buffer can neutralize the added KOH. Therefore, the answer is No.
Part C: For 1.25 g of HBr, we calculate the moles of HBr:
[tex]\[ \text{moles of HBr} = \frac{1.25 \times 10^{-3} \text{g}}{80.9 \text{g/mol}} = 1.55 \times 10^{-2} \text{mol} \][/tex]
Since the moles of HBr added (1.55 × 10^-2 mol) are greater than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer cannot neutralize the added HBr. Therefore, the answer is Yes.
Part D: For 1.35 g of HI, we calculate the moles of HI:
[tex]\[ \text{moles of HI} = \frac{1.35 \times 10^{-3} \text{g}}{127.9 \text{g/mol}} = 1.06 \times 10^{-2} \text{mol} \][/tex]
Since the moles of HI added (1.06 × 10^-2 mol) are greater than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer cannot neutralize the added HI. Therefore, the answer is Yes.
Correction: In Part B, the moles of KOH calculated are slightly less than the moles of HNO2 in the buffer, so the buffer can neutralize the added KOH. The initial answer provided was incorrect; it should be No, not Yes. The corrected answer is No for Part B."
The volume of the buffer solution is 500.0 mL, which is equivalent to 0.500 L. The moles of HNO2 in the buffer are calculated by:
[tex]\[ \text{moles of HNO2} = \text{concentration of HNO2} \times \text{volume of buffer} = 0.100 \text{M} \times 0.500 \text{L} = 5.00 \times 10^{-2} \text{mol} \][/tex]
Since the moles of NaOH added (6.25 × 10^-3 mol) are less than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer can neutralize the added NaOH. Therefore, the answer is No.
Part B: Similarly, for 350 mg of KOH, we calculate the moles of KOH:
[tex]\[ \text{moles of KOH} = \frac{350 \times 10^{-3} \text{g}}{56.1 \text{g/mol}} = 6.24 \times 10^{-3} \text{mol} \][/tex]
Since the moles of KOH are approximately equal to the moles of NaOH added in Part A and are less than the moles of HNO2 in the buffer, the buffer can neutralize the added KOH. Therefore, the answer is No.
Part C: For 1.25 g of HBr, we calculate the moles of HBr:
[tex]\[ \text{moles of HBr} = \frac{1.25 \times 10^{-3} \text{g}}{80.9 \text{g/mol}} = 1.55 \times 10^{-2} \text{mol} \][/tex]
Since the moles of HBr added (1.55 × 10^-2 mol) are greater than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer cannot neutralize the added HBr. Therefore, the answer is Yes.
Part D: For 1.35 g of HI, we calculate the moles of HI:
[tex]\[ \text{moles of HI} = \frac{1.35 \times 10^{-3} \text{g}}{127.9 \text{g/mol}} = 1.06 \times 10^{-2} \text{mol} \][/tex]
Since the moles of HI added (1.06 × 10^-2 mol) are greater than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer cannot neutralize the added HI. Therefore, the answer is Yes.
Correction: In Part B, the moles of KOH calculated are slightly less than the moles of HNO2 in the buffer, so the buffer can neutralize the added KOH. The initial answer provided was incorrect; it should be No, not Yes. The corrected answer is No for Part B."
A chemist must prepare 550.0 mL of nitric acid solution with a pH of 2.00 at 25 °C. He will do this in three steps: Fill a 550.0 mL volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (9.0M) stock nitric acid solution and add it to the flask Fill the flask to the mark with distilled water Calculate the volume of concentrated nitric acid that the chemist must measure out in the second step. Round your answer to 2 significant digits 0- mL.
Answer:
The correct answer is 0.61 ml
Explanation:
Nitric acid is a strong acid. That means that it dissociates completely in water as follows:
HNO₃ → H⁺ + NO₃⁻
As the dissociation is complete, the concentration of H⁺ ions is equal to the initial concentration of the acid (HNO₃). Thus, the pH can be calculated from the initial concentration of the acid:
pH= -log [H⁺] = -log [acid]
We want a nitric acid solution with a pH of 2.0. so we first calculate the concentration of acid we need:
2.0 = -log [acid]
10⁻²= [acid]
The chemist has a stock solution with C= 9.0 M and he/she wants a solution with C= 1 x 10⁻² M and V= 550 ml. We use the equation that relates the initial concentration and volume (Ci and Vi, respectively) of a solution with the final concentration and volume (Cf and Vf, respectively):
Ci x Vi = Cf x Vf
⇒ Vi= (Cf x Vf)/Ci = (1 x 10⁻² M x 550 ml)/9.0 M = 0.611 ml
Summarizing, the chemist must measure 0.611 ml of concentrated solution (9.0 M), add it to the flask and fill the flask to the mark until 550 ml in order to obtain a nitric acid solution with a pH of 2.0.
When aqueous solutions of sodium cyanide and hydrobromic acid are mixed, an aqueous solution of sodium bromide and hydrocyanic acid results. Write the net ionic equation for the reaction.
Answer:
CN⁻(aq) + H⁺(aq) → HCN(l)
Explanation:
The reactants are aqueous solutions:
NaCN(aq) and HBr(aq)
When you mix these compounds you make pure HCN (l)
The molecular equation is:
NaCN(aq) + HBr(aq) → NaBr(aq) + HCN(l)
When you dissociate the reactants, you have:
Na⁺(aq) +CN⁻(aq) + H⁺(aq) + Br⁻(aq) → Na⁺(aq) + Br⁻(aq) + HCN(l)
Sodium bromide, it is a salt, that can also be dissociated in the solution
To make, the net ionic equation you remove the repeated ions
CN⁻(aq) + H⁺(aq) → HCN(l)
The net ionic equation for the reaction between sodium cyanide and hydrobromic acid, resulting in sodium bromide and hydrocyanic acid, is CN-(aq) + H+(aq) → HCN(aq). This equation only includes the reacting species.
Explanation:The reaction between sodium cyanide (NaCN) and hydrobromic acid (HBr) results in the formation of sodium bromide (NaBr) and hydrocyanic acid (HCN). This can be written as a chemical equation: NaCN(aq) + HBr(aq) → NaBr(aq) + HCN(aq). The net ionic equation for the reaction is: CN-(aq) + H+(aq) → HCN(aq). This equation represents only the species that are involved in the reaction and doesn't include the spectator ions (in this case, Na+ and Br-).
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The quantity of antimony in an ore can be determined by an oxidation‑reduction titration with an oxidizing agent. The ore is dissolved in hot, concentrated acid and passed over a reducing agent so that all of the antimony is in the form of Sb3+(aq) . The Sb3+(aq) is completely oxidized by an aqueous solution of BrO−3(aq) . Complete and balance the equation for this reaction in acidic solution.
Answer:
4 Sb (+3) + 2 Br (2) O (3) ----> 4 SbBr + 3 O(-2)
In the equation, place the valences in parentheses and the stoichiometric balance numbers so that the reaction is balanced, place them in bold.
Explanation:
In this chemical equation it seems to me that the valences of BrO are wrong.
That is why I will complete the equation, demonstrating the chemical reaction, and also use the appropriate valences.
Assuming that Sb has a valence of +3 and that the Br that appears in the form of oxide with oxygen (valence -2) has a valence of +3 (because -3 does not exist as valence or oxidation state of bromine).