Answer:
a.
[tex]m_K=8.056gK\\ \\m_{Cl_2}=4.028gCl_2[/tex]
b.
[tex]m_{Cr}=10.51gCr\\ \\m_{O_2}=4.851gO_2[/tex]
c.
[tex]m_{Sr}=13.88gSr\\\\m_{N_2}=1.479gN_2[/tex]
Explanation:
Hello,
In this case, we proceed via stoichiometry in order to compute the masses of all the reactants as shown below:
a. [tex]2K+Cl_2\rightarrow 2KCl[/tex]
[tex]m_K=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{2molK}{2molKCl}* \frac{39.1gK}{1molK}=8.056gK\\ \\m_{Cl_2}=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{1molCl_2}{2molKCl}* \frac{70.9gCl_2}{1molCl_2}=4.028gCl_2[/tex]
b. [tex]4Cr+ 3O_2\rightarrow 2Cr_2O_3[/tex]
[tex]m_{Cr}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{4molCr}{2molCr_2O_3}* \frac{52gCr}{1molCr_2O_3}=10.51gCr\\ \\m_{O_2}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{3molO_2}{2molCr_2O_3}* \frac{32gO_2}{1molCr_2O_3}=4.851gO_2[/tex]
c. [tex]3Sr(s) + N_2(g) \rightarrow Sr_3N_2[/tex]
[tex]m_{Sr}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{3molSr}{1molSr_3N_2}* \frac{87.62gSr}{1molSr}=13.88gSr\\\\m_{N_2}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{1molN_2}{1molSr_3N_2}* \frac{28gN_2}{1molN_2}=1.479gN_2[/tex]
Regards.
How do covalent bonds form? A Sharing valence electrons between atoms. B Donating and receiving valence electrons between atoms. C Opposite slight charges attract each other between compounds. D Scientists are still not sure how they form.
Answer:
A. Sharing valence electrons between atoms.
Explanation:
This is the definition of a covalent bond. Option B describes ionic bonds, Option C describes intermolecular forces, and Option D is wrong because then there wouldn't be any mention of them in our high school chemistry textbooks :).
whts the ph of po4 9.78
Answer:
4.22
Explanation:
We know from the question, that the pOH of the solution is 9.78. Now the pOH is defined as -log [OH^-].
If the pOH of a solution is given, one may obtain the pH of such solution from the formula;
pH + pOH =14
Hence we can write;
pH = 14-pOH
pH = 14 - 9.78 = 4.22
Hence the pH of the solution is 4.22.
Account for the change when NO2Cl is added using the reaction quotient Qc. Match the words in the left column to the appropriate blanks in the sentences on the right.
1. decreases
2. loss
3. Increases
4. greater
A. Disturbing the equilibrium by adding NO2Cl______Qc to a value_____than Kc.
B. To reach a new state of equilibrium, Qc therefore______which means that the denominator of the expression for Qc______.
C. To accomplish this, the concentration of reagents______, and the concentration of products_______.
Answer:
A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc.
B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases.
C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases.
Explanation:
Hello,
In this case, for the equilibrium reaction:
[tex]NO_2Cl(g)+NO(g)\rightleftharpoons NOCl(g)+NO_2(g)[/tex]
Whose equilibrium expression is:
[tex]Kc=\frac{[NO_2][NOCl]}{[NO_2Cl][NO]}[/tex]
The proper matching is:
A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc, since the denominator becomes greater, therefore, Qc decreases.
B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases, since the lower the denominator, the higher Qc as it has the concentration of reactants.
C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases, since the reactants must be consumed in order to reestablish equilibrium by shifting the reaction towards the products.
Best regards.
which of the following is an example of precipitate a. liquid evaporating into gas b. a solid form from a frozen liquid c. a chunky solid form from 2 liquids combining d. a liquid solution that contains 2 substances
Answer:
B and C
Explanation:
The google definition of precipitate is "a solid formed by a change in a solution, often due to a chemical reaction or change in temperature that decreases solubility of a solid". So in this case, the "solid formed from a frozen liquid" and "a chunky solid form from 2 liquids combining" are both examples of a precipitate.
The correct answer is C, a chunky solid form from 2 liquids combining.
In the laboratory, we usually mix two chemicals. When we mix the chemicals, we expect them to interact in one way or another. The interaction of those chemicals is known as a chemical reaction.
When we mix two liquid chemicals and they interact with each other to yield a solid product, we say that a chemical reaction has occurred and that such chemical reaction has produced a precipitate.
https://brainly.com/question/24276721
Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25°C. (The equation is balanced.) 3 Cl2(g) + 2 Fe(s) → 6 Cl-(aq) + 2 Fe3+(aq) Cl2(g) + 2 e- → 2 Cl-(aq) E° = +1.36 V Fe3+(aq) + 3 e- → Fe(s) E° = -0.04 V
The cell potential for the electrochemical cell has been 1.40 V.
The standard reaction for the cell will be:
[tex]\rm 3\;Cl_2\;+\;2\;Fe\;\rightarrow\;6\;Cl^-\;+\;2\;Fe^3^+[/tex]
The half-reaction of the cells has been:
[tex]\rm Fe^3^+\;+\;3\;e^-\;\rightarrow\;Fe[/tex]
The potential for this reduction has been -0.04 V.
[tex]\rm Cl_2\;+\;2\;e^-\;\rightarrow\;2\;Cl^-[/tex]
The potential for the reduction has been 1.36 V.
The cell potential has been: Potential of reduction - Potential of oxidation
Cell potential = 1.36 - (-0.04) V
Cell potential = 1.40 V.
The cell potential for the electrochemical cell has been 1.40 V.
For more information about the electrochemical cell, refer to the link:
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How many grams of PtBr4 will dissolve in 250.0 mL of water that has 1.00 grams of KBr dissolved in it
Answer:
[tex]m_{PtBr_4}=0.306gPtBr_4[/tex]
Explanation:
Hello,
In this case, since the solubility product of platinum (IV) bromide is 8.21x10⁻⁹, and the dissociation is:
[tex]PtBr_4(s)\rightleftharpoons Pt^{4+}(aq)+4Br^-(aq)[/tex]
The equilibrium expression is:
[tex]Ksp=[Pt^{4+}][Br^-]^4[/tex]
Thus, since the salt is added to a solution initially containing 1.00 grams of potassium bromide, there is an initial concentration of bromide ions:
[tex][Br^-]_0=\frac{1.00gKBr*\frac{1molKBr}{119gKBr}*\frac{1molBr^-}{1molKBr} }{0.250L}=0.0336M[/tex]
Hence, in terms of the molar solubility [tex]x[/tex], we can write:
[tex]8.21x10^{-9}=(x)(0.0336+4x)^4[/tex]
In such a way, solving for [tex]x[/tex], we obtain:
[tex]x=0.00238M[/tex]
Which is the molar solubility of platinum (IV) bromide. Then, since its molar mass is 514.7 g/mol, we can compute the grams that get dissolved in the 250.0-mL solution:
[tex]m_{PtBr_4}=0.00238\frac{molPtBr_4}{1L}*0.250L *\frac{514.7gPtBr_4}{1molPtBr_4} \\\\m_{PtBr_4}=0.306gPtBr_4[/tex]
Best regards.
15. How many moles of carbon tetrachloride (CCI) is represented by 543.2 g of carbon tetrachloride? The atomic weight of carbon is 12.01
and the atomic weight of chlorine is 35.45.
O A. 11.4 moles
O B.3.53 moles
C. 5.42 moles
D. 8.35x10 moles
Answer:
well, first off. the formula for carbon tetrachloride is CCl4
We need to find the molar mass of carbon tetrachloride
1(Mass of C) + 4(mass of chlorine)
1(12) + 4(35.5)
12 + 142
154 g/mol
Number of moles of CCl3 in 543.2g CCl3
n = given mass / molar mass
n = 543.2/153
n = 3.53 moles
always remember to brainly the questions you find helpful
Answer:
3.53 moles
Explanation:
A gas mixture contains 3.50 moles of helium, 5.00 moles of krypton and 7.60 moles of neon. A) What is the mole fraction for each gas
Answer:
.217, .311, and .472, respectively.
Explanation:
The total number of moles of gas is 3.50 + 5.00 + 7.60 = 16.10 (to preserve significant digits).
X of helium=3.50/16.10 = .217
X of krypton=5.00/16.10 = .311
X of neon=7.60/16.10 = .472
1500 L has how many significants figures
Answer:
It has 2
Explanation:
The significant figures are 1 and 5!
Hope this helps:)
What will be formed when 2,2,3-trimethylcyclohexanone reacts with hydroxylamine?
Answer:
Following are the solution to this equation:
Explanation:
In the given-question, an attachment file of the choices was missing, which can be attached in the question and its solution can be defined as follows:
In the given question "Option (iii)" is correct, which is defined in the attachment file.
When 2,2,3-trimethylcyclohexanone reacts with hydroxylamine it will produce the 2,2,3-trimethylcyclohexanoxime.
In the experiment, you heated a sample of metal in a water bath and quickly removed and dried the sample prior to inserting it into the cold water "system". Why did you need to dry the sample prior to placing it into the cold water "system"
a. The extra water might react with the metal which would ruin the sample.
b. Any residual hot water would have added to the heat transferred from the metal to the cold water "system".
c. The wet metal would not transfer any heat thus causing inaccuracy in you measurements.
d. The metal would oxidize in the presence of water thus ruining the sample
Answer:
b
Explanation:
Provided the experiment is about the transfer of heat from one medium to another, any residual hot water on the metal would have added to the heat transferred from the metal to the cold water system.
The residual hot water on the metal posses its own heat and when such is transferred along with the metal into the cold water, the heat from the residual hot water will interfere with the measurement of the actual heat transferred to the cold water by the metal. Hence, the accuracy of the result would be impacted.
The correct option is b.
What is the density of a 10 kg mass that occupies 5 liters?
( pls need help)
Answer: d=2000 g/L
Explanation:
Density is mass/volume. The units are g/L. Since we are given mass and volume, we can divide them to find density. First, we need to convert kg to g.
[tex]10kg*\frac{1000g}{1kg} =10000 g[/tex]
Now that we have grams, we can divide to get density.
[tex]d=\frac{10000g}{5 L}[/tex]
d=2000g/L
What's the concentration of hydronium ions if a water-base solution has a temperature of 25°C (Kw = 1.0×10–14), with a concentration of hydroxide ions of 2.21×10–6 M? answer options: A) 2.8×10–8 M B) 4.52 ×10–9 M C) 1.6×10–9 M D) 3.1×10–6 M
Answer:
ITS NOT D. ITS B. 4.52x10^-9 M
Explanation:
Answer:
4.52 ×10–9 M
Explanation:
Without doing any calculations, determine the sign of ΔSsys for each of the following chemical reactions. Drag the appropriate items to their respective bins.
1. 2H30' (aq) + CO23- (aq) - CO2(g) +3H2O(1)
2. CH4(g) + 202,(g) - CO2(g) + 2H2O(l)
3. Mg (s) + Cl2(g) - MgCǐ2(s)
4. SO3(g) + H2O(I) - H2SO4(I)
A. ΔSsys greater than
B. ΔSsys smaller than
Answer:
Answers are in the explanation.
Explanation:
In a chemical reaction we can determine the sign of ΔSsys based on the states of products and reactants knowing that:
Entropy of gases >>> entropy of liquid > entropy of solids.
The entropy of solids is lower than entropy of liquids that is lower than entropy of gases.
In the reactions:
1. 2H₃O⁺(aq) + CO₃²⁻(aq) → CO₂(g) +3H₂O(l)
As 1 gas is produced, entropy of products is higher than entropy of reactants. That means ΔSsys > 0 (That because ΔSsys is ΔSProducts - ΔSReactants)
2. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
3 moles of gas are converted in 1 mole of gas in products. Entropy of reactants is higher than entropy of products, ΔSsys < 0.
3. Mg(s) + Cl₂(g) → MgCǐ₂(s)
You have 1 mole of gas in reactants and 1 mole of solid in products. ΔSProducts <<< ΔSReactants. ΔSsys < 0.
4. SO₃(g) + H₂O(I) → H₂SO₄(I)
1 mole of gas in reactants, a liquid in products. ΔSProducts <<< ΔSReactants. ΔSsys < 0.
Determining the sign of ΔSsys for each given chemical reaction, the following can be obtained without calculations:
1. ΔSsys > 0.
2. ΔSsys < 0
3. ΔSsys < 0
4. ΔSsys < 0
Recall:
The states of the reactants and the products in a chemical reaction determines the sign of ΔSsys of the reaction.The entropy of gasses is greater than the entropy of liquid and solids.The entropy of solids is less than the entropy of liquid and gasses.Gasses have the highest entropy, while solids have the least.Thus:
In the first chemical reaction, 1 mole of gas is produced, therefore: ΔSsys > 0.
In the second chemical reaction, 3 moles of gasses gives a products of 1 mole of gas, therefore: ΔSsys < 0.
In the third chemical reaction, 1 mole of gas gives 1 mole of solid as product, therefore: ΔSsys < 0.
In the fourth chemical reaction, 1 mole of gas gives 1 mole of liquid as product, therefore: ΔSsys < 0.
Learn more here:
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I need to name an ionic compound containing a transition metal cation and a halogen anion. Below are the rules I should follow to write the correct name for such compound, but one of the options is incorrect: identify and select it.
a. Identify the metal and write its name first
b. Use the periodic table to work out the charge (oxidation number) on the transition metal according to the group in the periodic table
c. From the charge of the anion work out the charge of cation as Roman number in parenthesis: specify this charge in the name as a Roman number in parenthesis.
d. Write the number of the anion after the name of the metal
Answer:
b. Use the periodic table to work out the charge (oxidation number) on the transition metal according to the group in the periodic table
Explanation:
The keyword in this problem us "transition metal". Transition metals are found between the group 2 and group 3 elements. They have the d sub shells and also exhibit variable oxidation numbers (valency).
Among the options, the incorrect option is option B.
This is because transition metals d not have a fixed oxidation number and they cannot be obtained by looking up the group in the periodic table.
The iconic compounds obtain a transition of metal caution and a halon anon. As per the rules the correct name of the compounds should be written as to identify the incorrect one.
Option B use the ability to check and to work out the charges (oxidation number) of the transition metal as per the group given in the table. The problem with the keyword is transition metal.Learn more about the ionic compound containing a transition metal.
brainly.com/question/21578354.
I think I'm typing it into my calculator wrong. I will give brainliest to whoever gets it right.
Answer:
36.7 mg
Explanation:
The following data were obtained from the question.
Original amount (A₀) = 65.1 mg
Rate constant (K) = 2.47×10¯² years¯¹
Time (t) = 23.2 years
Amount of substance remaining (A) =?
Thus, we can obtain the amount of substance remaining after 23.2 years as follow
ln A = lnA₀ – Kt
lnA = ln(65.1) – (2.47×10¯² × 23.2)
lnA = 4.1759 – 0.57304
lnA = 3.60286
Take the inverse of ln
A = e^3.60286
A = 36.7 mg
Therefore, the amount remaining after 23.2 years is 36.7 mg.
If a substance has a half-life of 55.6 s, and if 230.0 g of the substance are present initially, how many grams will remain after 10.0 minutes?
Answer:
[tex]m=0.127g[/tex]
Explanation:
Hello,
In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:
[tex]k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{55.6s}=0.0125s^{-1}[/tex]
In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:
[tex]m=m_0*exp(-kt)=230.0g*exp(-0.0125s^{-1}*600s)\\\\m=0.127g[/tex]
Best regards.
Consider the bond dissociation energies listed below in kcal/mol. CH3-Br 70 CH3CH2-Br 68 (CH3)2CH-Br 68 (CH3)3C-Br 65 These data show that the carbon-bromine bond is weakest when bromine is bound to a __________.
Answer:
The answer is "Tertiary carbon".
Explanation:
Accent to the results, the carbon-bromine bond is weak, whenever, the bromine is connected to tertiary carbon so, bonding energy is separation for methyl-carbon, which is connected to the bromine = 70 kcal/mol and for the primary energy to the secondary energy is= 68 kcal/mol, and for tertiary CO2 = 65 kcal/mol.
The stronger the energy dissociating connection and the weaker, its power dissociation connection and its weaker bond becomes connecting with a tertiary carbon, that's why "Tertiary carbon" is the correct answer.
If 100-mL of 1.0 M Sr(OH)2 is added to 100 mL of 1.0 M HCl, the pH of the mixture would be _____. Group of answer choices
Answer:
pH = 13.7
Explanation:
A strong acid (HCl) reacts with a strong base Sr(OH)₂ producing water and a salt, thus:
2HCl + Sr(OH)₂ → 2H₂O + SrCl₂
To solve this problem, we need to find initial moles of both reactants and, with the chemical equation find limiting reactant and moles in excess to find pH as follows:
The initial moles of HCl and Sr(OH)₂ are:
100mL = 0.1L ₓ (1.0mol / L) = 0.100 moles of both HCl and Sr(OH)₂
As 2 moles of HCl reacts per mole of Sr(OH)₂, moles of Sr(OH)₂ that reacts with 0.100 moles of HCl are:
0.100 moles HCl ₓ (1 mol Sr(OH)₂ / 2 mol HCl) = 0.050 moles Sr(OH)₂
That means HCl is limiting reactant and after reaction will remain in solution:
0.100 mol - 0.050mol =
0.050 moles of Sr(OH)₂
Find pH:
1 mole of Sr(OH)₂ contains 2 moles of OH⁻, 0.050 moles contains 0.050×2 = 0.100 moles of OH⁻. In 200mL = 0.2L:, molar concentration of OH⁻ is:
0.100 moles / 0.2L =
[OH⁻] = 0.5M
As pOH of a solution is -log[OH⁻],
pOH = -log 0.5M
pOH = 0.301
And knowing:
pH = 14 - pOH
pH = 14 - 0.301
pH = 13.7A rock has a mass of 15.8 g and causes the water level in a graduated cylinder to raise from 22.3 mL to 32.5 mL. What is the density of the rock in Kg/mL?
Answer:
Explanation:
mass - 15.8 g = 0.0158 kg
volume = 32.5 - 22.5 = 10.2 ml
density = mass / volume
= 0.0158 / 10.2
= 0.00154 kg/ml
hope this helps
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How does the spontaneity of the process below depend on temperature? PCl5(g)+H2O(g)→POCl3(g)+2HCl(g) ΔH=−126 kJ mol−1, ΔS=146 J K−1mol−
The given question is incomplete, the complete question is:
How does the spontaneity of the process below depend on temperature? PCI5(9)+H2O(g)POCI3(g) +2HCI(g) -126 kJ mol1, AS = 146 J K-'mol1 ΔΗ Select the correct answer below: nonspontaneous at all temperatures spontaneous at all temperatures spontaneous at high temperatures and nonspontaneous at low temperatures spontaneous at low temperatures and nonspontaneous at high temperatures
Answer:
The correct answer is spontaneous at all the temperatures.
Explanation:
Gibbs Free energy is an essential relation that determines the spontaneity of any reaction, that is, ΔG = ΔH - TΔS
When ΔG is less than zero, that is, negative, the reaction is considered to be in spontaneous state. Based on the given information, ΔH = -126 kJ/mol
= -126000 J/mol, it is negative
ΔS = 146 J/K/mol, it is positive
Now, ΔG = ΔH-TΔS
= (-ve) - T (+ve), Thus, when ΔH, is -ve, ΔS is +ve, -TΔS is -ve, the ΔG will be -ve. Therefore, reaction will be spontaneous at all the temperatures.
Zeros laced at the end of the significant number are...
Answer:
Zeros located at the end of significant figures are significant.
Explanation:
Hope it will help :)
At a given temperature the vapor pressures of benzene and toluene are 183 mm Hg and 59.2 mm Hg, respectively. Calculate the total vapor pressure over a solution of benzene and toluene with X Benzene
CHECK COMPLETE QUESTION BELOW
At a given temperature the vapor pressures of benzene and toluene are 183 mm Hg and 59.2 mm Hg, respectively. Calculate the total vapor pressure over a solution of benzene and toluene with Xbenzene = 0.580.
Answer:
The total vapor pressure is [tex]81.3 mmHg[/tex]
Explanation:
We will be making use of Dalton and Raoults equation in order to calculate the total pressure,
Which is [tex]PT= (PA × XA) +(PB ×XB)[/tex]
PT= total vapor pressure
From the question
Benene's Mole fraction = 0.580
then to get Mole fraction of toluene we will substract the one of benzene from 1. because total mole fraction is always 1.
= (1 - 0.580) = 0.420
Vapor pressure of benzene given = 183 mmHg
Vapor pressure of toluene given= 59.2 mmHg
If we substitute those value into above equation, we have
PT=(183×0.580)+(59.2×0.420)
=81.3mmHg
Therefore,, the total vapor pressure of the solution is 81.3 mmHg
What attractive force holds two hydrogen atoms and one oxygen atom
together to make the substance water?
A. Molecules
B. Chemical bonding
O C. Valence electrons
O D. Cations
Answer:
It is a hydrogen bond but if I had to coose one of thee answers it is b. chemical bonding
Explanation:
What is the pH of 10.0 mL solution of 0.75 M acetate after adding 5.0 mL of 0.10 M HCl (assume a Ka of acetic acid of 1.78x10-5)
Answer:
5.90
Explanation:
Initial moles of CH3COO- = 10.0/1000 x 0.75 = 0.0075 mol
Moles of HCl added = 5.0/1000 x 0.10 = 0.0005 mol
CH3COO- + HCl => CH3COOH + Cl-
Moles of CH3COO- left = 0.0075 - 0.0005 = 0.007 mol
Moles of CH3COOH formed = moles of HCl added = 0.0005 mol
pH = pKa + log([CH3COO-]/[CH3COOH])
= -log Ka + log(moles of CH3COO-/moles of CH3COOH)
= -log(1.78 x 10^(-5)) + log(0.007/0.0005)
= 5.90
Answer:
The correct answer is 5.895.
Explanation:
The reaction will be,
CHCOO⁻ + H+ ⇔ CH₃COOH
Both the HCl and the acetate are having one n factor.
The millimoles of CH₃COO⁻ is,
= Volume in ml × molarity = 10 × 0.75 = 7.5
The millimoles of HCl = Volume in ml × molarity = 5 × 0.1 = 0.5
Therefore, 0.5 will be the millimoles of CH₃COOH formed, now the millimoles of the CH₃COO⁻ left will be, 7.5-0.5 = 7.0
The volume of the solution is, 10+5 = 15 ml
The molarity of CH₃COO⁻ is, millimoles / volume in ml = 7/15
The molarity of CH₃COOH is 0.5/15
pH = pKa + log[CH₃COO⁻]/[CH₃COOH]
= 4.74957 + 1.146
= 5.895
Explain your reasoning. Match each explanation to the appropriate blanks in the sentences on the right.
1. the atomic radius decreases
2. the number of gas molecules decreases
3. molar mass and structure complexity decreases
4. structure complexity decreases
5. molar mass decreases
6. each phase (gas, liquid, solid) becomes more ordered
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as______.
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as_______.
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as_______.
Answer:
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases.
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases.
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases.
Explanation:
Hello.
In this case, we can understand a higher entropy when more disorder is present and a lower entropy when less disorder is present, thus:
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases since iodine has the greatest molar mass (254 g/mol) and fluorine the least molar mass (38 g/mol).
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases since hydrogen peroxide weights 34 g/mol as well as hydrogen sulfide but the peroxide has more bonds (more complex, higher entropy).
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases since diamond has a well-ordered structure and amorphous carbon has a very disordered one.
Best regards.
A student is performing a Benedict’s test on an unknown substance. He adds the reagent (the chemical required to make a color change), and nothing happens. What can he conclude? A- The substance is glucose-based. B- The substance is not glucose-based. C- The test was inconclusive because he needed to also test with iodine or vinegar. D- The test was inconclusive because he forgot to add heat.
Answer:
The correct answer is : option D. The test was inconclusive because he forgot to add heat.
Explanation:
Benedict's test is a test that is used to confirm the presence of the simple carbohydrates (mono saccharides and some disaccharides). It is a reagent made by mixture of solution of CuSO4 with sodium citrate and Na2CO3.
Benedict's reagent is added to the substance to test and then heated if it turns yellow to orange or red the presence of simple sugar is confirmed.
Thus, the correct answer is : option D. The test was inconclusive because he forgot to add heat.
Answer:
The test was inconclusive because the student forgot to add heat.
Explanation:
If the test revealed it was not glucose, then the student could run these tests. The student, however, does not need these substances to run the glucose test properly.
Determine the oxidation state for each of the elements below. The oxidation state of ... silver ... in ... silver oxide Ag2O ... is ... ___ . The oxidation state of sulfur in sulfur dioxide SO2 is ___ . The oxidation state of iron in iron(
Answer:
The oxidation state of silver in [tex]\rm Ag_2O[/tex] is [tex]+1[/tex].
The oxidation state of sulfur in [tex]\rm SO_2[/tex] is [tex]+4[/tex].
Explanation:
The oxidation states of atoms in a compound should add up to zero.
Ag₂OThere are two silver [tex]\rm Ag[/tex] atoms and one oxygen [tex]\rm O[/tex] atom in one formula unit of [tex]\rm Ag_2O[/tex]. Therefore:
[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].
The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is [tex]-2[/tex]. Silver oxide [tex]\rm Ag_2O[/tex] isn't an exception. Therefore:
[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}[/tex].
Solve this equation for the (average) oxidation state of [tex]\rm Ag[/tex]:
[tex]\text{Oxidation state of $\rm Ag$} = 1[/tex].
SO₂Similarly, because there are one sulfur [tex]\rm S[/tex] atom and two oxygen [tex]\rm O[/tex] atoms in each [tex]\rm SO_2[/tex] molecules:
[tex]\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].
The oxidation state of [tex]\rm O[/tex] in [tex]\rm SO_2[/tex] is also [tex]-2[/tex], not an exception, either.
Therefore:
[tex]\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}[/tex].
Solve this equation for the oxidation state of [tex]\rm S[/tex] here:
[tex]\text{Oxidation state of $\rm S$} = 4[/tex].
A battery is an example of a(n) _________. A. anode B. voltaic cell C. cathode D. electrolytic cell
Answer:
The answer is D) Electrolytic cell
Explanation:
An electrolytic cell is a device used for the decomposition by the electrical current of ionized substances called electrolytes.
When the two electrodes are connected by a wire, electrical energy is produced, and a flow of electrons takes place from the electrode.
These cells are the closest thing to a galvanic battery.
Answer:
b. voltaic cell
Explanation:
Founders Education answer. had to take this quiz 4 times
A student mixed 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH in a coffee cup calorimeter and calculate the molar enthalpy change of the acid-base neutralization reaction to be –54 kJ/mol. He next tried the same experiment with 100 ml of 1.0 M HCl and 100 ml of 1.0 M NaOH. The calculated molar enthalpy change of reaction for his second trial was:
Answer: The calculated molar enthalpy change of reaction for his second trial was -108 kJ.
Explanation:-
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
[tex]\text{no of moles}={\text{Molarity}\times {\text{Volume in L}}[/tex]
Thus [tex]\text{no of moles}of HCl={1.0M}\times {0.05L}=0.05moles[/tex]
Thus [tex]\text{no of moles}of NaOH={1.0M}\times {0.05L}=0.05moles[/tex]
[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)[/tex]
Given for second trial:
[tex]\text{no of moles}of HCl={1.0M}\times {0.1L}=0.1moles[/tex]
[tex]\text{no of moles}of NaOH={1.0M}\times {0.1L}=0.1moles[/tex]
0.05 moles of [tex]HCl[/tex] reacts with 0.05 moles of [tex]NaOH[/tex] to release heat = 54 kJ
0.1 moles of [tex]HCl[/tex] reacts with 0.05 moles of [tex]NaOH[/tex] to release heat =[tex]\frac{54}{0.05}\times 0.1=108kJ[/tex]
Thus calculated molar enthalpy change of reaction for his second trial was -108 kJ.
