Answer:
Explanation: Độ thẩm thấu của NaCl 0.9% và glucose 5% lần lượt là 308 và 278 ... Dung dịch natri clorid sử dụng trong pha thuốc tiêm truyền thường dùng
Gu
Magnesium metal is reacted with hydrochloric acid to produce
hydrogen gas. A sample of hydrogen gas is collected over water
in a eudiometer at 28.0°C. The atmospheric pressure is 636
mmHg. Determine the pressure (in atm) of the hydrogen gas
produced
Pressure =
atm
The pressure of hydrogen gas is 607.7 mmHg
According of Dalton's law of Partial pressure, the total pressure of a mixture of gases is the sum of the partial pressures of the individual gases in the mixture.
We can now write;
The for hydrogen collected over water, we have a mixture of hydrogen gas and water vapour.
Total pressure = pressure of hydrogen gas + vapour pressure of water
Pressure of hydrogen gas = Total pressure - vapour pressure of water
Pressure of hydrogen gas = 636 mmHg - 28.3 mmHg
Pressure of hydrogen gas = 607.7 mmHg
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A solution of the primary standard potassium hydrogen phthalate (KHP), KHC8H4O4 , was prepared by dissolving 0.4877 g of KHP in about 50 mL of water. Titration of the KHP solution with a KOH solution of unknown concentration required 38.91 mL to reach a phenolphthalein end point. What is the concentration of the KOH solution
Answer:
The concentration of KOH is 0.06137 M
Explanation:
Step 1: Data given
Molar mass of KHP = 204.22 g/mol
Mass of KHP = 0.4877 grams
Volume of water = 50 mL
Volume of KOH solution = 38.91 mL
Step 2: The balanced equation
C8H5KO4 + KOH ⇒ C8H4K2O4+ H2O
Step 3: Calculate number of moles of KHP
Moles = Mass / molar mass
Moles KHP = 0.4877 grams / 204.22 g/mol
Moles KHP = 0.002388 moles
Step 4: Calculate moles of KOH
For 1 mol KHP we need 1 mol KOH to produce 1 mol C8H4K2O4 and 1 mol H2O
For 0.002388 moles KHP we need 0.002388 moles KOH
Step 5: Calculate the concentration of KOH
Concentration = moles / volume
Concentration of KOH = 0.002388 moles / 0.03891 L
Concentration of KOH = 0.06137 M
The concentration of KOH is 0.06137 M
2. Write the chemical equation for the reaction NaOH Sodium Hydroxide AgNO3 Silver Nitrate
Answer:
AgNO3 + NaOH = AgOH + NaNO3.
Explanation:
Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.
Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.However, the equation balanced here is the initial reaction which produces AgOH and NaNO3.
You perform a distillation to separate a mixture of propylbenzene and cyclohexane, and you obtain 2.9949 grams of cyclohexane (density -0.779 g/mL, MW - 84.16 g/mol) and 1.6575 grams of propylbenzene (density = 0.862 g/mL, MW = 120.2 g/mol). What is the volume percent composition of cyclohexane in the mixture?
Answer:
66.67%
Explanation:
From the given information:
mass of cyclohexane = 2.9949 grams
density of cyclohexane = 0.779 g/mL
Recall that:
Density = mass/volume
∴
Volume = mass/density
So, the volume of cyclohexane = 2.9949 g/ 0.779 g/mL
= 3.8445 mL
Also,
mass of propylbenzene = 1.6575 grams
density of propylbenzene = 0.862 g/mL
Volume of propylbenzene = 1.6575 g/ 0.862 g/mL
= 1.9229 mL
The volume % composition of cyclohexane from the mixture is:
[tex]= (\dfrac{v_{cyclohexane}}{v_{cyclohexane}+v_{propylbenzene}})\times 100[/tex]
[tex]= (\dfrac{3.8445}{3.8445+1.9229})\times 100[/tex]
[tex]= (\dfrac{3.8445}{5.7674})\times 100[/tex]
= 66.67%
A sample of gas contains 0.1500 mol of CH4(g) and 0.1500 mol of H2O(g) and occupies a volume of 13.0 L. The following reaction
takes place:
CH_(g) + H2O(g) 3H2(g) + CO(g)
Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.
L
Answer:
26.0L is the volume of the sample after the reaction
Explanation:
Based on the reaction, 1 mole of CH4 reacts with 1 mole of H2O to produce 1 mole of CO and 3 moles of H2.
That is, 1 mole of each reactant produce 4 moles of gases
As in the reaction, 0.1500 moles of CH4 and 0.1500 moles of H2O are added, 0.1500 moles of CO and 0.4500 moles of H2 are produced.
Before the reaction, the moles of gas are 0.3000 moles and after the reaction the moles are 0.6000 moles of gas.
Based on Avogadro's law, the moles of a gas are directly proportional to the volume under temperatura and pressure constant. The equation is:
V1/n1 = V2/n2
Where V is volume and n are moles of 1, initial state and 2, final state.
Replacing:
V1 = 13.0L
n1 = 0.3000 moles
V2 = ?
n2 = 0.6000 moles
13.0L*0.6000 moles / 0.3000 moles = V2
V2 = 26.0L is the volume of the sample after the reaction
How does the number of molecules in one mole of carbon dioxide compare with the number of molecules in one mole of water?
ОА.
There are four times as many molecules in one mole of carbon dioxide as there are in one mole of water.
ОВ.
There are twice as many molecules in one mole of carbon dioxide as there are in one mole of water.
OC
There are three times as many molecules in one mole of carbon dioxide as there are in one mole of water.
OD
There are the same number of molecules in one mole of carbon dioxide as there are in one mole of water.
Answer:
d
Explanation:
The "Nutrition Facts" on a label of a 16 fluid ounce container of apple juice states that a serving size is 8 fluid ounces contains 176 Calories and 240 milligrams of potassium.
Nutrition Facts
Serving Size 8 fl. oz. (240mL)
Servings Per Container: 2
Amount Per Serving
Calories 176
% Daily Value*
Total Fat 0g 0 %
Sodium 32mg 1 %
Potassium 240mg 6 %
Total Carbohydrate 29g 10 %
Sugars 26g
Protein 0g
a) How many calories would 1 fluid ounce of apple juice contain?
b) How many milligrams of potassium would 1 fluid ounce of apple juice contain?
Answer:
a) 22 calories
b) 30 mg
Explanation:
Divide number of cal or mg of pot by 8 fl oz.
Why is the formation of fructose-1,6-bisphosphate a step in which control is likely to be exercised in the glycolytic pathway
How many moles of Fe contains 3.41 x 1023 Fe atoms?
Answer:
[tex]\boxed {\boxed {\sf 0.566 \ mol \ Fe}}[/tex]
Explanation:
We are asked to convert a number of atoms to moles.
We can convert atoms to moles using Avogadro's Number, which is 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this problem, the particles are atoms of iron (Fe). There are 6.022 ×10²³ atoms of iron in 1 mole of iron.
We use dimensional analysis to convert atoms to moles. This involves setting up ratios. Use Avogadro's Number and the underlined information to make a ratio.
[tex]\frac {6.022 \times 10^{23}\ atoms \ Fe}{1 \ mol \ Fe}[/tex]
We are converting 3.41 × 10²³ atoms of iron to moles, so we multiply by this value.
[tex]3.41 \times 10^{23} \ atoms \ Fe *\frac {6.022 \times 10^{23}\ atoms \ Fe}{1 \ mol \ Fe}[/tex]
Flip the ratio. It stays equivalent, but it allows the units of atoms of iron to cancel.
[tex]3.41 \times 10^{23} \ atoms \ Fe *\frac{1 \ mol \ Fe} {6.022 \times 10^{23}\ atoms \ Fe}[/tex]
[tex]3.41 \times 10^{23}*\frac{1 \ mol \ Fe} {6.022 \times 10^{23}}[/tex]
[tex]\frac{3.41 \times 10^{23}} {6.022 \times 10^{23}} \ mol \ Fe[/tex]
[tex]0.5662570575\ mol \ Fe[/tex]
The original measure ment of iron atoms ( 3.41 × 10²³ ) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 2 in the ten-thousandths place ( 0.5662570575) tells us to leave the 6 in the thousandths place.
[tex]0.566 \ mol \ Fe[/tex]
3.41 × 10²³ atoms of iron is equal to approximately 0.566 moles of iron.
Conversion Problem (show all work):
1. A patient required 3.0 pints of blood during surgery. How many liters does this correspond
to? Show all work. Use conversion factors available in the text or the exam packet. (4)
1.42liters, which is equivalent to 3pints, of blood is required for the surgery
Pints is a unit of measurement for volume in the United States. However, it can be converted to litres using the following equation:1 US pint = 0.473 liters
Hence, according to this question which states that a patient required 3.0 pints of blood during surgery. This means that the patient required:3 × 0.473
= 1.419 liters of blood for the surgery
1.42liters, which is equivalent to 3pints, of blood is required for the surgeryLearn more at: https://brainly.com/question/24168664
At constant temperature and pressure, if 0.4 mole of a gas. A occupies 220 ml, and x mole of B gas occupies 120 ml. what is the number of moles of gas B?
Answer:
0.218mol
Explanation:
Using Avogadro's law equation;
Va/na = Vb/nb
Where;
Va = volume of gas A
Vb = volume of gas B
na = number of moles of gas A
nb = number of moles of gas B
According to the information in this question,
na = 0.4mol
nb = x mol
Va = 220ml
Vb = 120ml
Using Va/na = Vb/nb
220/0.4 = 120/x
Cross multiply
0.4 × 120 = 220 × x
48 = 220x
x = 48/220
x = 0.218mol
7.7 cm
9.8 cm
0.00
0.162 m
Answer:
Volume = 1222.5cm³
Explanation:
If the question is about the volume of the rectangle:
The volume of a rectangle is obtained by the multiplication of its 3 dimensions: Length, width, height.
In the problem, the length of the rectangle is 0.162m = 16.2cm
The width is 7.7cm
And the height is 9.8cm
The volume is:
Volume = 16.2cm*7.7cm*9.8cm
Volume = 1222.5cm³An atom's first 2 energy levels are filled and there are 2 electrons in the third energy
level. It's atomic number is:
Answer:
12
Explanation:
2+8+2=12
atomic no is the No of protons
Answer:
Atomic number is 12.
Explanation:
Atomic number = electrons in filled shells + outermost electrons
= 2 + 8 +2
= 12
Calculate the pH of each solution.
A. 0.18 M CH3NH2
B. 0.18 M CH3NH3Cl
C. a mixture of 0.18 M CH3NH2 and 0.18 M CH3NH3Cl
Answer:
See Explanations
Explanation:
pH =-log[H₃O⁺] = -log[H⁺]
pOH = -log[OH⁻]
For weak acids [H⁺] = SqrRt(Ka·[Acid])
For weak bases [OH⁻] = SqrRt(Kb·[Base])
pH + pOH = 14
__________________________________________
A. Given 0.18M CH₃NH₂; Kb = (4.4 x 10⁻⁴)* => pH = 11.95
CH₃NH₂ + H₂O => CH₃NH₃OH ⇄ CH₃NH₃⁺ + OH⁻;
[OH⁻] = SqrRt(Kb·[weak base]) = SqrRt(4.4 x 10⁻⁴ x 0.18)M = 8.97 x 10⁻³M
=> pOH = -log[OH⁻] = -log(8.93x10⁻³) = -(-2.05) = 2.05
=> pH = 14 - pOH = 14 - 2.05 = 11.95.
*Kb values for most ammonia derivatives in water can be found online by searching 'Kb-values for weak bases'. Kb-values for methyl amine and methylammonium chloride are both 4.4x10⁻⁴.
___________________________________________________
B. Given 0.18M CH₃NH₃Cl
In water ... CH₃NH₃Cl => CH₃NH₃⁺ + Cl⁻; Kb(CH₃NH₃Cl) = 4.4 x 10⁻⁴
Cl⁻ + H₂O => No Rxn (i.e.; no hydrolysis occurs) ... Cl⁻ does not react with H₂O.
Hydrolysis Reaction of Methylammonium Ion:
CH₃NH₃⁺ + H₂O => CH₃NH₄OH ⇄ CH₃NH₄⁺ + OH⁻
Ka' x Kb = Kw => Ka' = Kw/Kb = 10⁻¹⁴/4.4 x 10⁻⁴ = 2.27 x 10⁻¹¹ Ka' = [CH₃NH₄⁺][OH⁻]/[CH₃NH₄OH] = (x)(x)/(0.18M) = (x²/0.18M) = 2.27 x 10⁻¹¹ => x = [OH⁻] = SqrRt(2.27x10⁻¹¹ x 0.18)M = 2.02 x 10⁻⁶M => pOH = -log(2.02 x 10⁻⁶) = -(-5.69) = 5.69 => pH = 14 - pOH = 14 - 5.69 = 8.31.
*note => the general nature of halide interactions would increase acidity (lower pH) of the halogenated compound.
C. A mixture of 0.18M CH₃NH₂ and 0.18M CH₃NH₃Cl
Mixture of 0.18M CH₃NH₂ + 0.18M CH₃NH₃Cl
In Water ...
=> 0.18M CH₃NH₃OH + 0.18M CH₃NH₃Cl
=> 0.18M CH₃NH₃⁺ + 0.1M OH⁻ + 0.18M CH₃NH₃⁺ + 0.18M Cl⁻
=> 0.36M CH₃NH₃⁺ + 0.18M OH⁻ + 0.18M Cl⁻
-----------------------------------------------------------
Ka'(CH₃NH₃⁺) x Kb(CH₃NH₂) = Kw => Ka'(CH₃NH₃⁺) = Kw/Kb(CH₃NH₂)
=> Ka'(CH₃NH₃⁺) = (10⁻¹⁴/4.4x10⁻⁴) = 2.27x10⁻¹¹
----------------------------------------------------------
From the 0.36M CH₃NH₃⁺
=> CH₃NH₃⁺ + H₂O ⇄ CH₃NH₄⁺ + OH⁻
C(eq) 0.36M ---- x x (<= at equilibrium after mixing)
Ka'(CH₃NH₃⁺) = [CH₃NH₄⁺][OH⁻]/[CH₃NH₃⁺] = x²/(0.36M)
=> x = [OH⁻] = SqrRt(Ka'(CH₃NH₃⁺)·0.36M) = SqrRt(2.27x10⁻¹¹/0.36) = 0.0126M
=> Total [OH⁻] = 0.0126M + 0.18M = 0.1926M from hydrolysis process
=> final solution mix is therefore, 0.1926M in OH⁻ + 0.18M in Cl⁻
--------------------------------------------------------
Cl⁻ + H₂O => No Rxn (Cl⁻ does not react with H₂O)The 0.1926M in OH⁻ => [H⁺] = Kw/[OH⁻] = (10⁻¹⁴/0.1926)M = 5.192 x 10⁻¹⁴M in H₃O⁺ ions (= H⁺ ions) ...∴pH = -log[H⁺] = -log(5.192x10⁻¹⁴) = -(-13.29) = 13.29 for solution mix
The acid and base dissociation constant and the 0.18 M of CH₃NH₂ and
CH₃NH₃Cl and the mixture give the following approximate values;
A. The pH value of the 0.18 M CH₃NH₂ is 11.93
B. The pH value of the 0.18 M CH₃NH₃Cl is 5.69
C. The pH value of the mixture is 10.644
Which method can be used to calculate the pH values?A. 0.18 M CH₃NH₂
The solution is presented as follows;
CH₃NH₂ + H₂O → CH₃NH₃⁺ + OH⁻
Let x represent the number of moles of CH₃NH₃⁺ and OH⁻ produced, we
have;
The number of moles of CH₃NH₂ remaining = 0.18 - x
Which gives;
[tex]K_b = \mathbf{\dfrac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}}[/tex]
[tex]K_b[/tex] for CH₃NH₂ = 4.167 × 10⁻⁴
Therefore;
[tex]4.167 \times 10^{-4} = \mathbf{\dfrac{x \times x}{0.18 - x}}[/tex]
4.167 × 10⁻⁴ × (0.18 - x) = x²
4.167 × 10⁻⁴ × (0.18 - x) - x² = 0
Which gives;
x = [OH⁻] = 8.455 × 10⁻³
pH = 14 + log[OH⁻]
Which gives;
pH = 14 + log(8.455 × 10⁻³) ≈ 11.93
B. 0.18 M CH₃NH₃Cl
The solution is presented as follows;
CH₃NH₃⁺ → CH₃NH₂ + H⁺
Let x represent the number of moles of CH₃NH₂ and H⁺ produced,
respectively, we have;
The number of moles of CH₃NH₃⁺ remaining = 0.18 - x
Which gives;
[tex]K_a = \mathbf{\dfrac{[CH_3NH_2][H^+]}{[CH_3NH_3^+]}}[/tex]
Kₐ for CH₃NH₃Cl = 2.27 × 10⁻¹¹
Therefore;
[tex]2.27\times 10^{-11} = \dfrac{x \times x}{0.18 - x}[/tex]
2.27 × 10⁻¹¹ × (0.18 - x) = x²
2.27 × 10⁻¹¹ × (0.18 - x) - x² = 0
Which gives;
x = [H⁺] ≈ 2.02 × 10⁻⁶
pH = -log[H⁺]
Which gives;
pH = -log(2.02 × 10⁻⁶) ≈ 5.69
C. For the mixture of 0.18 M CH₃NH₂ and 0.18 M of CH₃NH₃Cl, we have;
Based on the Henderson-Hasselbalch equation, we have;
[tex]pH = \mathbf{ pKa + log\dfrac{[Conjugate \ base]}{[acid ]}}[/tex]
Which gives;
[tex]pH = -log\left(2.27 \times 10^{-11} \right)+ log\dfrac{0.18}{0.18} \approx \underline{10.644}[/tex]
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Select the number of valence electrons for hydrogen.
Answer:
Vanlency of hydrogen - 11
Electrons of hydrogen - 1
Answer:
The answer is: 1
Hope this helps :) <3
Explanation:
Which of the following will affect the rate of a chemical reaction?
solution temperature
solution color
solute mass
solution volume
Answer:
Solution temperature.
Explanation:
Hello there!
In this case, considering this question about chemical kinetics, it will be possible for us to analyze two perspectives:
1. Molecular: here, we infer that the solution temperature will provide more energy to the molecules in order to undergo more effective crashes which will make more products and therefore, increase the rate constant.
2. Mathematical: by means of the Arrhenius equation, it will be possible to tell that the increase in the temperature of the system, the negative of the exponent present in such equation will increase and therefore turn the rate constant bigger.
In such way, we infer the answer is solution temperature.
Regards!
What is the percentage by mass of carbon in CH3(CH2)5COOH?
A. 48.6%
B. 9.2%
C. 55.4%
D. 64.6%
Answer:
F 64.6 percent of carbon may be
A system fitted with a piston expands when it absorbs 53.1 ) of heat from the surroundings. The piston is working against a pressure of 0.677 atm. The final volume is 63.2 L. What was the initial volume of the system if the internal energy of the system decreased by 108.3 J?
a. 65.6 L
b. 64.0 L
c. 70.8 L
d. 60.8 L
e. 54.4L
Answer:
d. 60.8 L
Explanation:
Step 1: Given data
Heat absorbed (Q): 53.1 JExternal pressure (P): 0.677 atmFinal volume (V2): 63.2 LChange in the internal energy (ΔU): -108.3 JStep 2: Calculate the work (W) done by the system
We will use the following expression.
ΔU = Q + W
W = ΔU - Q
W = -108.3 J - 53.1 J = -161.4 J
Step 3: Convert W to atm.L
We will use the conversion factor 1 atm.L = 101.325 J.
-161.4 J × 1 atm.L/101.325 J = -1.593 atm.L
Step 4: Calculate the initial volume
First, we will use the following expression.
W = - P × ΔV
ΔV = - W / P
ΔV = - 1.593 atm.L / 0.677 atm = 2.35 L
The initial volume is:
V2 = V1 + ΔV
V1 = V2 - ΔV
V1 = 63.2 L - 2.35 L = 60.8 L
how many electrons does tin have?
A. 50
B. 68
C. 118
how many electrons does tin have?
Answer : 2, 8, 18, 18, 4
Therefore, total electrons = 50
11 Explain how you would obtain solid lead carbonate from a mixture of lead carbonate and sodium chloride
Explanation:
Add water, Na2CO3 dissolves, filter, PbCO3 stays in the paper and dissolved Na2CO3 goes through as the solution. Dry the PbCO3 and you have the dry solid.
OR
Add water to dissolve then filter to obtain PbCo3 as you're residue and Na2Co3 as the filtrate. Dry the insoluble PbCo3 between filter papers and you obtain solid PbCo3
What is normality in chemistry?
Answer:
a measure of concentration equal to the gram equivalent weight per liter of solution.
Explanation:
Gram equivalent weight is the measure of the reactive capacity of a molecule. The solute's role in the reaction determines the solution's normality. Normality is also known as the equivalent concentration of a solution.
hope it helped
Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water . Suppose 2.73 g of methane is mixed with 6.7 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
3.8g of H2O are produced
Explanation:
The balanced reaction of the problem is:
CH4 + 2O2 → CO2 + 2H2O
Where 1 mole of CH4 reacts with 2 moles of O2
To solve this question we need to find, as first, the moles of each reactant in order to find limiting reactant. With limiting reactant we can find the moles of H2O produced and its mass as follows:
Moles CH4 - 16.04g/mol-
2.73g * (1mol/16.04g) = 0.170 moles CH4
Moles O2 -32g/mol-
6.7g (1mol/32g) = 0.209 moles O2
For a complete reaction of 0.170 moles of CH4 are needed:
0.170 moles CH4 * (2 mol O2 / 1mol CH4) = 0.340 moles O2
As there are just 0.209 moles of O2, oxygen is limiting reactant
The moles of water produced are:
0.209 moles O2 * (2mol H2O / 2mol O2) = 0.209 moles H2O
Mass water -Molar mass: 18.01g/mol-
0.209 moles H2O * (18.01g/mol) = 3.8g of H2O are produced
Choose the correct answer:
1.9 × 103 g
1.9 x 106 g
1.9 x 1010 g
Answer:
A. 1.9 × 103 g
(next one)
Which metric unit would be the best choice to report the result?
A. kg
Answer:
1. 2
2. 1.9 × 10^3 g
3. kg
Explanation:
Calculate [H3O+] for pH 1.86. Steps please.
Answer:
[H₃O⁺] = [H⁺] = 10^-pH = 10⁻¹°⁸⁶ = 0.0138M in [H⁺]
Explanation:
By definition pH = -log[H⁺] => [H⁺] = 10^-pH = 10⁻¹°⁸⁶ = 0.0138M in [H⁺]
Using your calculator ... I am using a TI-30XA scientific calculator.
=> start by entering the number 1.86 => then press the (+/-) function => this will insert a negative symbol => -1.86,=> next find button with "2nd" printed on face (on some calculators the button is in yellow); press this button to change to 'secondary mode',=> find the symbol (10ˣ) ... the button below this symbol is usually the 'log' button, then press it => the answer of interest will show in the display window. => ...Depending on the calculator, the answer may show as 0.0138, or 1.38x10⁻², or 1.38E-2 (=1.38 x 10⁻²). It is the user's job to insert dimensional units into answer of interest => 0.0138M, or 1.38 x 10⁻²M, or 1.38E-2M.
1.38E-2 which is 1.38 x 10⁻².
The elementary reaction 2H2O(g)↽−−⇀2H2(g)+O2(g) proceeds at a certain temperature until the partial pressures of H2O, H2, and O2 reach 0.0900 bar , 0.00100 bar , and 0.00350 bar respectively. What is the value of the equilibrium constant at this temperature?
Answer:
3.89 ×10^-5
Explanation:
Since they are gaseous reactants, we obtain the equilibrium constant from the given partial pressures;
p(H2O) = 0.0900 bar
p(H2) = 0.00100 bar
p(O2) = 0.00350 bar
The equation of the reaction is;2H2O(g)⇄2H2(g)+O2(g)
Kp= p(H2) . p(O2)/p(H2O)
Kp= 0.00100 × 0.00350/0.0900
Kp= 3.89 ×10^-5
As discussed in class, the Fischer esterification reactants and products are at equilibrium. How was the equilibrium of the reaction that you performed shifted towards the products
Answer:
See explanation
Explanation:
The particular reactants in the Fischer esterification reaction were not stated.
Generally, a Fischer esterification is a reaction that proceeds as follows;
RCOOH + R'OH ⇄RCOOR' + H2O
This reaction occurs in the presence of an acid catalyst.
We can shift the equilibrium of this reaction towards the products side in two ways;
I) use of a large excess of either of the reactants
ii) removal of one of the products as it is formed.
Any of these methods shifts the equilibrium of the Fischer esterification reaction towards the products side.
do you think that religion and science are in conflict ,or that people can both have both without any problems
When determining the amount of oxidant present by titration, you can use iodine/starch as an indicator. First, the oxidant, like hypochlorite, oxidizes Choose... When starch and iodine are both present, the solution is Choose... During the titration, a titrant like thiosulfate reduces the
The question is incomplete, the complete question is;
When determining the amount of an oxidant present by titration, you can use iodine and starch as an indicator.
First, the oxidant, like hypochlorite, oxidizes
Choose...
neutral iodine into iodide ion
iodide ion into neutral iodine
iodate polyatomic ion into iodide ion
When starch and iodine are both present, the solution is
Choose...
blue-black
brownish yellow
clear
During the titration, the titrant, like thiosulfate, reduces the
Choose...
iodide ion into iodate polyatomic ion
neutral iodine into iodide ion
iodide ion into neutral iodine
When the iodine has completely reacted at the endpoint of the titration, the solution should become
Choose...
clear
blue-black
brownish yellow
Answer:
1. iodide ion into neutral iodine
2. blue-black
3. neutral iodine into iodide ion
4. clear
Explanation:
Hypochlorite oxidizes the iodide ion to iodine molecule according to the reaction equation;
ClO-(aq) + 2H+(aq) + 2I-(aq) ---------> 6 I2(l) + Cl- (aq)+ H2O(l)
When iodine is added, the colour of the starch solution immediately changes to blue-black.
A reduction reaction occurs when the titrant, thiosulfate is added as follows;
I2 + 2S2O32- → 2I- + S4O62-
The solution at end point is found to become clear again.
A 2.00-mol sample of hydrogen gas is heated at constant pressure from 294 K to 414 K. (a) Calculate the energy transferred to the gas by heat. kJ (b) Calculate the increase in its internal energy. kJ (c) Calculate the work done on the gas. kJ
Answer:
a) The energy transferred is 6.91 kJ
b) The internal energy is 4.90 kJ
c) The work done on the gas is - 2.01 kJ
Explanation:
Step 1: Data given
Number of moles of hydrogen gas = 2.00 moles
Pressure = constant
Temperature is heated from 294 K to 414 K
Molar heat capacity of hydrogen gas = 28.8 J/mol*K
Step 2: Calculate the energy transferred to the gas by heat.
Q = n* Cp * ΔT
⇒with Q =the energy transferred
⇒with n = the number of moles = 2.00 moles
⇒with Cp = the Molar heat capacity of hydrogen gas = 28.8 J/mol*K
⇒ with ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K
Q = 2.00 * 28.8 * 120
Q = 6912 J = 6.91 kJ
Step 3: Calculate the increase in its internal energy.
ΔEint = n*Cv*ΔT
⇒with ΔEint = the increase in its internal energy.
⇒with n = the number of moles = 2.00 moles
⇒with Cv = The constant volume = 20.4 J/mol*K
⇒with ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K
ΔEint = 2.00 * 20.4 * 120
ΔEint =4896 J = 4.90 kJ
Step 4: Calculate the work done on the gas.
Work done on the gas = -Q + ΔEint
W = -6.91 kJ + 4.90 kJ
W = -2.01 kJ
What is the oxidation number of the metal ion in the coordinate complex [Fe(CN)6]3–?
A. NCO-
B. -OH
C. -CN
D. -SCN
Answer:
The options are incorrect.................
Explanation:
The Oxidation no. is +3
