Answer:
494.49 g of NaCl.
Explanation:
Data obtained from the question include the following:
Molality of NaCl = 3.140 m
Mass of water = 2.692 kg
Mass of NaCl =.?
Next, we shall determine the number of mole of NaCl in the solution.
Molality is simply defined as the mole of solute per unit kilogram of solvent. Mathematically, it is expressed as
Molality = mole of solute /Kg of solvent
With the above formula, we can obtain the number of mole NaCl in the solution as follow:
Molality of NaCl = 3.140 m
Mass of water = 2.692 kg
Mole of NaCl =..?
Molality = mole of solute /Kg of solvent
3.140 = mole of NaCl /2.692
Cross multiply
Mole of NaCl = 3.140 x 2.692
Mole of NaCl = 8.45288 moles
Finally, we shall covert 8.45288 moles of NaCl to grams. This can be obtained as follow:
Mole of NaCl = 8.45288 moles
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =.?
Mole = mass /Molar mass
8.45288 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 8.45288 × 58.5
Mass of NaCl = 494.49 g.
Therefore, 494.49 g of NaCl are present in the solution.
The mass of NaCl in 3.140 molal NaCl solution has been 494.493 grams.
Molality can be defined as the mass of solute present in 1000 grams of solvent.
Molality = [tex]\rm \dfrac{moles\;of\;solute}{mass\;of\;solvent\;(kg)}[/tex]
The moles of NaCl has been calculated as;
3.140 = [tex]\rm \dfrac{moles\;of\;NaCl}{mass\;of\;water\;(kg)}[/tex]
3.140 = [tex]\rm \dfrac{moles\;of\;NaCl}{2.692\;kg}[/tex]
Moles of NaCl = 3.140 [tex]\times[/tex] 2.692
Moles of NaCl = 8.45288 mol
The moles can be expressed as;
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Molecular weight of NaCl = 58.5 g/mol
The mass of NaCl can be calculated as:
8.45288 mol = [tex]\rm \dfrac{mass\;of\;NaCl}{58.5\;g/mol}[/tex]
Mass of NaCl = 58.5 g/mol [tex]\times[/tex] 8.45288 mol
Mass of NaCl = 494.493 grams.
The mass of NaCl in 3.140 molal NaCl solution has been 494.493 grams.
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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 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.
243
Am
95
1. The atomic symbol of americium-243 is shown. Which of the following is correct?
• A. The atomic mass is 243 amu, and the atomic number is 95.
B. The atomic mass is 338 amu, and the atomic number is 95.
• C. The atomic mass is 95 amu, and the atomic number is 243.
D. The atomic mass is 243 amu, and the atomic number is 338.
Answer:
A. The atomic mass is 243 amu, and the atomic number is 95.
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.
At a constant temperature, a sample of a gas in a balloon that originally had a volume of 5.00 L and pressure of 626 torr has its volume changed to 6.72 L. Calculate the new pressure in torr.
Answer:
466 torr
Explanation:
Step 1: Given data
Initial pressure (P₁): 626 torrInitial volume (V₁): 5.00 LFinal pressure (P₂): ?Final volume (V₂): 6.72 LConstant temperatureStep 2: Calculate the final pressure
Since we have a gas changing at a constant temperature, we can calculate the final pressure using Boyle's law.
P₁ × V₁ = P₂ × V₂
P₂ = P₁ × V₁ / V₂
P₂ = 626 torr × 5.00 L / 6.72 L
P₂ = 466 torr
g Does a reaction occur when aqueous solutions of barium hydroxide and aluminum sulfate are combined
Answer:
3BaO + Al₂(SO₄)₃ → Al₂O₃+ 3BaSO₄
Explanation:
Yes! A reactiin occurs between barium hydroxide and auminium sulphate.
barium sulfate (BaSO4) and aluminum hydroxide (Al(OH)3) are the products obtained in this reaction.
The reaction is given by the equation below;
3BaO + Al₂(SO₄)₃ → Al₂O₃+ 3BaSO₄
Calculate the molality of a solution containing 141.5 g of glycine (NH2CH2COOH) dissolved in 4.456 kg of H2O
Answer:
0.423 m.
Explanation:
The following data were obtained from the question:
Mass of glycine (NH2CH2COOH) = 141.5 g
Mass of water = 4.456 kg
Molality =.?
Next, we shall determine the number of mole in 141.5 g of glycine (NH2CH2COOH.
This is illustrated below:
Mass of glycine (NH2CH2COOH) = 141.5 g
Molar mass of glycine (NH2CH2COOH) = 14 + (2x1) + 12 + (2x1) + 12 + 16 + 16 + 1 = 75 g/mol
Mole of glycine (NH2CH2COOH) =.?
Mole = mass /Molar mass
Mole of glycine (NH2CH2COOH) = 141.5/75
Mole of glycine (NH2CH2COOH) = 1.887 moles
Finally, we shall determine the molality of the solution as follow:
Molality is simply defined as the mole of solute per kilogram of water. Mathematically it is expressed as:
Molality = mole / mass (kg) of water
With the above formula, we can obtain the molality of the solution as follow:
Mole of glycine (NH2CH2COOH) = 1.887 moles
Mass of water = 4.456 kg
Molality =.?
Molality = mole /mass (kg) of water
Molality =1.887/4.456
Molality = 0.423 m
Therefore, the molality of the solution is 0.423 m
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.
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.
A 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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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.
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.
Indicate the peptides that would result from cleavage by the indicated reagent: a. Gly-Lys-Leu-Ala-Cys-Arg-Ala-Phe by trypsin b. Glu-Ala-Phe-Gly-Ala-Tyr by chymotrypsin
Answer:
a. Gly-Lys + Leu-Ala-Cys-Arg + Ala-Phe
b. Glu-Ala-Phe + Gly-Ala-Tyr
Explanation:
In this case, we have to remember which peptidic bonds can break each protease:
-) Trypsin
It breaks selectively the peptidic bond in the carbonyl group of lysine or arginine.
-) Chymotrypsin
It breaks selectively the peptidic bond in the carbonyl group of phenylalanine, tryptophan, or tyrosine.
With this in mind in "peptide a", the peptidic bonds that would be broken are the ones in the "Lis" and "Arg" (See figure 1).
In "peptide b", the peptidic bond that would be broken is the one in the "Phe" (See figure 2). The second amino acid that can be broken is tyrosine, but this amino acid is placed in the C terminal spot, therefore will not be involved in the hydrolysis.
Write an equation to show how the base NaOH(s) behaves in water. Include states of matter in your answer. Click in the answer box to open the symbol palette.
Answer:
The reaction is given as:
[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]
Explanation:
Bases are defined as those chemical substances which give hydroxide ions in their aqueous solutions.
[tex]BOH(s)\rightarrow B^+(aq)+OH^-(aq)[/tex]
When sodium hydroxide is added to water it gets dissociated into two ions that are sodium ions and hydroxide ions. Along with this heat energy also releases during this reaction.
The reaction is given as:
[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]
The equation to show how NaOH behaves in water is NaOH → Na⁺ + (OH)⁻
The compound that produce negative hydroxide (OH−) ions when dissolved
in water are called bases .
This compounds NaOH (sodium hydroxide) is an example of a base.
When it dissolves in water it dissociate to form negative hydroxide (OH−)
ions and positive sodium (Na+) ions.
It can be represented by the following equation:
NaOH → Na⁺ + (OH)⁻
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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.
A chemist fills a reaction vessel with 0.978 g aluminum hydroxide AlOH3 solid, 0.607 M aluminum Al+3 aqueous solution, and 0.396 M hydroxide OH− aqueous solution at a temperature of 25.0°C.
Under these conditions, calculate the reaction free energy ΔG for the following chemical reaction:
Al(OH)3(s) = A1+ (aq) +30H (aq)
Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.
KJ
Answer: [tex]\Delta G^{0}[/tex] = 168.12 kJ
Explanation: Gibbs Free Energy, at any time, is defined as the enthalpy of the system minus product of temperature and entropy of the reaction, i.e.:
[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]
Enthalpy is defined as internal heat existent in the system. It is calculated as:
[tex]\Delta H^{0} = \Sigma H^{0}_{product} - \Sigma H^{0}_{reagent}[/tex]
Using Enthalpy Formation Table:
[tex]\Delta H^{0} = [3*(-299.9)+(-524.7)] - (-1277)[/tex]
[tex]\Delta H^{0} = 62,6 kJ[/tex]
Entropy is the degree of disorder in the system. It is found by:
[tex]\Delta S^{0} = \Sigma S^{0}_{products} - \Sigma S^{0}_{reagents}[/tex]
Calculating:
[tex]\Delta S^{0} = (-321.7) + 3(-10.8) - 0[/tex]
[tex]\Delta S^{0} = -354.1J[/tex]
And so, Gibbs Free energy will be:
[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]
[tex]\Delta G^{0} = 62600 - [298.(-354.1)][/tex]
[tex]\Delta G^{0} = 168121.8 J[/tex]
Rounding to the nearest kJ:
[tex]\Delta G^{0}[/tex] = 168.12 kJ
What type of bond would form between two atoms of phosphorus? A. Single covalent bond B. Single ionic bond C. Triple covalent bond D. Double covalent bond
Answer:
A double bond is formed when two pairs of electrons are shared between the two participating atoms. It is represented by two dashes (=). It is represented by two dashes (=). Double covalent bonds are much stronger than a single bond, but they are less stable
Explanation:
1500 L has how many significants figures
Answer:
It has 2
Explanation:
The significant figures are 1 and 5!
Hope this helps:)
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].
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.
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.
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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 :)
A compound is found to contain 11.21 % hydrogen and 88.79 % oxygen by mass. What is the empirical formula for this compound?
Answer:
H₂O
Explanation:
The empirical formular of the compound is obtained using the following steps;
Step 1: Divide the percentage composition by the atomic mass
Hydrogen = 11.21 / 1 = 11.21
Oxygen = 88.79 / 16 = 5.55
Step 2: Divide by the lowest number
Hydrogen = 11.21 / 5.55 = 2.02 ≈ 2
Oxygen = 5.55 / 5.55 = 1
This means the ratio of the elements is 2 : 1
The empirical formular (simplest formular of a compound) of the compound is;
H₂O
Answer:Empirical formula ======== H₂O
Explanation:The empirical formula of a compound shows the whole number ratio for each atom in a compound.
To find empirical formula. we follow the below steps
The total mass of the compound here is 100 grams, that is (11.21% of hydrogen + 88.79% of oxygen) we can then assume 11.21 grams of hydrogen and 88.79grams of oxygen
Hydrogen Oxygen
1.composition by mass 11.21 88.79
molecular weight 1.007g/mol 15.990g/mol
2.Divide composition by mass 11.21/1.007 88.79/15.99
by each molecular weight to get 11.13 5.553
no of moles
3 Divide by the least number of moles
to get atomic ratio 11.13/5.553 5.553/5.553
2.004 1.00
4.Convert to whole numbers 2 1
Empirical formula ======== H₂O
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
The insoluble salts below are put into 0.10 M hydrochloric acid solution. Do you expect their solubility to be more, less, or about the same as in a pure water solution?
1. Zinc sulfide
2. Silver chloride
3. Lead iodide
4. Silver hydroxide
Answer:
1. Zinc sulfide : about the same solubility, no common ion is found.
2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.
3. Lead iodide : about the same solubility, no common ion is found.
4. Silver hydroxide : about the same solubility, no common ion is found.
Explanation:
Hello,
In this case, we first must remember that adding a common ion (which is related with the dissolving solid) decreases the solubility of the insoluble solid due to the fact Le Chatelier's principle states the reaction will shift leftwards (reactants) to reestablish equilibrium, therefore, we have:
1. Zinc sulfide : about the same solubility, no common ion is found.
2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.
3. Lead iodide : about the same solubility, no common ion is found.
4. Silver hydroxide : about the same solubility, no common ion is found.
Best 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 :).
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:
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.
A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat. If the radiation wavelength is 12.5 cm, how many photons of this radiation would be required to heat a container with 0.250 L of water from a temperature of 20.0oC to a temperature of 99oC
Answer:
The total photons required for this radiation = 5.1938 × 10²⁸ photons
Explanation:
Given that:
A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat.
If the radiation wavelength is 12.5 cm,
density of water = 1g/cm³
volume of the container = 0.250 L = 250 cm³
density = mass/volume
mass of the water = density × volume
mass of the water = 1g/cm³ × 250 cm³
mass of the water = 250 g
specific heat capacity of water = 4.182 J/g°C
The change in temperature was from 20.0° C to 99° C
ΔT =( 99 -20.0)° C
ΔT = 79.0° C
The heat absorbed in the process is calculated by using the formula,
q = mcΔT
q = 250 g × 4.182 J/g°C × 79.0° C
q = 82594.5 Joules
Recall that the radiation wavelength λ = 12.5 cm = 0.125 m
The amount of energy of one photon of the radiation wavelength is determined by using the formula:
E = hv
since v = c/λ
E = hc/λ
where;
h = Planck's constant = 6.626 × 10⁻³⁴ J.s
c = velocity of light = 3.0 × 10⁸ m/s
∴
E = (6.626 × 10⁻³⁴ J.s × 3.0 × 10⁸ m/s)/ 0.125 m
E = 1.59024⁻²⁴ Joules
The total photons required for this radiation = total heat energy/energy of radiation
The total photons required for this radiation = 82594.5 Joules/1.59024⁻²⁴ Joules
The total photons required for this radiation = 5.1938 × 10²⁸ photons
