Answer:
10 ms-1
Explanation:
Kinetic energy = 1/2 × m × v^2
1 = 1/2× 20 ×10^ -3 × v^2
v ^ 2 = 100
v = 10 ms-1
note : convert grams in to kg before substitution as above
Given:
Kinetic energy,
K.E = 1.00 JMass,
m = 20.0 gWe know the formula,
→ [tex]K.E = \frac{1}{2} mv^2[/tex]
By putting the values, we get
[tex]1 = \frac{1}{2}\times 20\times 10^{-3}\times (v)^2[/tex]
[tex]v^2 = 100[/tex]
[tex]v = \sqrt{100}[/tex]
[tex]v = 10 \ m/s[/tex]
Thus the above response is correct.
Learn more about K.E here:
https://brainly.com/question/24997625
(a) (i) What is the name of apparatus used to measure conductivity of water
Answer:
An electrical conductivity meter (EC meter) measures the electrical conductivity in a solution. It has multiple applications in research and engineering, with common usage in hydroponics, aquaculture, aquaponics, and freshwater systems to monitor the amount of nutrients, salts or impurities in the water.
The Bohr effect:_____.
a. explains through the Bohr model of the atom why Fe2+ will bind O2 in heme but Fe3+ will not.
b. contributes to binding of O2 by hemoglobin in lungs and release of O2 from hemoglobin in tissues.
c. applies to both myoglobin and hemoglobin.
d. relates [H+] to [CO2].
Answer:
b. contributes to binding of O2 by hemoglobin in lungs and release of O2 from hemoglobin in tissues.
Explanation:
The Bohr effect is a phenomenon described by Christian Bohr. Is an affinity that binds oxygen and hemoglobin and is inversely related to the concentration of carbon dioxide. As CO2 reacts with water and an increase in CO2 results in a decrease in blood ph.A sample of Kr gas is observed to effuse through a pourous barrier in 8.15 minutes. Under the same conditions, the same number of moles of an unknown gas requires 4.53 minutes to effuse through the same barrier. The molar mass of the unknown gas is ____________ g/mol.
Answer:
25.88 g/mol
Explanation:
Graham's law is a famous law which states that the diffusion rate or the effusion rate of any gas varies inversely to the square root of the molecular weight the gas.
So from Graham's law, we have,
[tex]$\frac{\text{time}}{M^{1/2}}=\text{constant}$[/tex]
Using the sample of Kr gas having M = 83.8
[tex]$\frac{8.15}{(83.8)^{0.5}}= \frac{4.53}{M^{0.5}}$[/tex]
[tex]$M^{0.5}= 5.088$[/tex]
M = 25.88 g/mol
The solubility of lithium fluoride, LiF, is 1.6 g/L, or 6.2 x 10â2 M.
a. Write the balanced solubility equilibrium equation for LiF.
b. Determine the molar concentration of the lithium ion and the fluoride ion.
c. Write the Ksp expression for the reaction.
d. Calculate Ksp for lithium fluoride.
Answer:
a. LiF(s) ⇄ Li⁺(aq) + F⁻(aq)
b. [Li⁺] = [F⁻] = 6.2 x 10⁻² M
c. Ksp = [Li⁺] [F⁻]
d. Ksp = 3.8 × 10⁻³
Explanation:
The solubility (S) of lithium fluoride, LiF, is 1.6 g/L, or 6.2 x 10⁻² M.
a. The balanced solubility equilibrium equation for LiF is:
LiF(s) ⇄ Li⁺(aq) + F⁻(aq)
b. We will make an ICE chart.
LiF(s) ⇄ Li⁺(aq) + F⁻(aq)
I 0 0
C +S +S
E S S
Then, [Li⁺] = [F⁻] = S = 6.2 x 10⁻² M
c. The solubility product constant, Ksp, is the equilibrium constant for a solid substance dissolving in an aqueous solution.
Ksp = [Li⁺] [F⁻]
d.
Ksp = [Li⁺] [F⁻] = (6.2 x 10⁻²)² = 3.8 × 10⁻³
Critique this statement: Electrons can exist in any position
outside of the nucleus.
Answer:
However, there has to be 2 electrons on the first shell, and 8 on the others.
Explanation:
Hope this helps :)
Determine the total pressure of a mixture that contains 5.25 g of He and 3.25 g of N2 in a 7.75-L flask at a temperature of 27ºC.
Answer:
4.54 atm
Explanation:
Step 1: Calculate the total number of gaseous moles
We will calculate the moles of each gas using its molar mass.
He: 5.25 g × 1 mol/4.00 g = 1.31 mol
N₂: 3.25 g × 1 mol/28.01 g = 0.116 mol
The total number of moles is:
n = 1.31 mol + 0.116 mol = 1.43 mol
Step 2: Convert 27 °C to Kelvin
We will use the following expression.
K = °C + 273.15 = 27 + 273.15 = 300 K
Step 3: Calculate the total pressure of the mixture
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T / V
P = 1.43 mol × (0.0821 atm.L/mol.K) × 300 K / 7.75 L = 4.54 atm
Based on the reaction below:
[tex]N_2 + 3H_2[/tex] ↔ [tex]2NH_3 + heat[/tex]
If we decrease the temperature, equilibrium will shift towards the...
Please explain!
N₂ + 3H₂ ⇄ 2NH₃ + heat
In the given equilibrium, we notice that the heat is on the right. which means that if the heat requirements don't meet, the reactants on the right will no longer react due to the lack of heat
but because the reactants on the left don't have such weaknesses, they will keep reacting hence producing more and more ammonia until a new equilibrium is reached
where there will be more ammonia and less nitrogen and hydrogen as compared to the equilibrium we had initially
Answer:
Explanation:
heat is given out as 1 of the products, along w/ NH3 in the forward reaction. so its an exothermic reaction
decreasing temperature favors exothermic reaction as more heat can be absorbed by the environment
so equilibrium will shift towards the products
pplication)Using multiple models simultaneously: This FNT refers to a processinvolving three moles of a diatomic gas (which behaves as an ideal gas).The PV curve at right describes this process.Assume, as is typical near room temperature,vibrational modes are frozen out.a)Determine the energy transferred as work, the change in internal energy, and the energy transferred as heat in this process. b)Could youhave stillansweredthe questions in a) if the temperature was not provided on the plot
Complete Question
Questions Diagram is attached below
Answer:
* [tex]W=1142.86Joule[/tex]
* [tex]Q=997.7J[/tex]
* [tex]H=2140.5J[/tex]
Explanation:
From the question we are told that:
Temperature [tex]T=337K[/tex]
Pressure [tex]P=(60-55)Pa*10^5[/tex]
Volume[tex]V=(1.6-1.4)m^3*10^{-3}[/tex]
Generally the equation for gas Constant is mathematically given by
[tex]\frac{P_2}{P_1}=\frac{V_1}{V_2}^n[/tex]
[tex]\frac{55*10^5}{60*10^5}=\frac{1.4*10^{-3}}{1.6*10^{-3}}^n[/tex]
[tex]n=0.65[/tex]
Therefore
Work-done
[tex]W=\int{pdv}[/tex]
[tex]W=\frac{55*10^5*1.6*10^{-3}*60*10^5*1.4*10^{-3}}{1-0.65}[/tex]
[tex]W=1142.86Joule[/tex]
Generally the equation for internal energy is mathematically given by
[tex]Q=mC_vdT\\\\Q=\frac{3*1*3.314*16}{1.4-1}[/tex]
[tex]Q=997.7J[/tex]
Therefore
[tex]H=Q+W[/tex]
[tex]H=997.7J-11.42.9[/tex]
[tex]H=2140.5J[/tex]
A person slips over banana pills. Give reason
Answer:
We slip when we step on a banana peel because the inner side of banana peel being smooth and slippery reduces the friction between the sole of our shoe and the surface of road.
A person uses 400.8 kcal of energy to run a race. Convert the energy used for the race to the following energy units:
(provide an answer in 4 significant figures)
Calories
calories
Food Calories
Joules
Kilojoules
Hint: 1kcal=4.184kJ
Calories, we know that fat burn is calories.
A solution is prepared by dissolving 6.60 g of an nonelectrolyte in water to make 550 mL of solution. The osmotic pressure of the solution is 1.84 atm at 25 °C. The molecular weight of the nonelectrolyte is ________ g/mol.
Answer:
160 g/mol
Explanation:
Step 1: Calculate the molarity of the solution
We will use the following expression.
π = M × R × T
where,
π: osmotic pressure of a nonelectrolyteM: molarityR: ideal gas constantT: absolute temperature (25 °C = 298 K)M = π / R × T
M = 1.84 atm / (0.0821 atm.L/mol.K) × 298 K = 0.0752 mol/L
Step 2: Calculate the moles of solute in 550 mL (0.550 L)
0.550 L × 0.0752 mol/L = 0.0413 mol
Step 3: Calculate the molecular weight of the nonelectrolyte
0.0413 moles weigh 6.60 g.
6.60 g/0.0413 mol = 160 g/mol
Bond length is the distance between the centers of two bonded atoms. On the potential energy curve, the bond length is the internuclear distance between the two atoms when the potential energy of the system reaches its lowest value. Given that the atomic radii of H and Br are 37.0 pm and 115 pm , respectively, predict the upper limit of the bond length of the HBr molecule. Express your answer to three significant figures and include the appropriate units. View Available Hint(s)for Part C
Answer:
The answer is "152 pm".
Explanation:
The bond length from the values inside the atomic radii is calculated according to the query. This would be the upper limit of a molecule's binding length.
The atomic radius of [tex]H= 37.0 \ pm[/tex]
The atomic radius of [tex]Br = 115.0 \ pm[/tex]
[tex]\text{Bond length = Atomic radius of H + Atomic radius of Br}[/tex]
[tex]= 37.0\ pm + 115.0 \ pm\\\\= 152\ pm[/tex]
melting points of lipids are strongly influenced by the length and degree of unsaturation of hydrocarbon chain.justify this statement?
Answer
The properties of fatty acids and of lipids derived from them are markedly dependent on chain length and degree of saturation. Unsaturated fatty acids have lower melting points than saturated fatty acids of the same length. For example, the melting point of stearic acid is 69.6°C, whereas that of oleic acid (which contains one cis double bond) is 13.4°C. The melting points of polyunsaturated fatty acids of the C18 series are even lower. Chain length also affects the melting point, as illustrated by the fact that the melting temperature of palmitic acid (C16) is 6.5 degrees lower than that of stearic acid (C18). Thus, short chain length and unsaturation enhance the fluidity of fatty acids and of their derivatives.
By using photons of specific wavelengths, chemists can dissociate gaseous HI to produce H atoms with certain speeds. When HI dissociates, the H atoms move away rapidly, whereas the heavier I atoms move more slowly. If a photon of 231 nm is used, what is the excess energy (in J) over that needed for dissociation
Answer:
The excess energy over that needed for dissociation is 3.712 × 10⁻¹⁹ J
Explanation:
Given the data in the question;
wavelength of proton λ = 231 nm = 231 × 10⁻⁹ m
we determine the energy of the proton;
E = hc / λ
where h is plank constant ( 6.626 × 10⁻³⁴ JS )
and c is the speed of light ( 3 × 10⁸ m/s )
we substitute
E = [ ( 6.626 × 10⁻³⁴ JS ) × ( 3 × 10⁸ m/s ) ] / [ 231 × 10⁻⁹ m ]
E = 8.61 × 10⁻¹⁹ J
we know that, bond energy for H-I is 295 kJ/mol
so, H = 295 × 10³ J/mol
Now, energy to dissociate HI will be;
⇒ H / N
where N is the Avogadro's number ( 6.023 × 10²³ mol⁻¹ )
energy to dissociate HI = ( 295 × 10³ J/mol ) / ( 6.023 × 10²³ mol⁻¹ )
= 4.898 × 10⁻¹⁹ J
Therefore, Excess energy over dissociation will be;
⇒ ( 8.61 × 10⁻¹⁹ J ) - ( 4.898 × 10⁻¹⁹ J )
= 3.712 × 10⁻¹⁹ J
The excess energy over that needed for dissociation is 3.712 × 10⁻¹⁹ J
Explain why
when suger
is heated it
does not boil but rather break down
into carbon and water?
Answer:
When simple sugars such as sucrose (or table sugar) are heated, they melt and break down into glucose and fructose, two other forms of sugar. Continuing to heat the sugar at high temperature causes these sugars to lose water and react with each other producing many different types of compounds.
How many grams of NaCl (MM = 58.44g/mol) are in 250mL of a 0.75 molar solution?
Answer:
[tex]\boxed {\boxed {\sf 11 \ grams \ NaCl}}[/tex]
Explanation:
We are asked to find how many grams of sodium chloride are in a solution.
1. Moles of SoluteMolarity is a measure of concentration in moles per liter.
[tex]molarity= \frac{ moles \ of \ solute}{liters \ of \ solution}[/tex]
We know the molarity is 0.75 molar. 1 molar is the same as 1 mole per liter, so the solution contains 0.75 moles of sodium chloride per liter.
There are 250 milliliters of solution but molarity uses liters for volume. We must convert milliliters to liters. Remember that 1 liter contains 1000 milliliters. Set up a ratio and use dimensional analysis to convert.
[tex]250 \ mL * \frac{1 \ L} {1000\ mL} = \frac{ 250}{1000} \ L = 0.250 \ L[/tex]
Now we know the molarity and the liters of solution, but the moles of solute are unknown.
molarity = 0.75 mol NaCl/L moles of solute =x liters of solution = 0.25 LSubstitute the values into the formula.
[tex]0.75 \ mol \ NaCl/L = \frac{x}{0.250 \ L}[/tex]
We are solving for the moles of solute, so we must isolate the variable x. It is being divided by 0.250 liters. The inverse of division is multiplication, so multiply both sides of the equation by 0.250 L.
[tex]0.250 \ L *0.75 \ mol \ NaCl/L = \frac{x}{0.250 \ L} * 0.250 \ L[/tex]
[tex]0.250 \ L *0.75 \ mol \ NaCl/L = x[/tex]
The units of liters cancel.
[tex]0.250 * 0.75 \ mol \ NaCl[/tex]
[tex]\bold {0.1875 \ mol \ NaCl}[/tex]
2. Grams of SoluteNow that we have calculated the moles of solute, we must convert this to grams. We use the molar mass or the mass of 1 mole of a substance. Sodium chloride's molar mass is given and it is 58.44 grams per mole. This means there are 58.44 grams of sodium chloride in 1 mole of sodium chloride.
Set up a ratio so we can convert using dimensional analysis.
[tex]\frac {58.44 \ g \ NaCl}{1 \ mol \ NaCl}[/tex]
Multiply by the number of moles we calculated.
[tex]0.1875 \ mol \ NaCl *\frac {58.44 \ g \ NaCl}{1 \ mol \ NaCl}[/tex]
The units of moles of sodium chloride cancel.
[tex]0.1875 *\frac {58.44 \ g \ NaCl}{1}[/tex]
[tex]\bold {10.9575 \ g \ NaCl}[/tex]
3. Round using Significant FiguresThe original measurements of molarity and volume have 2 significant figures, so our answer must have the same. For the number we calculated, that is the ones place. The 9 in the tenths place tells us to round the 0 up to a 1.
[tex]11 \ g \ NaCl[/tex]
There are approximately 11 grams of sodium chloride in 250 mL of a 0.75 molar solution.
FILL IN THE BLANK:
The rate of a reaction is measured by how fast a (Product Or Reactant)
is used up or how fast a
(Reactant Or Product) is formed?
Answer:
the rate of a reaction is measured by how fast a REACTANT is used up or how fast a PRODUCT is formed
In the Bohr model of the hydrogen atom, the electron occupies distinct energy states.
a. True
b. False
Answer:
a. True
Explanation:
The Bohr model shows the n-2 as the first transitions between the energy states and hydrogen atom which is shown by changes in the electron structure n = 1 1. In this transition period, the electron moves from the n level to the next level, and energy d transmitted.write down the different uses of water that you know about
Answer:
The various uses of water :
1. Water is used for daily purpose like cooking , bathing , cleaning and drinking.
2. Water is used as a universal solvent.
3. water maintains the temperature of our body.
4. Water helps in digestion in our body.
5 .water is used in factories and industries.
6. Water is used to grow plants , vegetables and crops.
Answer:
The various use of water are;
I) Cooking.
ii) Drinking
III) Bathing
iv) Generating hydro- electricity
v) Construction work etc
Net ionic reaction of H2SO4 with Ba(OH)2
Answer:
This is an acid-base reaction (neutralization): Ba(OH) 2 is a base, H 2SO 4 is an acid. This is a precipitation reaction: BaSO 4 is the formed precipitate.
what is the difference between 25ml and 25.00ml
Answer:
There is no difference between the two.
Explanation:
They both show the same volume. But, adding decimal places shows the least count of the instrument used and is more acceptable when recording values in scientific experiments
Different steps of the oxidative decarboxylation of pyruvate by the pyruvate dehydrogenase PDH complex are given. Place these five steps in the correct order. Note that thiamine pyrophosphate, TPP, is sometimes called thiamine diphosphate, TDP.
1. FADH2 is reoxidized to FAD reducing NAD* to NADH.
2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.
3. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.
4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.
5. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.
Answer:
1. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.
2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.
3. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.
4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.
5. FADH2 is reoxidized to FAD reducing NAD+ to NADH.
Explanation:
The oxidation of pyruvate to AcetylCoA is catalyzed by the pyruvate dehydrogenase complex. The reaction is an irreversible oxidative decarboxylation process in which the carboxyl group of pyruvate is removed as a molecule of carbon dioxide, CO₂, while the remaining two carbons are attached to a CoASH molecule to form acetylCoA.
The pyruvate dehydrogenase complex contains three enzymes - Pyruvate dehydrogenase known as E₁, dihydrolipoyl transacetylase known as E₂, and dihydrolipoyl dehydrogenase known as E₃. It also requires five coenzymes namely: thiamine pyrophosphate (TPP), flavine adenine dinucleotide (NAD), coenzyme A (CoA-SH), nicotinamide adenine dinucleotide (NAD) and lipoate.
Oxidative decorbyxylation of pyruvate takes place in the pyruvate dehydrogenase complex in five steps:
1. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.
2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.
3. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.
4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.
5. FADH2 is reoxidized to FAD reducing NAD+ to NADH.
An analytical chemist is titrating of a solution of benzoic acid with a solution of . The of benzoic acid is . Calculate the pH of the acid solution after the chemist has added of the solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of solution added.
The question is incomplete. The complete question is :
An analytical chemist is titrating 148.9 mL of a 1.100 M solution of benzoic acid [tex]$HC_6H_5CO_2$[/tex] with a 0.3600 M solution of KOH. The [tex]pK_a[/tex] of benzoic acid is 4.20. Calculate the pH of the acid solution after the chemist has added 232.0 mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added.
Solution :
Number of moles of [tex]$C_6H_5OCOOH$[/tex] [tex]$=148.9 \ mL \times \frac{L}{1000\ mL} \times \frac{1.100 \ mol}{L}$[/tex]
= 0.16379 mol
Number of moles of NaOH added [tex]$=232.0 \ mL \times \frac{L}{1000\ mL} \times \frac{0.3600 \ mol}{L}$[/tex]
= 0.08352 mol
ICE table :
[tex]C_6H_5OCOOH \ \ \ + \ \ \ OH^- \ \ \ \rightarrow \ \ C_6H_5OCOO^- \ \ \ \ + \ \ H_2O[/tex]
I (mol) 0.16379 0.08352 0
C (mol) -0.08352 -0.08352 +0.08352
E (mol) 0.08027 0 0.08352
Total volume = (148.9 + 232) mL
= 380.9 mL
= 0.3809 L
Concentration of [tex]$C_6H_5OCOOH, [C_6H_5OCOOH]$[/tex] [tex]$=\frac{0.08027 \ mol}{0.3809 \ L}$[/tex]
= 0.211 M
Concentration of [tex]$C_6H_5OCOO^- , [C_6H_5OCOO^-] =\frac{0.08352 \ mol}{0.3809 \ L}[/tex]
= 0.219 M
[tex]pK_a[/tex] of [tex]C_6H_5OCOOH = 4.20[/tex]
According to Henderson equation,
[tex]$pH = pK_a + \log \frac{[C_6H_5OCOO^-]}{[C_6H_5OCOOH]}[/tex]
[tex]$=4.20 + \log \frac{0.219}{0.211}$[/tex]
= 4.22
Therefore, the pH of the acid solution is 4.22
Write a chemical equation for LiOH(aq) showing how it is an acid or a base according to the Arrhenius definition.
Answer:
LiOH(aq) → Li⁺(aq) + OH⁻(aq).
The mole fraction of NaCl in an
aqueous solution is 0.132. How
many moles of NaCl are present in
1 mole of this solution?
Molar Mass
NaCl: 58.44 g/mol
H2O: 18.016 g/mol
Answer:
Moles of water are 0.868
Explanation:
Phosphorus-32 is radioactive and has a half life of 14.3 days. Calculate the activity of a 3.5mg sample of phosphorus-32. Give your answer in becquerels and in curies. Round your answer to 2 significant digits.
Answer:
The activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.
Explanation:
The activity of P-32 can be calculated with the following equation:
[tex] A = \lambda N [/tex] (1)
Where:
N: is the number of atoms of P-32
λ: is the decay constant
We can find the number of atoms of P-32 as follows:
[tex] N = \frac{N_{A}*m}{M} [/tex] (2)
Where:
[tex]N_{A}[/tex]: is the Avogadro's number = 6.022x10²³ atoms/mol
m: is the mass of P-32 = 3.5x10⁻³ g
M: is the molar mass of the radionuclide (P-32) = 32 g/mol
Now, the decay constant is given by:
[tex] \lambda = \frac{ln(2)}{t_{1/2}} [/tex] (3)
Where:
[tex]{t_{1/2}} [/tex]: is the half-life of P-32 = 14.3 days
Finally, we can find the activity of P-32 by entering equations (2) and (3) into (1):
[tex] A = \lambda N = \frac{ln(2)}{t_{1/2}}*\frac{N_{A}*m}{M} = \frac{ln(2)}{14.3 d*\frac{24 h}{1 d}*\frac{3600 s}{1 h}}*\frac{6.022 \cdot 10^{23} mol^{-1}*3.5 \cdot 10^{-3} g}{32 g/mol} = 3.7 \cdot 10^{13} dis/s [/tex]
Since a becquerel (Bq) is defined as a disintegration (dis) per second, the activity in Bq is:
[tex] A = 3.7 \cdot 10^{13} Bq [/tex]
And, since a Curie (Ci) is 3.7x10¹⁰ Bq, the activity in Ci is:
[tex] A = 3.7 \cdot 10^{13} Bq*\frac{1 Ci}{3.7 \cdot 10^{10} Bq} = 1.0 \cdot 10^{3} Ci [/tex]
Therefore, the activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.
I hope it helps you!
Given the standard enthalpy changes for the following two reactions: (1) 2Fe(s) + O2(g)2FeO(s)...... ΔH° = -544.0 kJ (2) 2Zn(s) + O2(g)2ZnO(s)......ΔH° = -696.6 kJ what is the standard enthalpy change for the reaction:
Answer:
-76.3 kJ
Explanation:
Here is the complete question
Given the standard enthalpy changes for the following two reactions:
(1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ
(2) 2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ. What is the standard enthalpy change for the reaction:
(3) FeO(s) + Zn(s) → Fe(s) + ZnO(s)......ΔH° = ?
Solution
Since (1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ
reversing the reaction, we have
2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)
Adding reactions (2) and (3), we have
2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)
2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ (2)
This gives
2FeO(s) + 2Zn(s) → 2Fe(s) + 2ZnO(s)......ΔH° =
The enthalpy change for this reaction is the sum of enthalpy changes for reaction (2) and (3) = ΔH° = +544.0 kJ + (-696.6 kJ)
= +544.0 kJ - 696.6 kJ)
= -152.6 kJ
Since the required reaction is (3) which is FeO(s) + Zn(s) → Fe(s) + ZnO(s)
we divide the enthalpy change for reaction (4) by 2 to obtain the enthalpy change for reaction (3).
So, ΔH° = -152.6 kJ/2 = -76.3 kJ
So, the standard enthalpy change for the reaction
FeO(s) + Zn(s) → Fe(s) + ZnO(s) is -76.3 kJ
Aqueous hydrobromic acid HBr will react with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O. Suppose 55.8 g of hydrobromic acid is mixed with 17. g of sodium hydroxide. Calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. Round your answer to 2 significant digits.
Answer:
21.4g of HBr is the minimum mass that could be left over.
Explanation:
Based on the reaction:
HBr + NaOH → NaBr + H2O
1 mole of HBr reacts per mole of NaOH
To solve this question we need to find the moles of both reactants. If moles NaOH > moles HBr, the difference in moles represents the minimum moles of HBr that could be left over because this reaction is 1:1. Using the molar mass we can find the minimum mass of HBr that could be left over, as follows:
Moles NaOH -40.0g/mol-
17g * (1mol/40.0g) = 0.425 moles NaOH
Moles HBr -Molar mass: 80.91g/mol-
55.8g * (1mol/80.91g) = 0.690 moles HBr
The difference in moles is:
0.690 moles - 0.425 moles =
0.265 moles of HBr could be left over
The mass is:
0.265 moles * (80.91g/mol) =
21.4g of HBr is the minimum mass that could be left over.HBr can be added to an alkene in the presence of peroxides (ROOR). What function does the peroxide serve in this reaction
Answer:
Radical chain initiator
Explanation:
The peroxide here serves as a radical chain initiator. In the field of chemistry the radical initiatives are those substances that are used in industrial processes like polymer synthesis. These initiatives have weak bonds generally and they're mostly used to create free radicals. These radicals are atoms that have odd numbers of electrons. Peroxide is an example of such.
list two uses of H2SO4
