Calculate the mole fraction of KI in a solution made by dissolving 3.4 g of KI in 5.8 g of water. (Step by step)
A. 0.060
B. 0.064
C. 0.37
D. 0.59
E. 6.4

Answer:

Option d. 0.10 m Cr₂(SO₄)₃

Explanation:

Formula for the osmotic pressure is determined as:

π = M . R . T . i

So you have to take account the i (Van't Hoff factor, numbers of ions dissolved)

Urea is an organic compound, so the i value is 1

Zync acetate can be dissociated:

Zn(CH₃COO)₂  →  1Zn²⁺  + 2CH₃COO⁻

In this case, the i is 3. (you see, the stoichiometry of ions)

Cr₂(SO₄)₃ →  0.10 m

Chromium sulfate is dissociated:

Cr₂(SO₄)₃ →  2Cr³⁺  +  3SO₄⁻²

i = 5

BaI₂ → 0.16 m

BaI₂ →  1Ba²⁺  +  2I⁻

i = 3

Answer:

the number of milliliters of a 1M is 402mL

Explanation:

The computation of the number of milliliters could be determined by using the following formula

As we know that

[tex]V_1\times M_1 = V_2\times M_2[/tex]

where,

V_1 and V_2 are the starting and final volumes

And, the M_1 and M_2 are the starting and the final molarities

Now the V_1 is

[tex]V_1 \times 1M = 60mL \times 6.7M[/tex]

So, the V_1 is 402mL

Hence, the number of milliliters of a 1M is 402mL

Answer:

[tex]\mathbf{pZn ^{2+} =8.8569 }[/tex]

Explanation:

Using the approach of Henderson-HasselBalch equation, we have :

[tex]pH = pKa[NH^+_4] + log \dfrac{[NH_3]}{[NH_4^+]}[/tex]

where;

the pKa of [tex]NH^+_4[/tex] = 9.26

concentration of [tex]NH_3[/tex] = 0.100 M

concentration of [tex]NH_4Cl[/tex] = 0.176 M

∴

the pH of the buffered solution is :

[tex]pH = 9.26 + log \dfrac{[0.100]}{[0.176]}[/tex]

[tex]pH = 9.26 + log (0.5682)[/tex]

[tex]pH = 9.26 +(-0.2455)[/tex]

[tex]pH =9.02[/tex]

The Chemical equation for the reaction of [tex]Zn ^{2+}[/tex] and EDTA is :

[tex]Zn^{2+}_{(aq)} + Y^{4-}_{(aq)} \iff ZnY^{2-} _{(aq)}[/tex]

Here;

[tex]Y^{4-}_{(aq)}[/tex] denotes the fully deprotonated form of the EDTA

The formation constant [tex]K_f[/tex] of the equation for the reaction can be represented as:

[tex]K_f = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ][Y^{4-}]}[/tex]      ----- (1)

The logarithm of the formation constant of Zn - EDTA complex = 16.5

[tex]K_f[/tex]  = [tex]10^{16.5}[/tex]

[tex]K_f[/tex]  = [tex]3.16 \times 10^{16}[/tex]

Since the formation constant in the above equation signifies that the EDTA is present in  [tex]Y^{4-}[/tex],

Then:

[tex]\alpha _{Y^{4-} }= \dfrac{Y^{4-}}{C_{EDTA}}[/tex]

[tex]{Y^{4-}}= \alpha_ {Y^{4-}} \times {C_{EDTA}}[/tex]

From (1)

[tex]K_f = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ][Y^{4-}]}[/tex]  

[tex]K_f = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ \ \alpha_ {Y^{4-}} \times {C_{EDTA}}}[/tex]

∴

[tex]K_f' = K_f \times \alpha _Y{^4-} = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ C_{EDTA} }[/tex]

where;

[tex]K_f'[/tex] = conditional formation constant

[tex]\alpha _Y{^4-}[/tex] = the fraction of EDTA that exit in the form of the presences of the 4 charges .

So at equivalence point :

all the [tex]Zn^{2+}[/tex] initially in titrand  is now present in [tex]ZnY^{2-}[/tex]

[tex]K_f' = K_f \times \alpha _Y{^4-}[/tex]

Obtaining the data for the value of [tex]\alpha _Y{^4-}[/tex] at the reference table:

[tex]\alpha _Y{^4-}[/tex]  =  [tex]5.4 \times 10^{-12}[/tex]

∴

[tex]K_f' = 3.16 \times 10^{16} \times 5.4 \times 10^{-2}[/tex]

[tex]K_f' = 1.7064 \times 10^{15}[/tex]

To calculate the moles of  EDTA ,[tex]Zn^{2+}[/tex]  , [tex]ZnY^{2-}[/tex] ; we have:

moles of  EDTA = 0.0100 M × 0.025 L

moles of  EDTA = [tex]2.5 \times 10^{-4} \ mole[/tex]

moles of [tex]Zn^{2+}[/tex] = 0.00500 M  × 0.050 L

moles of [tex]Zn^{2+}[/tex] = [tex]2.5 \times 10^{-4} \ mole[/tex]

moles of  [tex]ZnY^{2-}[/tex]  =  [tex]\dfrac{initial \ mole}{total \ volume}[/tex]

moles of  [tex]ZnY^{2-}[/tex]  = [tex]\dfrac{2.5 \times 10^{-4}}{ 0.025 + 0.050 }[/tex]

moles of  [tex]ZnY^{2-}[/tex]  = [tex]\dfrac{2.5 \times 10^{-4}}{ 0.075 }[/tex]

moles of  [tex]ZnY^{2-}[/tex]  = 0.0033333 M

Recall that:

[tex]K_f' = K_f \times \alpha _Y{^4-} = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ C_{EDTA} }[/tex]

[tex]K_f' = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ C_{EDTA} }[/tex]

Assume Q² is the amount of complex dissociated in [tex]ZnY^{2-}[/tex]

[tex]ZnY^{2-} \iff Zn^{2+} + C_{EDTA}[/tex]  

i.e [tex]Q^2 = Zn^{2+} + C_{EDTA}[/tex]

[tex]1.707 \times 10^{15}= \dfrac{0.0033333}{Q}[/tex]

[tex]Q= \dfrac{0.0033333}{1.707 \times 10^{15}}[/tex]

[tex]Q^2= \dfrac{0.0033333}{1.707 \times 10^{15}}[/tex]

[tex]Q^2= 1.9527 \times 10^{-18}[/tex]

[tex]Q= \sqrt{1.9527 \times 10^{-18}}[/tex]

Q = [tex]1.397 \times 10^{-9}[/tex] M

[tex][Zn^{2+}]= 1.39 \times 10^{-9} \ M[/tex]

∴

[tex]pZn ^{2+} =- log [Zn^{2+}][/tex]

[tex]pZn ^{2+} = - log (1.39 \times 10^{-9} ) \ M[/tex]

[tex]\mathbf{pZn ^{2+} =8.8569 }[/tex]

Answer:

Boiling an egg is a chemical reaction

Vegetable rotting is a chemical reaction.

Chemical reaction: In this Chemical reaction takes place and new products is formed.

Explanation:

Answer:

A)

1. Reaction will shift rightwards towards the products.

2. It will turn green.

3. The solution will be cooler..

B) It will turn green.

Explanation:

Hello,

In this case, for the stated equilibrium:

[tex]heat + Cu(H_2O)_6 ^{+2} (blue) + 4Br^- \rightleftharpoons 6H_2O + CuBr_4^{-2} (green)[/tex]

In such a way, by thinking out the Le Chatelier's principle, we can answer to each question:

A)

1. If potassium bromide, which adds bromide ions, is added more reactant is being added to the solution, therefore, the reaction will shift rightwards towards the products.

2. The formation of the green complex is favored, therefore, it will turn green.

3. The solution will be cooler as heat is converted into "cold" in order to reestablish equilibrium.

B) In this case, as the heat is a reactant, if more heat is added, more products will be formed, which implies that it will turn green.

Regards.

Answer:

[tex]C_3H_7Cl[/tex] = Two structures

[tex]C_3H_6Cl_2[/tex] = Four structures

Explanation:

We must remember that in an isomer we have the same molecular formula but different structures. Thus, for the molecule [tex]C_3H_7Cl[/tex] we can draw a linear structure of 3 carbons and change the position of the chlorine atom, obtaining two different structures.

For the molecule [tex]C_3H_6Cl_2[/tex], we can use similar logic. Place a chain of 3 carbons and change the position of the chlorine atoms in such a way that for this formula we will have 4 different structures.

See figures 1 and 2 for further explanations.

Answer:

2.90

Explanation:

Any buffer system can be described with the reaction:

[tex]HA~->~H^+~+~A^-[/tex]

Where [tex]HA[/tex] is the acid and [tex]A^-[/tex] is the base. Additionally, the calculation of the pH of any buffer system can be made with the Henderson-Hasselbach equation:

[tex]pH=pKa~+~Log\frac{[A^-]}{[HA]}[/tex]

With all this in mind, we can write the reaction for our buffer system:

[tex]HF~->~H^+~+~F^-[/tex]

In this case, the acid is [tex]HF[/tex] with a concentration of 0.413 M and the base is [tex]F^-[/tex] with a concentration of 0.237 M. We can calculate the pKa value if we do the "-Log Ka", so:

[tex]pKa~=~-Log(7.2X10^-^4)=~3.14[/tex]

Now, we can plug the values into the Henderson-Hasselbach

[tex]pH=~3.14~+~Log(\frac{[0.237~M]}{[0.413~M]})~=~2.90[/tex]

The pH value would be 2.90

I hope it helps!

Answer:

a) Ka= 7.1 × 10⁻⁴; This is a weak acid because the acid is not completely dissociated in solution.

Explanation:

Step 1: Write the dissociation reaction for nitrous acid

HNO₂(aq) ⇄ H⁺(aq) and NO₂⁻(aq)

Step 2: Calculate the acid dissociation constant

Ka = [H⁺] × [NO₂⁻] / [HNO₂]

Ka = 0.022 × 0.022 / 0.68

Ka = 7.1 × 10⁻⁴

Step 3: Determine the strength of the acid

Since Ka is very small, nitrous acid is a weak acid, not completely dissociated in solution.

Answer:

One.

Explanation:

Hello,

In this case, the equilibrium condition is characterized by the equality of the rates at which a process happen and the contrary process happen, so its ratio is 1. For instance, a chemical reaction at equilibrium will have a contant ratio of the velocity at which the products are formed and the reactants consumed

However, for the described insoluble solid that is allowed the reach equilibrium, the ratio of the rate at which ions join the solution and the rate at which ions join the lattice will be one since it reaches an equilibrium state.

Best regards.

Answer:

prepare

creates

observe

ask

records

present

Answer:

Molality of the methanol solution = 8.33 m

Explanation:

Given:

Mass % = 21.1 %

Molar mass of methanol = 32.04 g / mol

Find:

Molality of the methanol solution?

Computation:

Moles of methanol = Mass / Molar mass

Moles of methanol =  = 21.1 / 32.04

Moles of methanol = 0.658

Assume.

Mass of solution = 100 g

Mass of solvent  = 100 -21.1 = 78.9 g  = 0.0789 kg

Molality of the methanol solution = 0.658 / 0.0789

Molality of the methanol solution = 8.33 m

Given values:

Now,

Moles of methanol:

= [tex]\frac{Mass}{Molar \ mass}[/tex]

= [tex]\frac{21.1}{31.04}[/tex]

= [tex]0.658[/tex]

then,

Mass of solution:

= [tex]100-21.1[/tex]

= [tex]78.9 \ g \ or \ 0.0789 \ kg[/tex]

hence,

The molality will be:

= [tex]\frac{0.658}{0.0789}[/tex]

= [tex]8.33 \ m[/tex]

Thus the answer above is correct.

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The given question is incomplete, the complete question is:

Balance the following chemical equations: A) P4+ O2 → P406 B) _P406+ LO2 → P4010 a) A reaction container holds 5.33 g of P4 and 3.77 g of O2 and reaction A occurs. If enough oxygen is available then the P406 reacts further to undergo reaction B. What is the limiting reactant for the formation of P406? b) What mass of P406 is produced (theoretical yield in grams)? c) If 7.12g of P406 were obtained what is the percent yield? d) Will reaction B occur? Why or why not? What mass of excess reactant is left in the reaction container?

Answer:

The balanced reaction will be,

A) P₄ + 3O₂ ⇒ P₄O₆

B) P₄O₆ + 2O₂ ⇒ P₄O₆

a) Based on the given information, the reaction container holds 5.33 grams of P₄ and 3.77 grams of oxygen. Thus, the moles of P₄ will be,

Moles = mass of P₄/Molar mass of P₄ = 5.33 grams/124 g/mole = 0.043 mole

Now the moles of O₂ will be,

Moles = mass of O₂/Molar mass of O₂ = 3.77 grams/32 g/mol = 0.112 mole

Now the moles of P₄O₆ formed when 0.043 moles of P₄ react completely will be = 1/1 × 0.043 = 0.043 mole of P₄O₆

Similarly, the moles of P₄O₆ formed, when 0.112 moles of O₂ react completely will be = 1/3 × 0.112 = 0.0373 mole of P₄O₆

Thus, from the analysis, the maximum moles of P₄O₆ formed will be 0.0373 moles. Therefore, oxygen will be the limiting reagent, which will react completely in the reaction.

b) From the above findings, the maximum moles of P₄O₆ produced is 0.0373 mole. Thus, the theoretical yield of P₄O₆ produced will be,

= Moles of P₄O₆ × Molar mass of P₄O₆

Theoretical yield = 0.0373 mole × 220 g/mole = 8.206 grams

c) Based on the given information, the actual mass of P₄O₆ produced is 7.12 grams.

Hence, percent yield = Actual yield/Theoretical yield * 100

= 7.12/8.206 × 100 = 86.77 %

d) In the given case, reaction B will not take place. This is due to the fact that oxygen is not left for reaction B, which was the limiting regent for reaction A. Here P₄ is the excess reactant, which was left in the reaction.

The initial moles of P₄ is 0.043, O₂ is 0.112, and P₄O₆ is O. The final moles of P₄ is 0.043 -1/3 × 0.112 = 0.0057 mole, O₂ is 0, and P₄O₆ is 0.0373 mole.

Thus, moles of P₄ left is 0.0057 mole. Hence, the mass of P₄ left will be,

= 0.0057 mole × Molar mass of P₄

= 0.0057 mole × 124 g/mole = 0.7068 grams.

Answer:

the solubility of BaCO₃ in pure  water  and in a solution is 4.472 × 10⁻⁵ M and 6.9204 × 10⁻⁹ M respectively.

Explanation:

To calculate the solubility of BaCO₃ in:

(a) pure water and (b) in a solution in which [CO₃²⁻] = 0.289 M

The [tex]ksp[/tex] (i.e solubility constant ) for BaCO₃= 2.0 × 10⁻⁹

BaCO₃   → Ba²⁺  + CO₃²⁻

ksp = s × s

s² = ksp

s = [tex]\sqrt{ksp}[/tex]

s = [tex]\sqrt{2.0 \times 10^{-9}}[/tex]

s = 4.472 × 10⁻⁵ M

(b) The solubility of BaCO₃ in a solution in which [CO₃²⁻] = 0.289 M

BaCO₃   → Ba²⁺  + CO₃²⁻

ksp = s × s

2.0 × 10⁻⁹ = s × 0.289

s = 2.0 × 10⁻⁹/0.289

s = 6.9204 × 10⁻⁹ M

Thus, the solubility of BaCO₃ in pure  water  and in a solution is 4.472 × 10⁻⁵ M and 6.9204 × 10⁻⁹ M respectively.

Answer:

physical projects

Journal entries

Materials

Lab reports

Artworks

Explanation:

Definition of a Portfolio:

Portfolio can be defined as a physical collection of student work that includes materials such as written assignments, completed tests, artwork, lab reports, physical projects and other material evidence of learning progress and academic accomplishment, including awards and honors,

A portfolio is a long-term form of self reflection and assessment that students do together.

Portfolios are a great way to demonstrate the competencies you would list on a resume or talk about in a science interview

Answer:

A.  NaOH

B.  20 g

C.  18.4 g

D.  108%

Explanation:

Al₂O₃  +  6 NaOH  +  12 HF  ⇒  2 Na₃AlF₆  +  9 H₂O

A.  Since you have excess HF, this is not the limiting reagent.  The two possibilities are Al₂O₃ and NaOH.  Find out how many moles you have of each.  Then, use the mole ratio in the chemical equation to find out how much product each reagent can produce.  The reagent the produces the least is the limiting reagent.  The molar mass of Al₂O₃ is 101.96 g/mol.

(6.55 g Al₂O₃)/(101.96 g/mol Al₂O₃) = 0.06424 mol Al₂O₃

(0.06424 mol Al₂O₃) × (2 mol Na₃AlF₆/1 mol Al₂O₃) = 0.12848 mol Na₃AlF₆

(1.75 L NaOH) × (0.15 M NaOH) = 0.2625 mol NaOH

(0.2625 mol NaOH) × (2 mol Na₃AlF₆/6 mol NaOH) = 0.0875 mol Na₃AlF₆

NaOH produces less product, making it the limiting reagent.

B.  The actual yield is 20 g.  This information is given in the problem.

C.  Since you know how much product you will get, convert moles of Na₃AlF₆ to grams to find actual yield.  Use the value found for Na₃AlF₆ that you got from the limiting reagent.  The molar mass is 209.94 g/mol.

(0.0875 mol Na₃AlF₆) × (209.94 g/mol Na₃AlF₆) = 18.4 g Na₃AlF₆

D.  To find the percent yield, divide the actual yield by the theoretical yield  and multiply by 100%.

(20 g)/(18.4 g) × 100% = 108%

If your percent yield is greater than 100% when performing a reaction, there has been a mistake somewhere.  Either you recorded the numbers incorrectly or there has been some human error in your experiment.  This might be the right answer for this problem, but you might want to double check to make sure the numbers you gave me were right.

Answer:

Butanoic acid.

Explanation:

Hello,

In this case, when a primary alcohol such as 1-butanol (OH is bonded to a primary carbon) is oxidized in the presence of a strong oxidizing media such as potassium dichromate (K2Cr2O7) and sulfuric acid, the stepwise oxidation goes to the corresponding aldehyde with a further oxidation to the corresponding carboxylic acid:

[tex]R-CH_2-OH\longrightarrow R-COH\longrightarrow R-COOH[/tex]

Therefore, on the attached picture you can find that the formed aldehyde is butanal and the inly organic product, due to the strong oxidizing media is finally butanoic acid.

Best regards.

Answer:

Van't Hoff factor for AlCl₃ = 3 (Approx)

Explanation:

Given:

Number of observed particular = 1.79 M

Number of theoretical particular = 0.56 M

Find:

Van't Hoff factor for AlCl₃

Computation:

Van't Hoff factor for AlCl₃ = Number of observed particular / Number of theoretical particular

Van't Hoff factor for AlCl₃ = 1.79 M / 0.56 M

Van't Hoff factor for AlCl₃ = 3.19

Van't Hoff factor for AlCl₃ = 3 (Approx)

The Van't Hoff factor for the AlCl₃ solution is 3

Van't Hoff factor describes the number of ions in a solution. The Van't Hoff factor for non-electrolyte is always 1 while it varies for ionic solutions.

From the question given above, the following data were obtained:

Van't Hoff factor = Actual molarity / Theoretical molarity

Van't Hoff factor for AlCl₃ = 1.79  / 0.56

Van't Hoff factor for AlCl₃ ≈ 3

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Answer:

1000 µL; 10 µL  

Explanation:

A p1000 micropipet is set to dispense 1000 µL.

A p10 micropipet set to dispense 10 µL.

Answer:

Activate glycolysis/Inhibit gluconeogenesis: Increased levels of fructose-2,6-bisphosphate, activation of PFK-2

Activate gluconeogenesis/Inhibit glycolysis: Increased levels of glucagon, Increased levels of cAMP, Activation of fructose-2,6-bisphosphatase (FBPase-2)

Note: The question is incomplete. The complete question is given below and in the attachment.

Fructose‑2,6‑bisphosphate is a regulator of both glycolysis and gluconeogenesis for the phosphofructokinase reaction of glycolysis and the fructose‑1,6‑bisphosphatase reaction of gluconeogenesis. In turn, the concentration of fructose‑2,6‑bisphosphate is regulated by many hormones, second messengers, and enzymes. How do the following affect glycolysis and gluconeogenesis?

Explanation:

Fructose-2,6-bisphosphate is an allosteric effector for the enzymes phosphofructokinase-1 (PFK-1) and fructose-1,6-bisphosphatase (FBPase-1). It increases the affinity of PFK-1 for fructose-6-phosphate thereby activating glycolysis. However, it reduces the affinity of FBPase-1 for its substrate, fructose-1,6-bisphosphate thereby inhibiting gluconeogenesis.

Activation of phosphofructokinase-2 activates glycolysis and inhibits gluconeogenesis by catalyzing the phosphorylation of fructose-6-phosphate to form fructose-2,6-bisphosphate.

Increased levels of glucagon stimulates the synthesis of cAMpP which activates cAMP-dependent ptrotein kinase which phosphorylates the bifunctional enzyme PFK-2/FBPase-2. The phosphorylation of this enzyme inhibits its PFK-2 activity and activates its FBPase-2 activity. This results in the activation of gluconeogenesis and inhibition of glycolysis.

Fructose-2,6-bisphosphatase breaks down fructose-2,6-bisphosphate to fructose-6-phospshate and a phosphoryl group. This results in the activation of gluconeogenesis and the inhibition of glycolysis.

Answer:

The formation of an aqueous solution of ammonium nitrate is An endothermic process

Explanation:

An exothermic process produce energy when occurs. As there is energy that is released, the temperature of the arounds increases.

In the other hand, an endothermic process absorb energy when occurs doing the temperature of the around colder than the initial temperature.

As the dissolution of NH₄NO₃ in water make the temperature of the water colder:

The formation of an aqueous solution of ammonium nitrate is an endothermic process

The question requires us to determine if reaction process is an endothermic or an exothermic reaction.

To do this,

First we will define the terms Endothermic reaction and exothermic reaction

Endothermic reactions are chemical reactions in which the reactants absorb heat energy from the surroundings to form products. These reactions lower the temperature of their surrounding area, thereby creating a cooling effect. They have a net positive standard enthalpy change.

Exothermic reactions are reactions or processes that release energy, usually in the form of heat or light. They have a net negative standard enthalpy change.

From the question,

When solid NH4NO3 is dissolved in water, the temperature of the water and beaker gets noticeably colder. This means it is an endothermic reaction.

Hence, the formation of an aqueous solution of ammonium nitrate is an endothermic process

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Answer:

yes

Explanation:

the number of protons in the nucleus of an atom determines the type element: the number of protons in an element is the atomic number on the periodic table (number at the top)

Each atom has characteristic number of protons present in it's nucleus which is unique for that element.

An element  is defined as a substance which cannot be broken down further into any other substance. Each element is made up of its own type of atom. Due to this reason all elements are different from one another.

Elements can be classified as metals and non-metals. Metals are shiny and conduct electricity and are all solids at room temperature except mercury. Non-metals do not conduct electricity and are mostly gases at room temperature except carbon and sulfur.

The number of protons in the nucleus is the defining property of an element and is related to the atomic number.All atoms with same atomic number are atoms of same element.

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Answer:

ZnS, Ksp=2.0×10−25

Explanation:

The Ksp is a constant that let us know the capacity of the compound to easily dissociates in water. The higher the Ksp, the more soluble the compound is. In the exercise giving, the highest Ksp is 2.0×10−25, that means ZnS will dilutes easily than others

Answer:

3Sb^3+(aq) + BrO3^-(aq) + 6H^+(aq)----->3Sb^5+(aq) + Br^-(aq) 3H2O(l)

Explanation:

When we want to balance redox reaction equations, we must ensure that the number of electrons lost in the oxidation half reaction equation is equal to the number of electrons gained in the reduction half reaction equation.

After we have done this, we can now write the overall balanced reaction equation without including the number of electrons lost or gained. Hence;

Oxidation half equation;

3Sb^3+(aq) -----> 3Sb^5+(aq) +6e

Reduction half equation;

BrO3^-(aq) + 6H^+(aq) + 6e ----> Br^-(aq) 3H2O(l)

Overall balanced reaction equation;

3Sb^3+(aq) + BrO3^-(aq) + 6H^+(aq)----->3Sb^5+(aq) + Br^-(aq) 3H2O(l)

Answer:

Explanation:

Density of a substance can be found by using the formula

[tex]Density = \frac{mass}{volume} [/tex]

From the question

mass = 145.8 g

volume = 91.75 mL

Substitute the values into the above formula and solve for the Density

That's

[tex]Density = \frac{145.8}{91.75} [/tex]

= 1.5891

We have the final answer as

Density = 1.6 g/mL

Hope this helps you

Answer:

Oxygen and glucose

Explanation:

____________

Answer:

how strong it is

Explanation:

Seismographs are not able to say when earthquake will happen, but they help humans to know how strong it is or if it is happening or not

Answer:

Mars

Explanation:

Mars is the planet with features that are really similar to the earth. As a matter of fact, Mars and the earth are referred to as "twin planets".

Mars and Earth are believed to have historically had even more similar attributes, however, that of mars was lost due to changes in the solar system.

The two planets are mostly similar because of their relatively similar size and proximity to the sun.

Answer:

[tex]\Huge \boxed{\mathrm{C_2H_5OH}}[/tex]

Explanation:

CH₃OCH₃ (Dimethyl ether) has a boiling point of -23 °C

C₂H₅OH (Ethanol) has a boiling point of 78.37 °C

CH₃CH₂CH₃ (Propane) has a boiling point of -42 °C

CH₃CHO (Acetaldehyde) has a boiling point of 20.2 °C

C₂H₅OH (Ethanol) has the highest boiling point.