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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A solution of acetic acid and water contains 205.0 g L-1 of acetic acid and 820.0 g L-1 of water. Compute the density of the solution ( report your answer in g per mL)
Answer:
[tex]\rho_t=1025000 gmL^{-1}[/tex]
Explanation:
From the question we are told that:
Density of acetic acid [tex]\rho_a=205 gL^{-1}[/tex]
Density of Water [tex]\rho_w=820 gL^{-1}[/tex]
Generally the equation for Solution Density is mathematically given by
[tex]\rho_t= \rho_w+\rho_a[/tex]
[tex]\rho_t=205+820[/tex]
[tex]\rho_t=1025 gL^{-1}[/tex]
[tex]\rho_t=1025000 gmL^{-1}[/tex]
Help me in this question!!!
Answer:
d. End product is that product with a ketone and carboxylic acid.
Explanation:
[tex]{ \sf{NaBH_{4} : }}[/tex]
Sodium borohydride is a reducing agent, it reduces the ketone to a primary alcohol.
[tex]{ \sf{H _{2} O \: and \: H {}^{ + } }}[/tex]
Then acidified water is an oxidising mixture which reverses the reduction reaction.
Explanation:
Option D is your answer
Hope it helps
A eudiometer is used to collect hydrogen gas in a chemical reaction, as in your Exp 7. The volume of the gas in the tube (when pressure is held
constant) is 479.10 mL. The pressure of the atmosphere during the experiment is 758.3 mmgHg, and the temperature of the water and gas is
19.0*C. The water vapor pressure at this temperature is 16.5 torr.
Calculate the mass of hydrogen, in mg, collected.
Answer:
39.29 mg
Explanation:
Step 1: Calculate the partial pressure of hydrogen
The pressure of the atmosphere is equal to the sum of the partial pressures of the water and the hydrogen. (1 Torr = 1 mmHg)
P = pH₂O + pH₂
pH₂ = P - pH₂O = 758.3 mmHg - 16.5 mmHg = 741.8 mmHg
We will convert it using the conversion factor 1 atm = 760.0 mmHg.
741.8 mmHg × 1 atm/760.0 mmHg = 0.9761 atm
Step 2: Convert 19.0 °C to Kelvin
We will use the following expression.
K = °C + 273.15 = 19.0 + 273.15 = 292.2 K
Step 3: Calculate the mass of hydrogen
First, we will calculate the moles of hydrogen using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 0.9761 atm × 0.47910 L / (0.08206 atm.L/mol.K) × 292.2 K = 0.01950 mol
The molar mass of hydrogen is 2.015 g/mol. The mass of hydrogen is:
0.01950 mol × 2.015 g/mol = 0.03929 g = 39.29 mg
Match each land resource to its use.
clay - used to make steel
iron ore - used to make batteries
salt - used to make pottery and tiles
aggregate - used in construction
graphite - used as a flavoring in food
i will give 10 points and brainliest!!!
Answer:
Explanation:
We are to match each land resource to what they are being used for.
Clay →→→ used to make pottery and tiles
iron ore →→→ used to make steel
Salt →→→ used as a flavoring in food
aggregate →→→ used in construction
graphite →→→ used to make batteries
Clay is a kind of soil particle that forms as a result of weathering processes. Examples include; pottery clays, glacial clays, and deep-sea clays e.t.c. The presence of one or more clay minerals, as well as variable quantities of organic and detrital components, characterizes all of them. Clay is usually sticky and moist when wet, but hard when dry. They are used in the making of tiles and potteries.
Iron ore: The iron ore deposits are found in the Earth's crust's sedimentary rocks. They're made up of iron and oxygen that mix during the chemical process in marine and freshwater. iron ores are used to produce almost every iron and steel product that we use today.
Aggregate: are utilized in construction activities. It is a material used to mix cement, gypsum, bitumen, or lime to produce concrete in the construction industry.
Graphite: Graphite is a mineral that occurs in both igneous and metamorphic rocks. It is generally generated on the earth's surface when carbon is exposed to high temperatures and pressures. It is mainly used in the production of batteries and electrodes,
A chemist dissolves 14.0 g of calcium hydroxide in one beaker of water, and 17.0 g of iron(III) chloride
in a second beaker of water. Everything dissolves.
When the two solutions are poured together, solid iron(III) hydroxide precipitates.
1. Write a balanced molecular equation.
2. Determine the identity of the limiting reactant.
3. Predict the mass of iron(III) hydroxide product.
Answer:
See detailed explanation.
Explanation:
Hello there!
In this case, for the given scenario, we will proceed as follows:
1. Here, we infer that the products are iron (III) hydroxide (precipitate) and calcium chloride:
[tex]3Ca(OH)_2+2FeCl_3\rightarrow 3CaCl_2+2Fe(OH)_3[/tex]
2. In this step we firstly calculate the moles of both reactants, by using their molar masses 74.093 and 162.2 g/mol respectively:
[tex]14.0gCa(OH)_2*\frac{1molCa(OH)_2}{74.093gCa(OH)_2}=0.189molCa(OH)_2 \\\\17.0gFeCl_3*\frac{1molFeCl_3}{162.2gFeCl_3}=0.105molFeCl_3[/tex]
Now, we calculate the moles of calcium hydroxide consumed by 0.105 moles of iron (III) chloride by using the 3:2 mole ratio between them:
[tex]0.105molFeCl_3*\frac{3molCa(OH)_2}{2molFeCl_3} =0.157molCa(OH)_2[/tex]
Thus, we infer that calcium hydroxide is in excess as 0.189 moles are available for it but just 0.157 moles react and therefore, iron (III) chloride is the limiting reactant.
3. Here, we use the moles of iron (III) chloride we've just computed, the 2:2 mole ratio with iron (III) hydroxide and its molar mass (106.867 g/mol) as shown below:
[tex]0.105molFeCl_3*\frac{2molFe(OH)_3}{2molFeCl_3} *\frac{106.867gFe(OH)_3}{1molFe(OH)_3} \\\\=11.2gFe(OH)_3[/tex]
Regards!
Nitric acid can be formed in two steps from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming natural gas. In the first step, nitrogen and hydrogen react to form ammonia: N2 (g) + 3H2 (g) â 2NH3 (g) =ÎHâ92.kJ In the second step, ammonia and oxygen react to form nitric acid and water:
NH3 (g) + 2O2 (g) â HNO3 (g) + H2O (g) =ÎHâ330.kJ
Required:
Calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these reactions.
Answer:
-376 kJ
Explanation:
The first step equation:
[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex] ---- (1)
The second step equation:
[tex]\mathsf{NH_{3(g)} + 2O_2{(g)} \to HNO_3{(g)} +H_2O_{(g)} \ \ \ \Delta H = -330\ kJ}[/tex] ---- (2)
To determine the enthalpy of formation for 1 mole of HNO₃ (nitric acid), we have the following.
From the above equations; let multiply equation (1) by 1 and equation (2) by 2.
[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex] ---- (3)
[tex]\mathsf{2NH_{3(g)} + 4O_2{(g)} \to 2HNO_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = 2(-330)\ kJ}[/tex] ----- (4)
adding the above two equations, we have:
[tex]\mathsf{N_{2(g)} + 3H_2{(g)}+ 2NH_{3(g)} + 4O_{2(g)} \to 2HNO_{3(g)} + 2NH_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-660 \ kJ -92\ kJ)}[/tex][tex]\mathsf{N_{2(g)} + 3H_2{(g)} + 4O_{2(g)} \to 2HNO_{3(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-752 \ kJ)}[/tex]
Now, from the recent equation, we have:
2 moles of nitric acid = -752 kJ
∴
1 mole of nitric acid will be: = (1 mole × (-752 kJ)) ÷ 2 moles
1 mole of nitric acid will be: = -376 kJ
Give the formula of each coordination compound. Include square brackets around the coordination complex. Do not include the oxidation state on the metal. Use parentheses only around polyatomic ligands.
a) potassium tetracyanonickelate(II)
b) sodium diamminedicarbonatoruthenate(III)
c) diamminedichloroplatinum(II)
Answer:
a) K2[Ni(CN)4]
b) Na3[Ru(NH3)2(CO3)2]
c) Pt(NH3)2Cl2
Explanation:
Coordination compounds are named in accordance with IUPAC nomenclature.
According to this nomenclature, negative ligands end with the suffix ''ato'' while neutral ligands have no special ending.
The ions written outside the coordination sphere are counter ions. Given the names of the coordination compounds as written in the question, their formulas are provided above.
Question 14
2 pts
A chemist wants to make 100 mL of a 0.500 M solution of NaCl. They have a
stock solution of 1.2 M NaCl. How much of the original stock solution do they
need to make their new dilute solution?
Explanation:
From the question given above, the following data were obtained:
Molarity of stock solution (M₁) = 1.2 M
Molarity of diluted solution (M₂) = 0.5 M
Volume of diluted solution (V₂) = 100 mL
Volume of stock solution needed (V₁) =?The volume of stock solution needed can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂1.2 × V₁ = 0.5 × 100
1.2 × V₁ = 50
Divide both side by 1.2
V₁ = 50 / 1.2
V₁ ≈ 42 mLThus, 42 mL of the stock solution is needed.
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Answer:
They need 41.7 mL of the original stock solution.
Explanation:
We can use the following equation for dilutions:
Cc x Vc = Cd x Vd
Where Cc and Vc are the concentration and volume values in the concentrated condition, whereas Cd and Vd are the concentration and volume values in the diluted condition.
The concentrated solution is the original stock solution, and it has:
Cc = 1.2 M
The diluted solution must be:
Cd = 0.500 M
Vd = 100 mL
So, we have to calculate Vc. For this, we replace the data in the equation:
[tex]V_{c} = \frac{C_{d} V_{d} }{C_{c} } = \frac{(0.500 M)(100 mL)}{1.2 M} = 41.7 mL[/tex]
Therefore, 41.7 mL of 1.2 M original stock solution are required to make 100 mL of a diluted solution with a concentration of 0.500 M.
the nutrition label on rice lists the amounts of protein, carbohydrates and fats in one serving. these substances are important for human nutrition
Answer:
Carbohydrates, proteins, and fats are biological macromolecules that are made up of chemical elements which are inherent to chemistry.
Chemistry explain how these macromolecules are bonded together at the molecular level and give an explanation for their behavior.
Explanation:
does anyone know how to solve this and what the answer would be?
Dynamic equilibrium is showed at the point at which solid liquid and gas intersect.
At the point at which solid liquid and gas intersect represents a system that shows dynamic equilibrium. There is equal amount of reactants and products at the point of dynamic equilibrium because the transition of substances occur between the reactants and products at equal rates, means that there is no net change. Reactants and products are formed at the rate that no change occur in their concentration.
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If 7 mol of copper reacts with 4 mol of oxygen, what amount of copper (II) oxide is produced? What amount of the excess reactant remains?
Answer:
7 mol CuO
0.5 mol O₂
Explanation:
Step 1: Write the balanced equation
2 Cu + O₂ ⇒ 2 CuO
Step 2: Identify the limiting reactant
The theoretical molar ratio (TMR) of Cu to O₂ is 2:1.
The experimental molar ratio (EMR) of Cu to O₂ is 7:4 = 1.75:1.
Since TMR > EMR, Cu is the limiting reactant
Step 3: Calculate the amount of CuO produced
7 mol Cu × 2 mol CuO/2 mol Cu = 7 mol CuO
Step 4: Calculate the excess of O₂ that remains
The amount of O₂ that reacts is:
7 mol Cu × 1 mol O₂/2 mol Cu = 3.5 mol O₂
The excess of O₂ that remains is:
4 mol - 3.5 mol = 0.5 mol
Three important nutritional additions to training for a long distance race are
O Protein
O Water
O Increase calories
O All of the answer choices
A sample of nitrogen gas occupies 117 mL at 100°C. At what
temperature would it occupy 234 mL if the pressure does not
change? (express answer in K and °C)
47
Page
8 I 8
- Q +
Answer:
The new temperature of the gas is 746 K.
Explanation:
Given that,
The volume of the gas, V₁ = 117 mL
Temperature, T₁ = 100°C = 373
Final volume of the gas, V₂ = 234 mL
We need to find the final temperature. The relation between temperature and volume is given by :
[tex]\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}\\\\T_2=\dfrac{T_1V_2}{V_1}\\\\T_2=\dfrac{373\times 234}{117}\\\\T_2=746\ K[/tex]
So, the new temperature of the gas is 746 K.
Consider the Fischer ester synthesis of methyl benzoate from benzoic acid and methanol in the presence of sulfuric acid as a catalyst. A reaction was performed in which 3.3 g of benzoic acid was reacted with excess methanol to make 1.7 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.
Answer:
46.2%
Explanation:
Number of moles benzoic acid reacts = 3.3g/122.12 g/mol = 0.027 moles
Since the reaction is 1:1, 0.027 moles of methyl benzoate is formed.
Hence;
Theoretical yield of methyl benzoate = 0.027 moles × 136.15 g/mol = 3.68 g
% yield = actual yield/theoretical yield × 100
% yield = 1.7 g/3.68 g × 100
% yield = 46.2%
Which statement best describes what happens during a chemical reaction?
A. Reactants change into products.
B. Reactants change into new reactants.
C. Products change into reactants.
D. Products change into new products.
Answer:
A. Reactants change into products
A given solution has 42.5 g NaNO3 in 1500 mL of water. What is the molarity of this solution? Show your work, rounding the atomic mass of each element to the nearest tenth and your final answer to the nearest hundredth
Explanation:
here's the answer to your question
methyl ether, a useful organic solvent, is prepared in two steps. In the first step, carbon dioxide and hydrogen react to form methanol and water:(g) (g) (l) (l)In the second step, methanol reacts to form dimethyl ether and water:(l) (g) (l)Calculate the net change in enthalpy for the format
This question is incomplete, the complete question is;
Dimethyl ether, a useful organic solvent, is prepared in two steps.
In the first step, carbon dioxide and hydrogen react to form methanol and water:
CO₂(g) + 3H₂(g) → CH₃OH(l) + H₂O(l) ΔH₁ = -131.kJ
In the second step, methanol reacts to form dimethyl ether and water:
2CH₃OH(l) → CH₃OCH₃(g) + H₂O(l) ΔH₂ = 8.kJ
Calculate the net change in enthalpy for the formation of one mole of dimethyl ether from carbon dioxide and hydrogen from these reactions. Round your answer to the nearest kJ.
Answer:
the net change in enthalpy for the formation of one mole of dimethyl ether is -254 kJ
Explanation:
Given the data in the question;
For the First Step;
CO₂(g) + 3H₂(g) → CH₃OH(l) + H₂O(l) ΔH₁ = -131.kJ
For the First Step;
2CH₃OH(l) → CH₃OCH₃(g) + H₂O(l) ΔH₂ = 8.kJ
Using Hess's Law of Constant Heat Summation;
" regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes "
we multiply step 1 reaction by the coefficient of 2
2CO₂(g) + 2×3H₂(g) → 2CH₃OH(l) + 2H₂O(l) ΔH₁ = 2 × -131.kJ
we have
2CO₂(g) + 6H₂(g) → 2CH₃OH(l) + 2H₂O(l) ΔH₁ = -262 kJ
2CH₃OH(l) → CH₃OCH₃(g) + H₂O(l) ΔH₂ = 8 kJ
{ 2CH₃OH cancels 2CH₃OH }
Hence, we have;
2CO₂ + 6H₂ → CH₃OCH₃(g) + 3H₂O(l)
So According to Hess's Law;
ΔH[tex]_{sum[/tex] = ΔH₁ + ΔH₂
we substitute
ΔH[tex]_{sum[/tex] = -262 kJ + 8 kJ
ΔH[tex]_{sum[/tex] = -254 kJ
Therefore, the net change in enthalpy for the formation of one mole of dimethyl ether is -254 kJ
All of the following statements concerning real cases is correct EXCEPT Group of answer choices molecules of real gases are attracted to each other. molecules of real gases occupy no volume. nonideal gas behavior is described by the Van der Waals Equation. the pressure of a real gas is due to collisions with the container. the pressure of a real gas at low temperatures is lower than for ideal gases.
Answer:
molecules of real gases occupy no volume.
Explanation:
As all the real gases are composed of particles that occupy the non-zero volume that is the excluded volume. If the gas is behaving in an ideal manner. The correction becomes negatable and is relative to the total volume. The extended volume is volume that is taken by the non ideal gas particles.A saturated solution of potassium iodide contains, in each 100 mL, 100 g of potassium iodide. The solubility of potassium iodide is 1 g in 0.7 mL of water. Calculate the specific gravity of the saturated solution
Answer:
Specific gravity of the saturated solution is 2
Explanation:
The specific gravity is defined as the ratio between density of a solution (In this case, saturated solution of potassium iodide, KI) and the density of water. Assuming density of water is 1:
Specific gravity = Density
The density is the ratio between the mass of the solution and its volume.
In 100mL of water, the mass of KI that can be dissolved is:
100mL * (1g KI / 0.7mL) = 143g of KI
That means all the 100g of KI are dissolved (Mass solute)
As the volume of water is 100mL, the mass is 100g (Mass solvent)
The mass of the solution is 100g + 100g = 200g
In a volume of 100mL, the density of the solution is:
200g / 100mL = 2g/mL.
The specific gravity has no units, that means specific gravity of the saturated solution is 2
Determine the number of water molecules in 0.2830g Na.
Answer:
7.38*10^21
Explanation:
2Na+2H20=2NaOH+H2
nNa=0.0123
number of water moles: 0.012*6*10^23=7.38*10^21
Diisopropyl ether reacts with concentrated aqueous HI to form two initial organic products
a. True
b. Fasle
Answer:
True
Explanation:
The reaction between Diisopropyl ether and concentrated aqueous HI forms two initial organic products as shown in the image attached.
The hydrogen of the HI becomes attached to the oxygen in the ether leading to a cleavage of the C-O bond to yield the first compound. The I^- become attached to the other moiety in the original molecule to yield the second compound as shown in the image attached.
Because the double bond in an alkene is rigid, alkenes can exist as geometric isomers. To clarify geometric isomers, IUPAC uses cis- and trans- as part of a compound name. If the substituents around the double bond are on the same side of the double bond, this is called
cis, cis.
cis.
cis, trans.
trans.
Answer:
cis
Explanation:
Cis isomers are formed when the substituents on the carbons of the double bond are on the same side of the double bond, forming a U. Trans isomers have substituents on opposite sides of the double bond, forming a sideways Z.
What is the difference between elimination and substitution reaction
Identify the key factors that will determine if a reaction undergoes elimination or substitution mechanism.
Use the following reagents to determine the type of reaction pathway expected and determine the products in each reaction.
a. Tert BuO- in tertbutanol and chlorobutane
b. KOH in water and bromobutane
c. NaI in acetone and bromobutane
Write a conclusion of no more than two paragraphs to summarize your results
Answer:
a) E2
b) SN2
c) SN2
Explanation:
A substitution reaction involves replacement of an atom or group in a molecule by another atom or group. An elimination reaction is the loss of two atoms from the same molecule leading to the formation of a multiple bond in the molecule.
We must note that primary alkyl halides never undergo SN1/E1 reactions. However, the presence of a strong bulky base such as tert BuO- , E2 reactions predominate. In the presence of strong bases such as OH^- and good nucleophiles such as I^-, SN2 mechanism predominates.
Complete the sentences by identifying the correct missing words. Alph and beta particles originate from the Choose... . Protection from radiation is necessary because if radiation passes through the body it can damage Choose... . Exposure to radiation can be limited by increasing the Choose... from the radioactive source.
Answer:
Alpha and beta particles originate from the nucleus, protection from radiation is important because if the radiation passes through the body it can damage cells. Exposure to radiation is often limited by increasing the distance from the radioactive source.
Explanation:
Alpha and beta particles come from unstable atoms during their decay. This radiation is extremely harmful which may damage DNA, causing a high rate of mutation. If we increase the distance of the source of radioactive exposure we will prevent damage.which primitive organic molecule was essential to form lipid bilayer?
a)protenoid
b)phospholipid
c)autocatalytic RNA
d)aminoacids
Answer:
c) autocatalytic RNA is the primitive organic molecules was essential to form lipid bilayer.
Write the complete ground-state electron configuration of I⁻.
Explanation:
here's the answer to your question
Iodine's neutral atom has the following electronic configuration: 1 s² 2 s² 2 p⁶ 3 s² 3 p⁶ 4 s² 3 d¹⁰ 4 p⁶ 5 s² 4 d¹⁰ 5 p⁵. The complete ground-state electron configuration of I⁻ is [Kr]5s² 4d¹⁰ 5p⁶.
What is electronic configuration?Electronic configuration is defined as each electron moves individually within an orbital while being surrounded by an average field produced by all other orbitals. The electron configuration is used to describe an atom's ground state orbitals, but it may also be used to depict an atom that has ionized into a cation or anion by making up for any lost or gained electrons in the orbitals after it.
The ground state electronic configuration is defined as the configuration of lower energy electrons surrounding an atom's nucleus. Ground state is defined as the least energetic state feasible for a physical system. It is important because permits us to determine the location of the excited electrons' departure and return when they release a photon.
Thus, iodine's neutral atom has the following electronic configuration: 1 s² 2 s² 2 p⁶ 3 s² 3 p⁶ 4 s² 3 d¹⁰ 4 p⁶ 5 s² 4 d¹⁰ 5 p⁵. The complete ground-state electron configuration of I⁻ is [Kr]5s² 4d¹⁰ 5p⁶.
To learn more about electronic configuration, refer to the link below:
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FORMULAS OF IONIC COMPOUNDS
FIND: POSITIVE ION, NEGATIVE ION AND FORMULA IN:
NAME:
Sodium chloride
Magnesium chloride
Calcium oxide
Lithium phosphide
Aluminum sulfide
Calcium nitride
Iron(III)chloride
Iron(II)oxide
Copper(I)sulfide
Copper(II)nitride
Zinc oxide
Silver sulfide
Potassium carbonate
Sodium nitrate
Calcium bicarbonate
Aluminum hydroxide
Lithium phosphate
Potassium sulfate
Answer:
NaCl, Na⁺,Cl⁻.
MgCl₂, Mg²⁺, Cl⁻.
CaO, Ca²⁺, O²⁻.
Li₃P, Li⁺, P³⁻.
Al₂S₃, Al³⁺, S²⁻.
Ca₃N₂, Ca²⁺, N³⁻.
FeCl₃, Fe³⁺, Cl⁻.
FeO, Fe²⁺, O²⁻.
Cu₂S, Cu⁺, S²⁻.
Cu₃N₂, Cu²⁺, N³⁻.
ZnO, Zn²⁺, O²⁻.
Ag₂S, Ag⁺, S²⁻.
K₂CO₃, K⁺, CO₃²⁻.
NaNO₃, Na⁺, NO₃⁻.
Ca(HCO₃)₂, Ca²⁺, HCO₃⁻.
Al(OH)₃, Al³⁺,OH⁻.
Li₃PO₄, Li⁺, PO₄³⁻.
K₂SO₄, K⁺, SO₄²⁻.
Explanation:
Sodium chloride. NaCl, formed by the cation Na⁺ and the anion Cl⁻.
Magnesium chloride. MgCl₂, formed by the cation Mg²⁺ and the anion Cl⁻.
Calcium oxide. CaO, formed by the cation Ca²⁺ and the anion O²⁻.
Lithium phosphide. Li₃P, formed by the cation Li⁺ and the anion P³⁻.
Aluminum sulfide. Al₂S₃, formed by the cation Al³⁺ and the anion S²⁻.
Calcium nitride. Ca₃N₂, formed by the cation Ca²⁺ and the anion N³⁻.
Iron(III)chloride. FeCl₃, formed by the cation Fe³⁺ and the anion Cl⁻.
Iron(II)oxide. FeO, formed by the cation Fe²⁺ and the anion O²⁻.
Copper(I)sulfide. Cu₂S, formed by the cation Cu⁺ and the anion S²⁻.
Copper(II)nitride. Cu₃N₂, formed by the cation Cu²⁺ and the anion N³⁻.
Zinc oxide. ZnO, formed by the cation Zn²⁺ and the anion O²⁻.
Silver sulfide. Ag₂S, formed by the cation Ag⁺ and the anion S²⁻.
Potassium carbonate. K₂CO₃, formed by the cation K⁺ and the anion CO₃²⁻.
Sodium nitrate. NaNO₃, formed by the cation Na⁺ and the anion NO₃⁻.
Calcium bicarbonate. Ca(HCO₃)₂, formed by the cation Ca²⁺ and the anion HCO₃⁻.
Aluminum hydroxide. Al(OH)₃, formed by the cation Al³⁺ and the anion OH⁻.
Lithium phosphate. Li₃PO₄, formed by the cation Li⁺ and the anion PO₄³⁻.
Potassium sulfate. K₂SO₄, formed by the cation K⁺ and the anion SO₄²⁻.
PLEASE HELP ASAP
A total of 132.33g C3H8 is burned in 384.00 g O2. Use the following questions to determine the amounts of products formed.
• How many grams of CO2 and H2O will be produced? (2 points)
b. If the furnace is not properly adjusted, the products of combustion can include other gases, such as CO and unburned hydrocarbons. If only 269.34 g of CO2 were formed in the above reaction, what would the percent yield be? (2 points)
Answer:
See explanation
Explanation:
We must first write the equation of the reaction as follows;
C3H8 + 5O2 ----> 3CO2 + 4H2O
Now;
We obtain the number of moles of C3H8 = 132.33g/44g/mol = 3 moles
So;
1 mole of C3H8 yields 3 moles of CO2
3 moles of C3H8 yields 3 × 3/1 = 9 moles of CO2
We obtain the number of moles of oxygen = 384.00 g/32 g/mol = 12 moles
So;
5 moles of oxygen yields 3 moles of CO2
12 moles of oxygen yields 12 × 3/5 = 7.2 moles of CO2
We can now decide on the limiting reactant to be C3H8
Therefore;
Mass of CO2 produced = 9 moles of CO2 × 44 g/mol = 396 g of CO2
Again;
1 moles of C3H8 yields 4 moles of water
3 moles of C3H8 yields 3 × 4/1 = 12 moles of water
Hence;
Mass of water = 12 moles of water × 18 g/mol = 216 g of water
In order to obtain the percentage yield from the reaction, we have;
b) Actual yield = 269.34 g
Theoretical yield = 396 g
Therefore;
% yield = actual yield/theoretical yield × 100/1
Substituting values
% yield = 269.34 g /396 g × 100
% yield = 68%
draw styrene
draw the structure of cyrene
Why is the formation of fructose-1,6-bisphosphate a step in which control is likely to be exercised in the glycolytic pathway
