Answer:
a) 101%
b)59.7%
Explanation:
The equation for the thermal decomposition of baking soda is shown;
2NaHCO3 → Na2CO3 + H2O + CO2
Number of moles of baking soda= mass/molar mass= 1.555g/84.007 g/mol = 0.0185 moles
From the reaction equation;
2 moles of baking soda yields 1 mole of sodium carbonate
0.0185 moles of baking soda will yield = 0.0185 moles ×1 /2 = 9.25 ×10^-3 moles of sodium carbonate.
Therefore, mass of sodium carbonate= 9.25 ×10^-3 moles × 106gmol-1= 0.9805 g of sodium carbonate. This is the theoretical yield of sodium carbonate.
%yield = actual yield/theoretical yield ×100
% yield = 0.991/0.9805 ×100
%yield = 101%
Since ;
2NaHCO3 → Na2CO3 + H2O + CO2
And H2O + CO2 ---> H2CO3
Hence I can write, 2NaHCO3 → Na2CO3 + H2CO3
Molar mass of H2CO3= 62.03 gmol-1
Molar mass of baking soda= 84 gmol-1
Therefore, mass of baking soda=
0.325/62.03 × 2 × 84 = 0.88 g of NaHCO3
% of NaHCO3= 0.88/1.473 × 100 = 59.7%
The decomposition reaction of baking soda is a reaction in which water and carbon dioxide ae given off as gaseous products.
5. The theoretical yield of Na₂CO₃ is approximately 0.9809 gramsThe percentage yield of sodium carbonate is approximately 101.02%.6. Percentage of NaHCO₃ in the mixture is approximately 59.76%.Reasons:
Mass of baking soda = 1.555 g
Mass of Na₂CO₃ produced = 0.991 g
Required:
Calculation for the theoretical yield
Solution:
Theoretical yield (mass) of Na₂CO₃ produced is found as follows;
Molar mass of Na₂CO₃ = 105.9888 g/mol
Molar mass of NaHCO₃ = 84.007 g/mol
[tex]\displaystyle 1.555 \, g \, NaHCO_3 \times \frac{1 \, mol \, NaHCO_3}{84.007 \, g \, NaHCO_3} \times \frac{1 \, mol \, Na_2CO_3}{2 \, mol \, NaHCO_3} \times 105.9888 \ g \approx 0.9809 \, g \, Na_2CO_3[/tex]
The theoretical yield of Na₂CO₃ ≈ 0.9809 grams.
The percentage yield is given as follows;
[tex]\displaystyle Percentage \ yield = \mathbf{\frac{Actual \, Yield}{Theorectical \, Yield} \times 100 \%}[/tex]
The percentage yield of Na₂CO₃ is therefore;
[tex]\displaystyle Percentage \ yield \ of \ Na_2CO_3= \frac{0.991}{0.9809} \times 100 \% \approx \underline{ 101.02 \%}[/tex]
(Some baking soda may remain if the reaction is not completed)
6. Mass of the unknown mixture of baking soda = 1473 g
Mass loss from the mixture = 0.325 g
Required:
The percentage of NaHCO₃ in the mixture.
Solution:
The chemical in the mass loss from heating the NaHCO₃ = H₂CO₃
Molar mass of H₂CO₃ = 62.03 g/mol
[tex]\displaystyle \mathrm{Number \ of \ moles \ of \ H_2CO_3 \ produced} = \frac{0.325 \, g}{62.03 \, g/mol} \approx 5.2394 \times 10^{-3} \ moles[/tex]
The chemical reaction is presented as follows;
2NaHCO₃(s) [tex]\underrightarrow {\Delta \ Heated}[/tex] Na₂CO₃(s) + H₂CO₃(g)2 moles of NaHCO₃ produces 1 mole of H₂CO₃The number of moles of NaHCO₃ in the mixture is therefore;
2 × 5.2394 × 10⁻³ moles ≈ 1.04788 × 10⁻² molesMass of NaHCO₃ in the mixture is therefore
Mass of NaHCO₃ = 1.04788 × 10⁻² moles × 84.007 g/mol = 0.88029 g[tex]\displaystyle Percentage \ of \ NaHCO_3 \ in \ the \ mixture \ = \mathbf{ \frac{Mass \ of \ NaHCO_3}{Mass \ of \ mixture} \times 100}[/tex]
Which gives;
[tex]\displaystyle Percentage \ of \ NaHCO_3 \ in \ the \ mixture \ = \ \frac{0.88029 \, g}{1.473 \, g} \times 100 \approx \underline{ 59.76 \%}[/tex]Learn more here:
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Answer these questions, please.
Answer:
1a. 0.89 gcm¯³
1b. Yes.
1c. Tetrahydrofuran.
2. 0.54 g/mL
Explanation:
1. Data obtained from the question include:
Volume = 0.988 L = 988 cm³
Mass = 879 g
1a. Determination of the density
Density = mass /volume
Density = 879/ 988
Density = 0.89 gcm¯³
Therefore, the density of the liquid is 0.89 gcm¯³
1b. From the given data, it is possible to determine the identity of the liquid.
1c. The density of the liquid is 0.89 gcm¯³. Comparing the density of the liquid obtained with those given in the table, the liquid is tetrahydrofuran
2. Data obtained from the question include:
Mass of empty cylinder = 5.25 g
Mass of cylinder and sodium thiosulfate = 75.82 g
Volume = 130.63 mL
Next, we shall determine the mass of sodium thiosulfate. This can be obtain as follow:
Mass of empty cylinder = 5.25 g
Mass of cylinder and sodium thiosulfate = 75.82 g
Mass of sodium thiosulfate =.?
Mass of sodium thiosulfate = Mass of cylinder and sodium thiosulfate – Mass of empty cylinder
Mass of sodium thiosulfate = 75.82 – 5.25
Mass of sodium thiosulfate = 70.57 g
Finally, we shall determine the concentration of the sodium thiosulfate as follow:
Mass = 70.57 g
Volume = 130.63 mL
Concentration =?
Concentration = mass /volume
Concentration = 70.57/130.63
Concentration = 0.54 g/mL
The concentration of the solution is 0.54 g/mL
A 10.0 mL sample of calcium hydroxide solution required 26.85 mL of 0.225 M hydrochloric acid for neutralization. The balanced equation is:
Answer:
[tex]C_{base}=0.302M[/tex]
Explanation:
Hello,
In this case, we can evidence that when calcium hydroxide solution reacts with hydrochloric acid solution, the balanced neutralization reaction turns out:
[tex]2HCl(aq)+Ca(OH)_2\rightarrow CaCl_2(aq)+2H_2O(l)[/tex]
Moreover, the concentration of neutralized calcium hydroxide can be computed by using the 2:1 mole ratio between the base and the acid:
[tex]C_{acid}V_{acid}=2*C_{base}V_{base}\\\\C_{base}=\frac{C_{acid}V_{acid}}{2*V_{base}} =\frac{0.225M*26.85mL}{2*10.0mL}\\ \\C_{base}=0.302M[/tex]
Regards.
Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction.
H2(g) → 2H+(aq) + 2e- oxidation
Cu2+(aq) → + 2e-Cu(s) reduction
Write a balanced equation for the overall redox reaction. Use smallest possible integer coefficients.
Answer:
H₂(g) + Cu²⁺(aq) → 2H⁺(aq) + Cu(s)
Explanation:
In a redox reaction, one half-reaction is the oxidation (where the atom loss electrons) whereas the other reaction is the reduction (Where the atom is gaining electrons.
In the reactions:
H₂(g) → 2H⁺(aq) + 2e⁻ oxidation
Here, the reaction is written as the oxidation because the hydrogen H₂ is in oxidation state 0 and H⁺ in +1. That means each atom is loosing one electron.
Cu²⁺(aq) + 2e⁻ → Cu(s) reduction
And here, the Cu²⁺ is in +2 oxidation state and after the reaction is in Cu(s) 0 state. Thus, each atom is gaining 2 electrons.
The sum of both reactions is:
H₂(g) + Cu²⁺(aq) + 2e⁻ → 2H⁺(aq) + 2e⁻ + Cu(s)
Subtracting the electrons in both sides of the reaction:
H₂(g) + Cu²⁺(aq) → 2H⁺(aq) + Cu(s)The molecular weight of table salt, NaCl, is 58.5 g/mol. A tablespoon of salt weighs 6.37 grams. Calculate the number of moles of salt in one tablespoon.
Finally, solve (remember significant figures):
Answer:
0.109 mol/tablespoon
Explanation:
6.37 g/ 58.5 mol = 0.10888888 mol (0.109 significantly)
Answer:
A: 0.109
Explanation:
Edge 2020
If 3.10 moles of P4010 reacted with excess water, how many grams of H3PO4
would be produced?
P4010 +6H20 + 4H3PO4
You Answered
126 g
0 0.007918
Correct Answer
O 1220 g
0.1278
75.98
Answer:
1.22 × 10³ g
Explanation:
Step 1: Write the balanced equation
P₄O₁₀ + 6 H₂O ⇒ 4 H₃PO₄
Step 2: Calculate the moles of H₃PO₄ produced by 3.10 moles of P₄O₁₀
The molar ratio of P₄O₁₀ to H₃PO₄ is 1:4. The moles of H₃PO₄ produced are 4/1 × 3.10 mol = 12.4 mol
Step 3: Calculate the mass corresponding to 12.4 moles of H₃PO₄
The molar mass of H₃PO₄ is 97.99 g/mol.
[tex]12.4 mol \times \frac{97.99g}{mol} = 1.22 \times 10^{3} g[/tex]
"Calculate the pH during the titration of 30.00 mL of 0.1000 M C3H7COOH(aq) with 0.1000 M LiOH(aq) after 29.9 mL of the base have been added. Ka of butanoic acid
Answer:
pH = 7.29
Explanation:
Ka of butanoic acid is 1.54x10⁻⁵
To obtain the pH of the solution you must use H-H equation for butanoic acid:
pH = pKa + log₁₀ [C₃H₇COO⁻] / [C₃H₇COOH]
Where pKa is defined as -log Ka = 4.81
Now, you need to find [C₃H₇COO⁻] and [C₃H₇COOH] concentrations (Also, you can find moles of each substance and replace them in the equation.
Butanoic acid reacts with LiOH, producing C₃H₇COO⁻, thus:
C₃H₇COOH + LiOH → C₃H₇COO⁻ + H₂O + Li⁺
Moles of both reactants, C₃H₇COOH and LiOH are:
C₃H₇COOH = 0.0300L ₓ (0.1000mol / L) = 0.003000moles of C₃H₇COOH
LiOH = 0.0299L ₓ (0.1000mol / L) = 0.00299 moles of LiOH.
That means moles of C₃H₇COO⁻ produced are 0.00299 moles.
And moles of C₃H₇COOH that remains in solution are:
0.00300 - 0.00299 = 0.00001 moles of C₃H₇COOH
Replacing in H-H equation:
pH = pKa + log₁₀ [C₃H₇COO⁻] / [C₃H₇COOH]
pH = 4.81 + log₁₀ [0.00299moles] / [0.00001moles]
pH = 7.29Homolysis, or homolytic bond dissociation, produces a very specific type of product under certain reaction conditions. In Part 1, select all the products (in formulae and general chemical terms) that could result from homolysis. In Part 2, select the reaction conditions that are most likely to promote homolysis.
Part 1. Choose all that may occur as possible products of a homolysis reaction.
Choose one or more:_______.
a. hydride ion
b. R3CO
c. Br2
d. H
e. a carbocation
f. H3C
g. H3CO-
h. hydrogen ion
i. a carbon free radical
Part 2. Choose the conditions under which homolysis is likely to occur.
Choose one or more:_______.
a. strong base
b. ultraviolet irradiation
c. high temperature
d. strong acid
e. infrared irradiation
f. low temperature
Answer:
1) R₃CO , H, H₃C, a carbon free radical
2) high temperature, ultraviolet irradiation
Explanation:
1) Homolysis leads to the formation of free radicals (species having a free electron). Thus, answer is :
R₃CO
H
H₃C
a carbon free radical
2) Homolysis require high temperature, ultraviolet irradiation.
When Carl Woese developed the modern system of classification, he broke the previous kingdom of into the two kingdoms of Bacteria and Archaea
Answer:
the answer is monerans
Explanation:
When Carl Woese developed the modern system of classification, he broke the previous kingdom of Monera into the two kingdoms of Bacteria and Archaea.
What kingdom of Monera ?Some biologists believed it made sense to classify prokaryotes as belonging to their own kingdom, the Monera. That served as the foundation for Richard Whittaker and Lynn Margulis's five-kingdom proposal, which enhanced the Haeckel plan by include a kingdom of fungus.
Protists, protozoa, monera, fungi, and viruses have long been proposed as belonging to different kingdoms, but traditional evolutionists during the majority of the 20th century had given none of them any thought.
Later, the Monera kingdom was split into Eubacteria and Archaebacteria by Carl Woese . Moreover, he divided the five kingdoms into three domains: Eukaryotes, Archaea, and Bacteria.
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Your question is incomplete. But your complete question is as follows:
When Carl Woese developed the modern system of classification, he broke the previous kingdom of into the two kingdoms of _____ into Bacteria and Archaea.
The element nitrogen forms a(n) _______ with the charge . The symbol for this ion is , and the name is . The number of electrons in this ion is .
Answer:
The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10.
Explanation:
The element nitrogen is in the Group 15 in the Periodic Table, so it tends to gain 3 electrons (3 negative charges) to fill its valance shell with 8 electrons.
The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10 (the original 7 plus the 3 gained). It is isoelectronic with the gas Neon, which accounts for its stability.
Argon gas has a boiling point of -197 °C. Which of the following diagrams best represents the
distribution of argon atoms in a steel sphere at -190 °C?
Answer:
(a)
Explanation:
Hello,
In this case, the temperature required to boil argon, it means, transform it from liquid to gas is -197 °C. In such a way, since the temperature inside the steel sphere is -190 °C, which is greater than the boiling point, we realize argon is gaseous, therefore, the molecules will be spread inside the sphere as they will be moving based on the kinetic theory of gases.
For that reason, answer is scheme (a).
Best regards.
If you start with 512 grams of aluminum and 1147 grams of copper chloride to make aluminum chloride and copper, what is the limiting reagent? 2Al + 3CuCl -> 2AlCl3 + 3Cu
Answer:
Copper (II) chloride.
Explanation:
Hello,
In this case, considering the described reaction which is also given as:
[tex]2Al + 3CuCl_2 \rightarrow 2AlCl_3 + 3Cu[/tex]
For us to identify the limiting reactant we first compute the available moles of aluminium:
[tex]n_{Al}=512gAl*\frac{1molAl}{27gAl}=19.0molAl[/tex]
Next, we compute the consumed moles of aluminium by the 1147 grams of copper (II) chloride by using their 2:3 molar ratio:
[tex]n_{Al}^{consumed}=1147gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2}*\frac{2molAl}{3molCuCl_2} =5.69molAl[/tex]
Thereby, we can infer aluminium is in excess since less moles are consumed than available whereas the copper (II) chloride is the limiting reactant.
Best regards.
A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of CO2 and 0.01961 mol of H2O. The empirical formula of the compound was found to be C3H6O2. Show how this was calculated.
What does the empirical formula tell you about the compound?
The molar mass of the actual compound was found to be 222.27g/mol. Find the molecular formula of this compound. What does the molecular formula tell you about the compound?
Can you see what type of functional group this compound could have?
Answer:
See explanation.
Explanation:
Hello,
In this case, we can show how the empirical formula is found by following the shown below procedure:
1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:
[tex]n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC[/tex]
2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:
[tex]n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH[/tex]
3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:
[tex]m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH} =0.1900gO[/tex]
4. Compute the moles of oxygen by using its molar mass:
[tex]n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO[/tex]
5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:
[tex]C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1[/tex]
6. Search for the closest whole number (in this case multiply by 2):
[tex]C_3H_6O_2[/tex]
Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:
[tex]M=12*3+1*6+16*2=74g/mol[/tex]
Which is about three times in the molecular formula, for that reason, the actual formula is:
[tex]C_9H_{18}O_6[/tex]
It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.
Best regards.
A transition in the balmer series for hydrogen has an observed wavelength of 434 nm. Use the Rydberg equation below to find the energy level that the transition originated. Transitions in the Balmer series all terminate n=2.
Delta E= -2.178 x10-18J ( 1/n2Final - 1/n2Initial )
The number is 5.
What is the energy of this transition in units of kJ/mole? ( hint: the anser is NOT 4.58x10-22kJ/mole or -4.58x10-22kJ/mole)
Answer:
i. n = 5
ii. ΔE = 7.61 × [tex]10^{-46}[/tex] KJ/mole
Explanation:
1. ΔE = (1/λ) = -2.178 × [tex]10^{-18}[/tex]([tex]\frac{1}{n^{2}_{final} }[/tex] - [tex]\frac{1}{n^{2}_{initial} }[/tex])
(1/434 × [tex]10^{-9}[/tex]) = -2.178 × [tex]10^{-18}[/tex] ([tex]\frac{n^{2}_{initial} - n^{2}_{final} }{n^{2}_{final} n^{2}_{initial} }[/tex])
⇒ 434 × [tex]10^{-9}[/tex] = (1/-2.178 × [tex]10^{-18}[/tex])[tex]\frac{n^{2}_{final} *n^{2}_{initial} }{n^{2}_{initial} - n^{2}_{final} }[/tex]
But, [tex]n_{final}[/tex] = 2
434 × [tex]10^{-9}[/tex] = (1/2.178 × [tex]10^{-18}[/tex])[tex]\frac{2^{2} n^{2}_{initial} }{n^{2}_{initial} - 2^{2} }[/tex]
434 × [tex]10^{-9}[/tex] × 2.178 × [tex]10^{-18}[/tex] = [tex](\frac{4n^{2}_{initial} }{n^{2}_{initial} - 4 })[/tex]
⇒ [tex]n_{initial}[/tex] = 5
Therefore, the initial energy level where transition occurred is from 5.
2. ΔE = hf
= (hc) ÷ λ
= (6.626 × 10−34 × 3.0 × [tex]10^{8}[/tex] ) ÷ (434 × [tex]10^{-9}[/tex])
= (1.9878 × [tex]10^{-25}[/tex]) ÷ (434 × [tex]10^{-9}[/tex])
= 4.58 × [tex]10^{-19}[/tex] J
= 4.58 × [tex]10^{-22}[/tex] KJ
But 1 mole = 6.02×[tex]10^{23}[/tex], then;
energy in KJ/mole = (4.58 × [tex]10^{-22}[/tex] KJ) ÷ (6.02×[tex]10^{23}[/tex])
= 7.61 × [tex]10^{-46}[/tex] KJ/mole
The initial energy level is 5 and the energy of this transition in units of kJ/mole is 7.57 * 10^-43 kJ/mole
We must first calculate ΔE as follows;
ΔE = hc/λ
h = Plank's constant = 6.6 * 10^-34 Js
c = speed of light = 3 * 10^8 m/s
λ = wavelength = 434 * 10^-9
ΔE = 6.6 * 10^-34 * 3 * 10^8/434 * 10^-9
ΔE = 0.0456 * 10^-17 J
ΔE = [tex]ΔE = -2.178 x10^-18 (\frac{1}{n^2final} - \frac{1}{n^2initial}) \\ΔE = -2.178 x10^-18 (\frac{1}{2^2} - \frac{1}{n^2initial} )\\\\4.56 * 10^-19/2.178 x10^-18 = (\frac{1}{2^2} - \frac{1}{n^2initial})\\0.210 = (\frac{1}{2^2} - \frac{1}{n^2initial})\\\frac{1}{n^2initial} = 0.25 - 0.210\\\frac{1}{n^2final} = 0.04\\n = (\sqrt{(0.04)^-1} \\n = 5[/tex]
Energy of this transition in units of kJ/mole = 4.56 * 10^-19/ 6.02 * 10^23
= 7.57 * 10^-43 kJ/mole
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What is a ‘control’ in an experiment?
A. A version of the experiment that is unchanged to make sure the experimental data is not due to chance.
B. A person who oversees the experiment to make sure it is following proper procedures.
C. The variable controlled by the scientist to affect the dependent variable.
D. The name for the set of independent and dependent variables that will be controlled by the scientist.
need help asap got 1 minute
D. The name for the set of independent and dependent variables that will be controlled by the scientist.
The statement, that describes the ‘control’ in an experiment is "the name for the set of independent and dependent variables that will be controlled by the scientist."
What is a control in experiment?A control is an element in an experiment that remains intact or unaffected by other variables. An experiment or observation aiming to minimise the influence of variables other than the independent variable is referred to as a scientific control. It serves as a standard or point of reference against which other test findings are measured.
In a scientific experiment, an independent variable is the variable that is modified or manipulated in order to assess the effects on the dependent variable. In a scientific experiment, the dependent variable is the variable that is being tested and measured. The designation given to the set of independent and dependent variables that the scientist will regulate.
Hence the correct option is D.
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Silver crystallizes in a face-centered cubic structure. What is the edge length of the unit cell if the atomic radius of silver is 144 pm?
Answer:
Edge length of the unit cell is 4.07x10⁻¹⁰m
Explanation:
In a face-centered cubic structure, the edge, a, could be obtained using pythagoras theorem knowing the hypotenuse of the unit cell, b, is equal to 4r:
a² + a² = b² = (4r)²
2a² = 16r²
a = √8 r
That means edge lenght is = √8 r
adius
As radius of Silver is 144pm = 144x10⁻¹²m:
a = √8 r
a = √8 ₓ 144x10⁻¹²m
a = 4.07x10⁻¹⁰m
Edge length of the unit cell is 4.07x10⁻¹⁰mThe average bond length in C-C in benzene (C6H6) is between single and double bond lengths. This is due to which of the following effect?
1. Due to its refractive index2. Due to the possession of resonance3. Due to its H atoms4. Due to the H-bonds
Answer:
2. Due to the possession of resonance
Explanation:
In the benzene ring, the electrons that results in the bonds between the carbon atoms are delocalized. That is, they do not belong to a specific carbon atom. It is this unique feature that enables them to have a bond length between single and double bond lengths.
This feature is as a result of resonance.
The correct option is 2.
neeeeed helpppppppppp
Answer:
Option C. Will always.
Explanation:
A spontaneous reaction is a reaction that occurs without an external supply of heat.
This implies that spontaneous reaction will always occur as no external supply of heat is needed.
Note the dynamic equilibrium in the opening photo which solution changes color when the pH of both solutions is increased explain?
Answer:
The colour of the orange solution becomes yellow.
Explanation:
1. Before adding NaOH
Assume the picture showed a beaker of potassium chromate and one of potassium dichromate.
Both solutions are involved in the same equilibrium:
[tex]\rm\underbrace{\hbox{2CrO$_{4}^{2-}$(aq)}}_{\text{yellow}} +2H^{+}(aq) \rightleftharpoons \, \underbrace{\hbox{Cr$_{2}$O$_{7}^{2-}$}}_{\text{orange}} + H_{2}O[/tex]
The first beaker contains mostly chromate ions with a few dichromate ions.
The position of equilibrium lies to the left and the solution is yellow.
The second beaker contains mostly dichromate ions with a few chromate ions.
The position of equilibrium lies to the right and the solution is orange.
2. After adding NaOH
According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.
Beaker 1
If you add OH⁻ to the equilibrium solution, it removes the H⁺ (by forming water).
The system responds by having the dichromate react with water to replace the H⁺.
At the same time, the system forms more of the yellow chromate ion.
The position of equilibrium shifts to the left.
However, the solution is already yellow, so you see no change in colour.
Beaker 2
The reaction is the same as in Beaker 1.
This time, however, as the dichromate ion disappears, do does its orange colour.
Also, the yellow chromate is being formed and its yellow colour appears .
The colour changes from orange to yellow.
If the vinegar were measured volumetrically (e.g., a pipet), what additional piece of data would be needed to complete the calculations for the experiment?
Answer:
the density if vinegar will also be needed
Explanation:
Because this is an experiment of volumetric analysis
The thermochemical equation is for the reaction of hydrogen bromide gas to form hydrogen gas and bromine liquid. 2HBr(g) = H 2 (g)+ Br 2 (l) 72.6 kJ How many grams of HBr (g) would be made to react if 11.4 energy were provided?
Answer:
the mass of HBr that would react is 25.41 g of HBr
Explanation:
attached is the calculations.
How many valence electrons are in the electron dot structures for the elements in group 3A(13)?
Answer:
here, as we have known the elements of group 3A(13) such as aluminium , boron has three valance electron and in perodic table the elements are kept with similar proterties in same place so, their valance electron is 3.
hope it helps...
The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.
What are Groups in the Periodic Table?The periodic table is organized into groups (vertical columns), periods (horizontal rows), and families (groups of elements that are similar). Elements in the same group have the same number of valence electrons.
Groups are the columns of the periodic table, and periods are the rows. There are 18 groups, and there are 7 periods plus the lanthanides and actinides.
There are two different numbering systems that are commonly used to designate groups, and you should be familiar with both.
The traditional system used in the United States involves the use of the letters A and B. The first two groups are 1A and 2A, while the last six groups are 3A through 8A. The middle groups use B in their titles.
Therefore, The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.
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merits of modern periodic table?
Answer:
Merits of modern periodic table:The wrong position of some elements like argon, potassium, cobalt and nickel due to atomic weights have been solved by arranging the elements in the order of increasing atomic number without changing their own places.The isotopes of some element have the same atomic numbers. Therefore, they find the same position in periodic table.It separates metals from non-metals.The groups of the table are divided into sub groups A and B due to their dissimilar properties which make the study of elements specific and easier.The representative and transition elements have been separated.Hope this helps...
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Which of the following is NOT one of the types of bonds? A. Ionic B. Metallic C. Covalent D. Valence
Considering the definition of bond and the different type of bonds, valence is not one of the types of bonds.
What is a chemical bondA chemical bond is defined as the force by which the atoms of a compound are held together. These are electromagnetic forces that give rise to different types of chemical bonds.
In other words, a chemical bond is the force that joins atoms to form chemical compounds and confers stability to the resulting compound.
Covalent bondThe covalent bond is the chemical bond between atoms where electrons are shared, forming a molecule. Covalent bonds are established between non-metallic elements, such as hydrogen H, oxygen O and chlorine Cl. These elements have many electrons in their outermost level (valence electrons) and have a tendency to gain electrons to acquire the stability of the electronic structure of noble gas. The shared electron pair is common to the two atoms and holds them together.
Ionic bondAn ionic bond is produced between metallic and non-metallic atoms, where electrons are completely transferred from one atom to another. During this process, one atom loses electrons and another one gains them, forming ions.
Metallic bondMetallic bonds are a type of chemical bond that occurs only between atoms of the same metallic element. In this way, metals achieve extremely compact, solid and resistant molecular structures, since the atoms that share their valence electrons.
SummaryIn summary, valence is not one of the types of bonds. The types of bonds are covalent, ionic and metallic.
Learn more about chemical bonds:
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What is the absolute magnitude of the rate of change for [NH3] if the
rate of change for [Hz] is 9.00 M/s in the reaction 2 NH3(g) → N2(g) +
3 H2(g)?
Answer:
[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]
Explanation:
Hello,
In this case, we can write the law of mass action for the undergoing chemical reaction, based on the rates and the stoichiometric coefficients:
[tex]\frac{1}{-2}r_{NH_3} =\frac{1}{1} r_{N_2}=\frac{1}{3}r_{H_2}[/tex]
In such a way, knowing the rate of formation hydrogen (H₂), we can know the rate of change of ammonia, that must be negative for consumption:
[tex]r_{NH_3} =\frac{-2}{3}r_{H_2}=\frac{-2}{3}*9.00\frac{M}{s} \\\\r_{NH_3} =-6.00\frac{M}{s}[/tex]
Nevertheless, the absolute magnitude will be positive:
[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]
Best regards.
Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the other product is magnesium fluoride. Write and balance the equation.
Answer:
2ErF3 + 3Mg → 2Er + 3MgF2
Explanation:
Erbium metal is a member of the lanthaniod series. It reacts with halogens directly to yield erbium III halides such as erbium III chloride, Erbium III fluoride etc.
Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the products are erbium metal and magnesium fluoride. This is a normal redox process in which the Erbium metal is reduced while the magnesium is oxidized. The balanced reaction equation of this process is; 2ErF3 + 3Mg → 2Er + 3MgF2
Calculate the combustion of gaseous dimethyl ether CH 3 OCH 3 (g)+3O 2 (g) 2CO 2 (g)+3H 2 O(l) using standard molar enthalpies of formation Molecule AH H l ^ 0 (k)/mol) CH 3 OCH 3 (g) - 184.1
Answer:
[tex]\Delta _cH=-1328.3kJ/mol[/tex]
Explanation:
Helllo,
In this case, for the given chemical reaction in gaseous state:
[tex]CH_3OCH_3+3O_2\rightarrow 2CO_2+3H_2O[/tex]
We comoute the combustion enthalpy as the reaction enthalpy for one mole of fuel (dimethyl ether) considering the formation enthalpy of each given substance and whether they are reactants (subtracting) or products (adding), therefore we write:
[tex]\Delta _cH=2*\Delta _fH_{CO_2}+3*\Delta _fH_{H_2O}-\Delta _fH_{CH_3OCH_3}-3*\Delta _fH_{O_2}[/tex]
Whereas the formation enthalpies for carbon dioxide, water, dimethyl ether and oxygen are -393.5, -241.8, -184.1 and 0 kJ/mol respectively, thereby, the combustion enthalpy turns out:
[tex]\Delta _cH=2(-393.5)+3*(-241.8)-(-184.1)-3(0)\\\\\Delta _cH=-1328.3kJ/mol[/tex]
Notice that enthalpy of formation of oxygen is zero since forming an element has no chemical sense, it just exists as it has been early demonstrated.
Regards.
Chemistry question. Image attached.
Answer:
The answer to your question is given below
Explanation:
The balanced equation for the reaction is given below:
CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)
1. Writing an expression for the equilibrium constant, K.
The equilibrium constant, K for a reaction is simply the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.
Thus, we can write the equilibrium constant, K for the reaction as follow:
CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)
K = [CaCO3] [H2]⁴ / [CaO] [CH4] [H2O]²
2. Based on the value of K, more products will be in the equilibrium mixture since the value of K is a positive large number.
Which example involves a phase change in which heat energy is released by the substance?
Ofreezing ice cream
O cooking a pot of soup
O melting ice under sunlight
O watching frost disappear into air
Answer:
Cooking a pot of soup
Explanation:
id say that because when you freeze ice cream, its already frozen, so no heat is being released. melting ice wouldn't be the answer because, once again, it is already frozen, and no heat is being released.
Answer:
the correct answer is freezing ice cream
Explanation:
i took the test & got this question correct. also, heat energy is released when freezing because there is no heat energy involved.
When 75.5 grams of phosphorus pentachloride react with an excess of water, as shown in the unbalanced chemical equation below, how many moles of hydrochloric acid will be produced? Please show all your work for the calculations for full credit. PCl5 + H2O --> H3PO4 + HC
Answer:
Explanation: M(PCL5)= 31 + 5(35.5)
=208.5g/mol
M(H20)= 18g/mol
n(PCL5) = 75.5÷208.5
= 0.362mol
n(HCl)/n(PCL5)= 5/1
n(HCl)= 5×0.362
=1.81mol of HCl
How many minutes would be required to electroplate 25.0 grams of chromium by passing a constant current of 4.80 amperes through a solution containing CrCl3
Answer:
483.27 minutes
Explanation:
using second faradays law of electrolysis