Be sure to answer all parts. The concentration of Cu2 ions in the water (which also contains sulfate ions) discharged from a certain industrial plant is determined by adding excess sodium sulfide (Na2S) solution to 0.700 L of the water. The molecular equation is
This is an incomplete question, here is a complete question.
The concentration of Cu²⁺ ions in the water (which also contains sulfate ions) discharged from a certain industrial plant is determined by adding excess sodium sulfide (Na₂S)solution to 0.700 L of the water.
The molecular equation is:
[tex]Na_2S(aq)+CuSO4(aq)\rightarrow Na_2SO_4(aq)+CuS(s)[/tex]
Write the net ionic equation and calculate the molar concentration of Cu²⁺ in the water sample if 0.0177 g of solid CuS is formed.
Answer :
The net ionic equation will be,
[tex]Cu^{2+}(aq)+S^{2-}(aq)\rightarrow CuS(s)[/tex]
The concentration of [tex]Cu^{2+}[/tex] is, [tex]2.65\times 10^{-4}M[/tex]
Explanation :
In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
The molecular equation is:
[tex]Na_2S(aq)+CuSO4(aq)\rightarrow Na_2SO_4(aq)+CuS(s)[/tex]
The ionic equation in separated aqueous solution will be,
[tex]2Na^+(aq)+S^{2-}(aq)+Cu^{2+}(aq)+SO_4^{2-}(aq)\rightarrow CuS(s)+2Na^+(aq)+SO_4^{2-}(aq)[/tex]
In this equation, [tex]Na^+\text{ and }SO_4^{2-}[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]Cu^{2+}(aq)+S^{2-}(aq)\rightarrow CuS(s)[/tex]
Now we have to calculate the mass of [tex]CuSO_4[/tex]
Molar mass of [tex]CuSO_4[/tex] = 159.5 g/mol
Molar mass of CuS is = 95.5 g/mol
From the balanced chemical reaction we conclude that,
As, 95.5 g of CuS produces from 159.5 g [tex]CuSO_4[/tex]
As, 0.0177 g of CuS produces from [tex]\frac{159.5}{95.5}\times 0.0177=0.0296g[/tex] [tex]CuSO_4[/tex]
Now we have to calculate the concentration of [tex]CuSO_4[/tex]
[tex]\text{Concentration}=\frac{\text{Mass of }CuSO_4}{\text{Molar mass of }CuSO_4\times \text{Volume of solution (in L)}}[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Concentration}=\frac{0.0296g}{159.5g/mol\times 0.700L}=2.65\times 10^{-4}M[/tex]
Concentration of [tex]Cu^{2+}[/tex] = [tex]2.65\times 10^{-4}M[/tex]
Therefore, the concentration of [tex]Cu^{2+}[/tex] is, [tex]2.65\times 10^{-4}M[/tex]
Which aqueous solution will theoretically have the highest boiling point? A) 0.001 M NaCl B) 0.001 M C6H12O6 C) 0.001 M CaCl2 D) 0.001 M AlCl3
Answer:
D)
Explanation:
I just answered the question and got it right.
True or False
During Solar Maximums, the thermosphere is bigger because it is warmer.
Arrange the ions N3-, O2-, Mg2 , Na , and F- in order of increasing ionic radius, starting with the smallest first. Arrange the ions N3-, O2-, Mg2 , Na , and F- in order of increasing ionic radius, starting with the smallest first. N3-, O2-, Mg2 , F-, Na Mg2 , Na , F-, O2-, N3- N3-, Mg2 , O2-, Na , F- N3-, O2-, F-, Na , Mg2
Answer:
Order of increasing ionic radius starting from the smallest is
Mg2+ < Na+ < F- < O2- < N3-
Explanation:
Ionic radius is the distance between the nucleus and the electrons in the outermost shell of an ion. In the ions given about, it can be deducted that they all have the same number of electrons and are said to be isoelectronic. This shows us that they all have the same electrons and we can only seperate them by checking the respective charges on the ions. A more positive charge will have a smaller radius while a more negative will have a larger radius. This is because when an atom loses an electron and form a cation (+), the remaining electrons move closer to the nucleus and hence a smaller radius of the atom occurs. But when an atom gains n electron,it forms anion (-), the new electron leads to the shielding of the remaining electrons from the nucleus and a large radius of the atom occurs.
So therefore, the above ions all have 10 electrons in their shells and magnesium with atomic number 12 and having lost 2 electrons to become positively charge has the smallest radius compared to others.
Mg2+ = 12 protons, 10 electrons
Na+ = 11 protons, 10 electrons
F- = 9 protons, 10 electrons
O2- = 8 protons, 10 electrons
N3- = 7 protons, 10 electrons
N3- has the largest ionic radius and Mg2+ has the lowest ionic radius.
The order is thus, Mg2+ , Na+, F-, O2-, N3-
The order of increasing ionic radius is N3-, O2-, F-, Na, Mg2+
The ionic radius is a measure of the size of an ion. In general, ions with more electrons have larger radii. Let's arrange the ions in order of increasing ionic radius:
N3-O2-F-NaMg2+The larger the negative charge on an ion, the larger its ionic radius. Therefore, N3- has the smallest radius, followed by O2-, F-, Na, and finally Mg2+, which has the largest radius among the given ions.
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At 25 oC the solubility of iron(II) hydroxide is 1.59 x 10-5 mol/L. Calculate the value of Ksp at this temperature. Give your answer in scientific notation to 2 SIGNIFICANT FIGURES (even though this is strictly incorrect). [a]
The Ksp of the solution is [tex]1.607*10^-^1^4[/tex]
Data;
Temperature = 25°solubility = 1.59*10^-5 mol/LKsp = ?Solubility ConstantThis is the point or temperature in which a solute is completely soluble in a solvent.
The solubility of iron(ii) hydroxide can be calculated by using the equation of reaction.
[tex]Fe(OH)_2 \to Fe^2^+ + 2OH^-\\[/tex]
But the solubility of the solution is given as 1.59*10^-5 mol/L
[tex][Fe^2^+] = s\\\\[/tex]
[tex][OH^-] = 2s\\K_s_p = s * (2s)^2 = 4s^3[/tex]
Let's substitute the values and solve
[tex]K_s_p = 4 *( 1.59*10^-^5)^3 = 1.607*10^-^1^4[/tex]
The Ksp of the solution is [tex]1.607*10^-^1^4[/tex]
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Final answer:
The Ksp of iron(II) hydroxide at 25 °C is calculated to be 1.61 x 10−14, based on its given solubility and the stoichiometry of its dissociation in water.
Explanation:
The solubility product constant, Ksp, for iron(II) hydroxide, Fe(OH)2, at 25 °C can be calculated using the mol/L solubility given and the stoichiometry of the dissolution reaction:
Fe(OH)2 (s) → Fe2+ (aq) + 2OH− (aq)
For iron(II) hydroxide, the solubility is 1.59 x 10−5 mol/L. This is the concentration of Fe2+. Since the ratio of Fe2+ to OH− is 1:2, this means the concentration of OH− is twice this value, thus 3.18 x 10−5 mol/L. The Ksp is the product of these ion concentrations:
Ksp = [Fe2+][OH−]2 = (1.59 x 10−5)(3.18 x 10−5)2
Calculating the above expression:
Ksp = 1.59 x 10−5 x (1.01 x 10−9)
Ksp = 1.61 x 10−14
This is the Ksp of iron(II) hydroxide at 25 °C, expressed in scientific notation to two significant figures.
175 mL of Cl2 gas is held in a flexible vessel at STP. If the
vessel is transported to the bottom of the impact basin
Hellas Planitia on the surface of Mars where the pressure is
1.16 kPa and the temperature is -5.0°C. What is the new
volume of Cl2 gas in liters?
The new volume of the Cl2 gas under the conditions at the bottom of Hellas Planitia on Mars would be approximately 4.2 litres, as calculated using the ideal gas law.
Explanation:The question is asking about the adjustment of gas volume under different conditions of pressure and temperature, so the ideal gas law is applicable here which states that: PV = nRT, where P is pressure, V is volume, T is temperature, n is number of moles of the gas and R is gas constant.
At Standard Temperature and Pressure (STP), the conditions are 1 atm pressure and 0°C (273 K). Please note that the volume given is already at STP, so to find the new volume, we should rearrange the ideal gas law to V2 = V1 * P1/P2 * T2/T1.
However, note that all the measurements need to be in the same standard units. Here, the initial pressure (P1) is 1 atm, converted to kPa becomes approximately 101.3 kPa, the final pressure (P2) is given as 1.16 kPa. The initial temperature (T1) is 273K and the final temperature (T2) needs to be converted from Celsius to Kelvin, making it 268 K (-5°C + 273). Thus, applying the equation, we get:
V2 = 0.175L * 101.3 kPa / 1.16 kPa * 268 K / 273 K = 4.2 L approximately
Therefore, the new volume of the Cl2 gas, if it is transported to the conditions at the bottom of Hellas Planitia on Mars, would be approximately 4.2 litres.
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Using the combined gas law, the new volume of Cl₂ gas at 1.16 kPa pressure and -5.0°C temperature at the bottom of Hellas Planitia on Mars is calculated to be approximately 15.06 liters.
To determine the new volume of Cl₂ gas at the conditions provided, we need to use the combined gas law which is given by the equation: [tex]\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}[/tex], where P is pressure, V is volume and T is temperature in Kelvin.
At STP (Standard Temperature and Pressure), we have a volume of 175 mL (or 0.175 liters), a pressure of 101.3 kPa, and a temperature of 273.15 K. The conditions at the bottom of the Hellas Planitia are 1.16 kPa and -5.0°C or 268.15 K.
First, we convert the initial volume into liters: 175 mL = 0.175 L.
Next, we rearrange the combined gas law to solve for V₂:
Therefore, the new volume of Cl₂ gas at the bottom of Hellas Planitia will be approximately 15.06 liters.
A 19.13 gram sample of chromium is heated in the presence of excess bromine. A metal bromide is formed with a mass of 77.92 g. Determine the empirical formula of the metal bromide.
Answer:
The empirical formula of the compound is [tex]CrBr_2[/tex].
Explanation:
Mass of chromium = 19.13 g
Mass metal bromide formed = 77.92 g
Mass of bromine in metal bromide = 77.92 g - 19.13 g = 58.79 g
Moles of chromium metal :
[tex]=\frac{19.13 g}{52 g/mol}=0.3679 mol[/tex]
Moles of bromine:
[tex]=\frac{58.79 g}{80 g/mol}=0.7349 mol[/tex]
For empirical formula divide the smallest number of moles of element from the all the moles of elements:
chromium : [tex]\frac{0.3679 mol}{0.3679}=1[/tex]
bromine : [tex]\frac{0.7349 mol}{0.3679}=2[/tex]
The empirical formula of the compound is [tex]CrBr_2[/tex].
To find the empirical formula, the moles of each element involved must be determined. When the sample weights are converted to moles, the ratio of chromium (Cr) to bromine (Br) in the bromide compound is found to be 1:2, making the empirical formula CrBr2.
To determine the empirical formula of the compound, we first need to identify the moles of each element. The chromium sample weighs 19.13g. Using the atomic weight of chromium, which is 52.00g/mol, the number of moles of chromium is 19.13g / 52.00g/mol = 0.368 mol.
Considering the entire mass of the metal bromide, which is 77.92 g, the mass of bromine in the compound must be 77.92 g - 19.13 g = 58.79 g. As the atomic weight of bromine is approximately 79.90g/mol, we can find that the moles of bromine are 58.79 g / 79.90 g/mol = 0.736 mol.
The ratio of Cr to Br in the compound is 0.368 : 0.736, which is approximately 1 : 2 when we divide both numbers by the smallest to obtain the simplest whole number ratio. Therefore, the empirical formula of the compound is CrBr2.
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What is the freezing point of a solution of ethylene glycol, a nonelectrolyte, that contains 59.0 g of (CH2OH)2 dissolved in 543 g of water? Use molar masses with at least as many significant figures as the data given.
Answer:
The expected freezing point of a 1.75 m solution of ethylene glycol is -3.26°C.
Explanation:
[tex]\Delta T_f=T-T_f[/tex]
[tex]\Delta T_f=K_f\times m[/tex]
[tex]m=\frac{\text{mass of solute}}{\text{Molar mass of solute}\times {Mass of solvent in kg}}[/tex]
where,
[tex]T[/tex] = Freezing point of solvent
[tex]T_f[/tex] = Freezing point of solution
[tex]\Delta T_f[/tex] =depression in freezing point
[tex]K_f[/tex] = freezing point constant
m = molality
we have :
Mass of ethylene glycol = 59.0 g
Molar mass of ethylene glycol = 62.1 g/mol
Mass of solvent i.e. water = 543 g = 0.543 kg ( 1 g = 0.001 kg)
[tex]K_f[/tex] =1.86°C/m ,
[tex]m =\frac{59.0 mol}{62.1 g/mol\times 0.543 kg}=1.75 m[/tex]
[tex]\Delta T_f=1.86^oC/m \times 1.75m[/tex]
[tex]\Delta T_f=3.26^oC[/tex]
Freezing point of pure water = T = 0°C
Freezing point of solution = [tex]T_f[/tex]
[tex]\Delta T_f=T-T_f[/tex]
[tex]T_f=T-\Delta T_f=0^oC-3.26^oC=-3.26^oC[/tex]
The expected freezing point of a 1.75 m solution of ethylene glycol is -3.26°C.
A microbiologist hypothesized that
Staphylococcus aureus is more susceptible to
antibiotics than Escherichia coli. She tested her
hypothesis by exposing the two species of
bacteria to three different antibiotics. The closer
the bacteria were able to grow to antibiotic-soaked
disks, the more resistant they were to treatment
by that antibiotic.
Which bacterium would you expect to be more
responsive to antibiotic treatment?
MANE
Answer: S. Auresus
Explanation:I just took it
A microbiologist studies the microbes and the disease they cause. The bacterium that will be more responsive to antibiotic treatment will be Staphylococcus aureus.
What are antibiotics?Antibiotics are said to be the medication and drug class that works against bacterial infections. They are used to kill and slow the rate of growth or reproduction of microbes like bacteria to stop diseases or infections like flu, cold, cough, etc.
The bacteria can be resistant or susceptible to the antibiotic and can be tested by exposing them to antibiotics. As Staphylococcus aureus species are vulnerable they will not be able to grow under the antibiotic treatment. Whereas, Escherichia coli being resistant will show some growth.
Therefore, Staphylococcus aureus will be more responsive to antibiotics.
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How is a mixture different from a chemical reaction?
Answer:
A mixture is made when two or more substances are combined, but they are not combined chemically. A chemical reaction is a transformation from one set of chemicals into another set.
Explanation:
A mixture can be homogeneous or heterogeneous, when homo, you cannot see each variable but hey are there. Hetero, means you can see each variable in the mixture. It is considered a mixture and not a chemical reaction because it can be reversed. It will always be able to go back to its separate forms and keep its original composition. When there is a chemical reaction, it is hard to separate and go back to the original variables because there is a molecular bond.
What conversion factor is used to convert between grams and moles?
The molar mass or gram formula mass is the conversion factor used to convert between grams and moles in chemistry.
Explanation:The conversion factor used to convert between grams and moles in chemistry is called the molar mass or the gram formula mass. It is defined as the mass of one mole of a substance. To convert grams to moles, you divide the given mass by the molar mass of the substance. To convert moles to grams, you multiply the given moles by the molar mass.
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Calculate the pH for the following weak acid. A solution of HCOOH has 0.12M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4. What is the pH of this solution at equilibrium?
Answer:
the pH of HCOOH solution is 2.33
Explanation:
The ionization equation for the given acid is written as:
[tex]HCOOH\leftrightarrow H^++HCOO^-[/tex]
Let's say the initial concentration of the acid is c and the change in concentration x.
Then, equilibrium concentration of acid = (c-x)
and the equilibrium concentration for each of the product would be x
Equilibrium expression for the above equation would be:
[tex]\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}[/tex]
[tex]1.8*10^-^4=\frac{x^2}{c-x}[/tex]
From given info, equilibrium concentration of the acid is 0.12
So, (c-x) = 0.12
hence,
[tex]1.8*10^-^4=\frac{x^2}{0.12}[/tex]
Let's solve this for x. Multiply both sides by 0.12
[tex]2.16*10^-^5=x^2[/tex]
taking square root to both sides:
[tex]x=0.00465[/tex]
Now, we have got the concentration of [tex][H^+] .[/tex]
[tex][H^+] = 0.00465 M[/tex]
We know that, [tex]pH=-log[H^+][/tex]
pH = -log(0.00465)
pH = 2.33
Hence, the pH of HCOOH solution is 2.33.
Answer:
The correct answer is 2.34
Explanation:
HCOOH is formic acid. It is a weak acid so it does not dissociates completely in water. At the beggining (I) the initial concentration is 0.12 M. In water it will dissociate in a certain grade x as follows:
HCOOH → H⁺ + HCOO⁻
I 0.12 M 0 0
C - x x x
E (0.12 M - x) x x
The mathematical expression for the equilibrium constant (Ka) is the following:
[tex]K_{a} = \frac{[H^{+} ][HCOO^{-} ]}{[HCOOH]}[/tex]
[tex]1.8 x 10^{-4} = \frac{(x x)}{(0.12 M -x)}[/tex]
As the value of Ka is too small in comparison with the initial concentration 0.12 M, we can approximate: 0.12 M - X ≅ 0.12 M. Then, we calculate x:
1.8 x 10⁻⁴ = x²/0.12 M
⇒ x= [tex]\sqrt{0.12 x 1.8 x 10^{-4} }[/tex]= 4.65 x 10⁻³
Since x = 4.65 x 10⁻³ , from the equilibrium we have:
[H⁺] = x = 4.65 x 10⁻³
From the definition of pH, we have:
pH = -log [H⁺] = -log (4.65 x 10⁻³)= 2.34
A 0.881 g sample of a diprotic acid is dissolved in water and titrated with 0.160 M NaOH . What is the molar mass of the acid if 35.8 mL of the NaOH solution is required to neutralize the sample
Final answer:
To find the molar mass of the diprotic acid, divide the mass of the sample by the number of moles. The molar mass of the acid is 307.03 g/mol.
Explanation:
To find the molar mass of the diprotic acid, we can use the information from the titration. First, we need to determine the number of moles of NaOH used in the titration. The volume of NaOH solution used is 35.8 mL, which is equal to 0.0358 L. The molarity of the NaOH solution is 0.160 M.
Therefore, the number of moles of NaOH used is 0.0358 L x 0.160 mol/L = 0.005728 mol.
Since the diprotic acid is diprotic, it can donate two moles of H+ ions per mole of acid. Therefore, the number of moles of the diprotic acid is half of the number of moles of NaOH used, which is 0.005728 mol / 2 = 0.002864 mol.
To calculate the molar mass of the acid, we need to divide the mass of the sample by the number of moles. The mass of the acid sample is 0.881 g. Therefore, the molar mass of the acid is 0.881 g / 0.002864 mol = 307.03 g/mol.
The reaction A + 2B occurs in one step in the gas phase. In each blank below, write the exponent of the concentration in the FORWARD rate law or write none if that concentration does not appear in the rate law.
[A] ...........
[B] ..........
[C] ...........
Answer:
see below
Explanation:
for A + 2B => Products ...
Rate Law => Rate =k[A][B]ˣ
As shown in expression, A & B are included, C is not.
Answer:
Exponent of a is 1, exponent of b is 2 and exponent of c = 0
Explanation:
or the rate equation is the expression which relates the rate of the reaction with the concentration of pressure of the reactants. The rate law is expressed in terms of the molar concentration of the reactants with each term raised to power its stoichiometric coefficient.
A lab group was supposed to make mL of a acid solution by mixing a solution, a solution, and a solution. However, the solution was mislabeled, and was actually a solution, so the lab group ended up with mL of a acid solution, instead. What are the volumes of the solutions that should have been mixed?
Question: The question is not complete. Find below the complete question and the answer.
Alab group was supposed to make 14 mL of a 36% acid solution by mixing a 20% solution, a 26% solution, and a 42% solution. However, the 20% solution was mislabeled, and was actually a 10% solution, so the lab group ended up with 14 mL of a 34% acid solution, instead. If the augmented matrix that represents the system of equations is given below, what are the volumes of the solutions that should have been mixed? mL
Volume of 20% solution= ?
Volume of 26% solution = ?
Volume of 42% solution= ? Round to the nearest whole number ml
Answer:
Volume of 20% solution= 3 mL
Volume of 26% solution = 1 mL
Volume of 42% solution= 10 mL
Explanation:
Find attached of the calculations.
Suppose of potassium bromide is dissolved in of a aqueous solution of silver nitrate. Calculate the final molarity of bromide anion in the solution. You can assume the volume of the solution doesn't change when the potassium bromide is dissolved in it. Round your answer to significant digits.
To find the final molarity of bromide ions in a solution, we need the mass of the dissolved potassium bromide and the volume of the solution. Then we do a simple molar calculation. The molarity of KBr and Br- are equivalent as KBr disassociates fully in solution.
Explanation:In order to determine the final molarity of the bromide anion, we first need to know the amount of potassium bromide (KBr) dissolved and the volume of the solution it's dissolved in. However, these key data are missing in the question.
Regardless, the steps to calculate molarity (M) are as follows:
Multiply the mass of KBr by its molar mass to convert to moles.Subtract the used amount from the total massDivide the moles of KBr by the volume of the total solution in liters, giving the molarity. M=mass(KBr) in moles/volume of solution in litersIn this particular scenario, we also know that one mole of KBr results in one mole of bromide ions (Br-), as it disassociates fully upon dissolution. So, the molarity of KBr would be equal to the molarity of Br-.
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Let’s say you have 3 UV active spots in the crude material and co-spot TLC plate. One has the same Rf value as the starting material and the other 2 are very different. What could these 2 nonstarting material spots be? Use structures and words. Hint: think about the nucleophilic addition of the hydride.
Answer:
Answer: (1R,2S) / (1S, 2R) , (1R,2R) / (1S, 2S)
Explanation:
Sodium borohydride reduction of benzoin will give four possible stereo isomers out of which are (1R,2S) - (1S, 2R) isomers and (1R,2R) - (1S, 2S) isomers which are known as enantiomers.
In general enantiomers show single spot in the TLC as they do not show any difference in Rf value (i.e) (1R,2S) - (1S, 2R) isomers show only one spot although they are two compounds and also (1R,2R) - (1S, 2S) isomers also show one spot. That is the reason why you are observing two spots in the TLC ( of reaction mixture) other than starting materilal.
Answer:
Explanation:
find the solution below
1. A sample of hydrogen gas is collected over water in a flask at 23.5 C. The pressure in the flask is equilibrated with the atmospheric pressure. The atmospheric (barometric) pressure is reported as 746.7 torr. Determine the pressure of hydrogen gas in the flask.
Answer:
the pressure of hydrogen gas in the flask is 725 torr
Explanation:
the solution is in the attached Word file
The pressure of hydrogen gas in the flask is [tex]\( 725.6 \, \text{torr} \)[/tex].
To determine the pressure of hydrogen gas collected over water, we need to account for the vapor pressure of water at the given temperature. The total pressure in the flask is the sum of the partial pressures of hydrogen gas and water vapor.
The steps to solve this are:
1. Find the vapor pressure of water at the given temperature (23.5°C):
The vapor pressure of water at 23.5°C is approximately 21.1 torr (this value can be found in tables of water vapor pressures).
2. Calculate the pressure of hydrogen gas:
The total pressure in the flask is the atmospheric pressure, which is 746.7 torr. The pressure of hydrogen gas is the total pressure minus the vapor pressure of water.
[tex]\[\text{Pressure of hydrogen gas} = \text{Total pressure} - \text{Vapor pressure of water}\][/tex]
[tex]\[P_{\text{H}_2} = 746.7 \, \text{torr} - 21.1 \, \text{torr}\][/tex]
[tex]\[P_{\text{H}_2} = 725.6 \, \text{torr}\][/tex]
1. Is energy absorbed or released when chemical bonds are broken during a chemical reaction?
2. Is energy absorbed or released when chemical bonds are formed during a chemical reaction?
3. What are some real life examples of an exothermic reaction?
4. What are some real life examples of a endothermic reaction?
5. Compare and contrast Endothermic vs. a Exothermic reaction
6. Analyze the folloiwng scenario:
In chemistry class after adding two chemical substances in a test tube ( which is the system) you noticed when you touch the tube that it begin to get very cold. What type of chemical reaction is this? Endothermic or Exothermic ... Use the scenario to explain your response.
7. Define ΔH Enthalpy
8. Compare and Contrast a positive ΔH(Enthalpy) to a negative ΔH( Enthalpy)
9. Explain the relationship between Enthalpy and a Exothermic reaction, and a Endothermic reaction
10. Explain how the Potential Energy curve's of a Endothermic different from a Exothermic reaction
A particular type of fundamental particle decays by transforming into an electron e- and a positron e . Suppose the decaying particle is at rest in a uniform magnetic field B of magnitude 3.64 mT and the e- and e move away from the decay point in the paths lying in a plane perpendicular to B. How long after the decay do the e- and e collide
Answer:
Check the explanation
Explanation:
Lorenz force exerted on electron by magnetic field is
F=e*v*H, where e=1.602E-19 C, H=3.64E-3 T, v is speed of electron;
? meanwhile Lorenz force is centripetal force F=m*v^2/r, where mass of electron m=9.11E-31 kg, r is radius of the path of electron;
? therefore F=F; e*v*H = m*v^2/r; eH=m*(v/r), hence
v/r = eH/m =w is angular speed of electron, hence
T=2pi/w =2pi*m/(eH) is period of rotation of electron;
e- and e+ starting at the same point are moving in the same circular path and in opposite directions, and should meet in T/2 time;
T/2 = pi*m/(eH) = pi*9.11E-31 /(1.602E-19 *3.48E-3) =5.13E-9 s;
Assume that Aluminum and Silver Sulfide are the starting substances (reactants) in the reaction: a. Write a balanced chemical equation describing the "re-creation" of silver, using the information in the case study. b. State the names of the products that are produced from this reaction. c. What type of reaction(s) is/are being represented by the chemical reaction you wrote in part (a)? d. Is the reaction in part (a) an oxidation-reduction (redox) reaction? e. If this is a redox reaction, then identify the following: What is undergoing oxidation (what is being oxidized)? Support your answer using oxidation numbers. What is undergoing reduction (what is being reduced)? Support your answer using oxidation numbers. What is the reducing agent? What is the oxidizing agent?
Answer:
Check Explanation
Explanation:
a) The balanced chemical reaction between Aluminium and Silver Sulfide is represented below
2Al + 3Ag₂S → Al₂S₃ + 6Ag
Aluminium displaces Silver from Silver sulfide because it is higher than Silver in the electrochemical series.
b) Names of the products from this reaction
Ag is called Silver metal. Free Silver metal.
Al₂S₃ is called Aluminium Sulfide.
c) This reaction is a single-displacement reaction because an element directly displaces and replaces another element in a compound.
It is also a redox reaction (reduction-oxidation reaction) because there are species being oxidized and reduced simultaneously!
d) Yes, this reaction is an oxidation-reduction (redox) reaction because there are species being oxidized and reduced simultaneously!
e) The specie that is being oxidized is said to undergo oxidation. And oxidation is defined as the loss of electrons, thereby leading to an increase in oxidation number.
In this reaction, it is evident that Aluminium undergoes oxidation as its oxidation number increases from 0 in the free state to +3 when it displaces Silver and becomes Aluminium sulfide.
Al → Al³⁺
0 → +3 (Oxidation)
The specie that is being reduced is said to undergo reduction. And reduction is defined as the gain of electrons, thereby leading to a decrease in oxidation number.
In this reaction, it is evident that the silver ion undergoes reduction as its oxidation number decreases from +1 in the Silver Sulfide compound to 0 when it is displaced and becomes Silver in free state.
Ag⁺ → Ag
+1 → 0 (Reduction)
The reducing agent is the specie that brings about reduction. Since Silver ion in Silver sulfide is reduced, Aluminium is the reducing agent that initiates the reduction process.
The oxidation agent is the specie that brings about oxidation. Since, Aluminium is the specie that undergoes oxidation, the oxidizing agent is the Silver Sulfide that brings about the oxidation.
Hope this Helps!!!
PLEASE HELP I HAVE A TIME LIMIT
A chemist prepared a solution of KOH by completely dissolving 24.0 grams of solid KOH in 2.25 liters of water at room temperature. What was the pH of the solution that the chemist prepared, to the nearest thousandth?
Answer:
13.279
Explanation:
24g/56.1056g/mol= 0.42 (mol KOH)
0.42 mol KOH/ 2.25 L= 0.19 M KOH
-log(0.19)= pOH= 0.72
pH=14-pOH
pH=14-0.72=13.279
sorry it's not the best explanation but since you're on a time crunch i hope this helps
A chemist prepared a solution of KOH by completely dissolving 24.0 grams of solid KOH in 2.25 liters of water at room temperature. The pH of the solution that the chemist prepared, to the nearest thousandth is 13.279.
pH is a numerical indicator of how acidic or basic aqueous or some other liquid solutions are. The phrase, which is frequently used during chemistry, biology, or agronomy, converts hydrogen ion concentrations.
The hydrogen ion concentration in pure water, which has a pH of 7, is 107 gram-equivalents per liter, making it neutral.
24g/56.1056g/mol= 0.42 (mol KOH)
0.42 mol KOH/ 2.25 L= 0.19 M KOH
-log(0.19)= pOH= 0.72
pH=14-pOH
pH=14-0.72
=13.279
To know more about pH, here:
https://brainly.com/question/27945512
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Solid potassium chromate is slowly added to 175 mL of a lead(II) nitrate solution until the concentration of chromate ion is 0.0366 M. The maximum amount of lead ion remaining in solution is
Answer:
[Pb⁺²] remaining = 9.0 x 10⁻¹²M in Pb⁺² ions.
Explanation:
K₂CrO₄(aq) + Pb(NO₃)₂(aq) => 2KNO₃(aq) + PbCrO₄(s)
PbCrO₄(s) ⇄ Pb⁺²(aq) + CrO₄⁻²(aq)
The K₂CrO₄(aq) + Pb(NO₃)₂(aq) is 1:1 => moles K₂CrO₄ added = moles of Pb(NO₃)₂ converted to PbCrO₄(s) in 175 ml aqueous solution.
Given rxn proceeds until [CrO₄⁻] = 0.0336 mol/L which means that moles Pb(NO₃)₂ converted into PbCrO₄ is also 0.0336 mol/L.
Assuming all PbCrO₄ formed in the 175 ml solution remained ionized, then [Pb⁺²] = [CrO₄⁻²] = 0.0336mol/0.175L = 0.192M in each ion.
To test if saturation occurs, Qsp must be > than Ksp ...
=> Qsp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = (0.0.192)² = 0.037 >> Ksp(PbCrO₄) = 3 x 10⁻13 => This means that NOT all of the PbCrO₄ formed will remain in solution. That is, the ppt'd PbCrO₄ will deliver Pb⁺² ions into solution until saturation is reached. However, one should note that determination of the concentration of Pb⁺² delivered back into solution will be in the presence of 0.0366M CrO⁻² ions and the problem then becomes a common ion type calculation. That is ...
PbCrO₄(s) ⇄ Pb⁺²(aq) + CrO₄⁻²(aq); Ksp = 3 x 10⁻¹³
C(i) --- 0.00M 0.0336M
ΔC --- +x +x
C(eq) --- x 0.0336 + x ≅ 0.0336M
Ksp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = [Pb⁺²(aq)](0.0336M) = 3 x 10⁻¹³
∴[Pb⁺²(aq)] = (3 x 10⁻¹³/0.0336)M = 8.93 x 10⁻¹²M ≅ 9.0 x 10⁻¹²M in Pb⁺² ions.
Answer:
[Pb⁺²] remaining = 9.0 x 10⁻¹²M in Pb⁺² ions.
Explanation:
K₂CrO₄(aq) + Pb(NO₃)₂(aq) => 2KNO₃(aq) + PbCrO₄(s)
PbCrO₄(s) ⇄ Pb⁺²(aq) + CrO₄⁻²(aq)
The K₂CrO₄(aq) + Pb(NO₃)₂(aq) is 1:1 => moles K₂CrO₄ added = moles of Pb(NO₃)₂ converted to PbCrO₄(s) in 175 ml aqueous solution.
Given rxn proceeds until [CrO₄⁻] = 0.0336 mol/L which means that moles Pb(NO₃)₂ converted into PbCrO₄ is also 0.0336 mol/L.
By assumption if we say, all PbCrO₄ formed in the 175 ml solution remained ionized,
Then [Pb⁺²] = [CrO₄⁻²] = 0.0336mol/0.175L = 0.192M in each ion.
To test if saturation occurs,
Qsp must be > than Ksp ...
=> Qsp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = (0.0.192)² = 0.037 >> Ksp(PbCrO₄) = 3 x 10⁻13 =>
This means that NOT all of the PbCrO₄ formed will remain in solution. That is, the ppt'd PbCrO₄ will deliver Pb⁺² ions into solution until saturation is reached. However, one should note that determination of the concentration of Pb⁺² delivered back into solution will be in the presence of 0.0366M CrO⁻² ions and the problem then becomes a common ion type calculation.
i.e .........
PbCrO₄(s) ⇄ Pb⁺²(aq) + CrO₄⁻²(aq); Ksp = 3 x 10⁻¹³
C(i) --- 0.00M 0.0336M
ΔC --- +x +x
C(eq) --- x
0.0336 + x ≅ 0.0336M
Ksp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = [Pb⁺²(aq)](0.0336M) = 3 x 10⁻¹³
Therefore;
[Pb⁺²(aq)] = (3 x 10⁻¹³/0.0336)M = 8.93 x 10⁻¹²M ≅ 9.0 x 10⁻¹²M in Pb⁺² ions.
Aqueous hydrochloric acid reacts with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . If of sodium chloride is produced from the reaction of of hydrochloric acid and of sodium hydroxide, calculate the percent yield of sodium chloride. Round your answer to significant figures.
The given question is incomplete. The complete question is:
Aqueous hydrochloric acid (HCl) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (H2O). If 1.60 g of sodium chloride is produced from the reaction of 1.8 g of hydrochloric acid and 1.4 g of sodium hydroxide, calculate the percent yield of sodium chloride. Be sure your answer has the correct number of significant digits in it.
Answer: Thus the percent yield of sodium chloride is 78.0%
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} HCl=\frac{1.8g}{36.5g/mol}=0.049moles[/tex]
[tex]\text{Moles of} NaOH=\frac{1.4g}{40g/mol}=0.035moles[/tex]
[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)[/tex]
According to stoichiometry :
1 mole of [tex]NaOH[/tex] require = 1 mole of [tex]HCl[/tex]
Thus 0.035 moles of [tex]NaOH[/tex] will require=[tex]\frac{1}{1}\times 0.035=0.035moles[/tex] of [tex]HCl[/tex]
Thus [tex]NaOH[/tex] is the limiting reagent as it limits the formation of product and [tex]HCl[/tex] is the excess reagent.
As 1 mole of [tex]NaOH[/tex] give = 1 mole of [tex]NaCl[/tex]
Thus 0.035 moles of [tex]NaOH[/tex] give =[tex]\frac{1}{1}\times 0.035=0.035moles[/tex] of [tex]NaCl[/tex]
Mass of [tex]NaCl=moles\times {\text {Molar mass}}=0.035moles\times 58.5g/mol=2.05g[/tex]
[tex]{\text {percentage yield}}=\frac{\text {Experimental yield}}{\text {Theoretical yield}}\times 100\%[/tex]
[tex]{\text {percentage yield}}=\frac{1.60g}{2.05g}\times 100\%=78.0\%[/tex]
Thus the percent yield of sodium chloride is 78.0%
AgNO3 to solutions of each and determined the moles of AgCl precipitated. He also measured and
compared the conductances of solutions of each of these compounds with solutions of simple
electrolytes. Which response below gives all true statements concerning his conclusions from these
experiments?
I. From the moles of AgCl, he determined the total number of Cl ions in each
compound.
II. From the moles of AgCl, he determined the number of uncoordinated Cl ions in each
compound.
III. From the conductance of each solution, he determined the total number of ions present
upon dissolution of each compound.
a. I
b. II
c. III
d. I and II
e. II and III
Answer:
d. I and II
Explanation:
In chemistry, it is an old analytical technique to determine the amount of free chloride ions by silver chloride precipitation. This precipitation method could both be used to determine the initial chloride ion concentration in different analyte solutions or to determine the number of uncoordinated chloride ions in a complex. The number of moles of chloride ions in the silver chloride, shows the number of uncoordinated ions in a chloride ion containing complex.
The concentration of chloride ions in an analyte could be gravimetrically determined by silver chloride precipitation.
The reaction A2 + 2 B → 2 BA is thought to occur by the following mechanism:
Step 1: A2 + Z → ZA2
Step 2: ZA2 + B → BA + Z + A
Step 3: A + B → BA
Which of the following statements is correct? Select one:
a. ZA2 and A are catalysts, and Z is a reaction intermediate.
b. There are no catalysts or reaction intermediates.
c. Z is a catalyst, and ZA2 and A are reaction intermediates.
d. Z, ZA2, and A are all catalysts.
e. Z, ZA2, and A are all reaction intermediates.
Answer:
c. Z is a catalyst, and ZA2 and A are reaction intermediates.
Explanation:
Overall reaction is given as;
A2 + 2 B → 2 BA
Mechanism:
Step 1: A2 + Z → ZA2
Step 2: ZA2 + B → BA + Z + A
Step 3: A + B → BA
The options given are centered upon catalysts and reaction intermediates. So before proceeding, we have to understand the difference between the two and how to identify them.
A catalyst is basically a reaction booster to speed up the rate of the reaction and the reaction intermediate is an unstable, temporl species formed from the reactants before getting to the products.
The difference is given as;
Catalysts are present as reactants in the very beginning and products at the end of the reaction.
Intermediates, on the other hand, are not present in the initial reaction but are produced within one of the steps and then consumed within another step.
Following the above, we can deduce that;
Z is a catalyst because it is present as a reactant in the beginning.
ZA2 and A are a reaction intermediates because they are not present in the overall reaction but are produced and consumed in one of the steps.
Correct option is given as;
c. Z is a catalyst, and ZA2 and A are reaction intermediates.
If you prepared a 3.25 M solution of sucrose (molar mass 342 g/mole) ,
a.How many moles are in 0.25 L of this solution?
b.How many grams is this?
Answer:
a. 0.8125 moles of sucrose
b. 277.8 g of sucrose
Explanation:
Consider these relation's value:
Molarity = Mol / Volume(L)
Mol = Molarity . Volume(L)
Volume(L) = Mol / Molarity
So, Molarity = 3.25M
Volume(L) = 0.25L
Molarity . Volume(L) = Mol → 3.25mol/L . 0.25L = 0.8125 mol
Let's convert the moles to mass → 0.8125 mol . 342 g /1mol = 277.8 g of sucrose
Answer:
We have 0.8125 moles sucrose in this solution. This is 277.9 grams of sucrose
Explanation:
Step 1: Data given
Molarity of a sucose solution = 3.25 M
Molar mass of sucrose = 342 g/mol
Step 2: Calculate moles sucrose
Moles sucrose = molarity sucrose solution * volume solution
Moles sucrose = 3.25 M * 0.25 L
Moles sucrose = 0.8125 moles sucrose
Step 3: Calculate mass of sucrose
Mass sucrose = moles sucrose * molar mass sucrose
Mass sucrose = 0.8125 moles * 342 g/mol
Mass sucrose = 277.9 grams
We have 0.8125 moles sucrose in this solution. This is 277.9 grams of sucrose
An Erlenmeyer flask containing 20.0 mL of sulfuric acid of an unknown concentration was titrated with exactly 15.0 mL of 0.25 M NaOH solution to the second equivalence point. What was the concentration of sulfuric acid in the flask? H2SO4 (aq) + 2 NaOH (aq) → 2 H2O (l) + Na2SO4 (aq)
Answer:
the concentration of sulfuric acid in the flask is 0.375 M
Explanation:
H2SO4 (aq) + 2 NaOH (aq) → 2 H2O (l) + Na2SO4 (aq)
Moles of NaOH = 15 x 0.25 /1000
= 0.00375
Moles of H2SO4 needed to neutralize = 0.00375 /2
= 0.001875
Molarity of H2SO4
= 0.001875 x 1000 /20
= 0.09375 M
concentration of sulfuric acid in the flask 0.09375 M
Moles
V2 = 20 ml
N2 =?
Substitute in the equation and find N2
N2 = 2 x 15 x 0.25 / 20
= 0.375 M
Thus, the concentration of sulfuric acid in the flask is 0.375 M
A chemist adds 435.0mL of a 2.28 M zinc nitrate ZnNO32 solution to a reaction flask. Calculate the millimoles of zinc nitrate the chemist has added to the flask. Round your answer to 3 significant digits.
Answer:
He added 992 milimoles of zinc nitrate (Zn(NO3)2)
Explanation:
Step 1: Data given
Volume of zinc nitrate = 435.0 mL =0.435 L
Molarity of zinc nitrate = 2.28 M
Step 2: Calculate moles zinc nitrate
Moles = molarity* volume
Moles Zn(NO3)2 = 2.28 M * 0.435 L
Moles Zn(NO3)2 = 0.9918 moles
Step 3: Convert moles to milimoles
Moles Zn(NO3)2 = 0.9918 moles
Moles Zn(NO3)2 = 0.9918 * 10^3 milimoles
Moles Zn(NO3)2 = 991.8 milimoles ≈ 992 milimoles
He added 992 milimoles of zinc nitrate (Zn(NO3)2)
Final answer:
To determine the millimoles of zinc nitrate added, convert 435.0 mL to liters, use the molarity of 2.28 M to find moles, and then multiply by 1000 to get 992.4 millimoles.
Explanation:
To calculate the millimoles of zinc nitrate a chemist has added to the reaction flask, you first need to know the concentration of the solution and the volume of the solution used. Here, we have a 2.28 M zinc nitrate solution, and the volume used is 435.0 mL. To find the millimoles, first convert the volume from milliliters to liters (since molarity is moles per liter), then use the molarity to find the moles of zinc nitrate, and finally convert that to millimoles.
Convert volume to liters: 435.0 mL × (1 L / 1000 mL) = 0.435 L
Calculate moles of Zn(NO3)2: 2.28 moles/L × 0.435 L = 0.9924 moles
Convert moles to millimoles: 0.9924 moles × 1000 mmol/mole = 992.4 mmol
Therefore, the chemist has added 992.4 millimoles of zinc nitrate to the flask.
ANSWER ASAP
Suppose that 10.00 HCl of unknown concentration is neutralized by 20.00 mL of a 1.50 M NaOH solution. Determine the concentration of the HCl solution?
a. 0.0750 M HCl
b. 0.150 M HCl
c. 3.00 M HCl
d. 1.50 M HCl
Answer:
c. 3.00 M HCl
Explanation:
From dilution formula
C1V1 = C2V2
C1=?, V1= 10.0ml, C= 1.5, V2= 20.0ml
Substitute and Simplify
C1×10= 1.5×20
C1= 3.00M