Explanation:
The balanced chemical equation of the reaction is:
[tex]C_3H_8(g)+ 5O_2 (g)->3CO_2(g)+4H_2O(g)[/tex]
From the balanced chemical equation,
1 mole of propane forms ------ 3 mol. of [tex]CO_2[/tex] gas.
The molar mass of propane is 44.1 g/mol.
One mole of any gas at STP occupies --- 22.4 L.
Hence, 44 g of propane forms (3x22.4 L=) 67.2 L of CO2 gas at STP.
Answer:
Thus, 67.2 L of CO2 is formed at STP.
tính ΔH° của phả ứng sau ở 200°C
CO+1÷2O=CO2
ΔH°
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
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
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.
list two uses of H2SO4
Please guys please answer this
Answer:
1. Because the rules will keep you safe it prevents you from getting hurt.
2.i) don't taste chemical
ii) Always wear protective gears
iii) be careful with tool
iv) wear protective gloves
Explanation:
help everyone get out quickly
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.
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.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 :)
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]
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.
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.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
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.
Radon-220 undergoes alpha decay with a half-life of 55.6 s.?
Assume there are 16,000 atoms present initially and calculate how many atoms will be present at 0 s, 55.6 s, 111.2 s, 166.8 s, 222.4 s, and 278.0 s (all multiples of the half-life). Express your answers as integers separated by commas.
Calculate how many atoms are present at 50 s, 100 s, and 200 s (not multiples of the half-life).
The half life of a radioactive isotope refers to the time taken for half of the number of original number of atoms present in the sample to decay.
The equation below gives the number of atoms present at time t
[tex]N=Noe^-kt[/tex]
N = Number of atoms present at time t
No = Number of atoms initially present
k = decay constant
t = time taken
Given that;
t1/2 = 0.693/k
where t1/2 = half life
k = 0.693/t1/2
k = 0.693/ 55.6 s
k = 0.0125 s-1
Substituting values;
N = 16,000 e^-0.0125(0)
N = 16,000 atoms
At 50 s
N = 16,000 e^-0.0125(50)
= 8564 atoms
At 100 s
N = 16,000 e^-0.0125(100)
= 4584 atoms
At 200 s
N = 16,000 e^-0.0125(200)
= 1313 atoms
https://brainly.com/question/2998270
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
Kc is 1.67 × 1020 at 25°C for the formation of iron(III) oxalate complex ion: Fe3+(aq) + 3 C2O42-(aq) ⇌ [Fe(C2O4)3]3-(aq). If 0.0800 M Fe3+ is initially mixed with 1.00 M oxalate ion, what is the concentration of Fe3+ ion at equilibrium?
Answer:
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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
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
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
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.
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.
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:
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.
When hydrogen gas reacts with oxygen gas, water vapour is formed according to the
reaction 2H2 + O2 2H2O. If 3.00 mol of hydrogen gas react with 3.00 mol
of oxygen gas, which reactant will be the reactant in excess?
Explanation:
here's the answer to the question
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).
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
The mass of an empty flask plus stopper is 64.232g. When the flask is completely filled with water the new mass is 153.617 g. The flask is emptied and dried, and a piece of metal is added. The mass of the flask, stopper and metal is 143.557. Next, water is added to the flask containing the metal and the mass is found to be 226.196. What is the density of the metal
Answer:
11.76 g/cm^3
Explanation:
Mass of empty flask and stopper = 64.232g
Mass of flask filled with water = 153.617 g
Mass of water = 153.617 g - 64.232g = 89.385 g
Mass of flask, stopper and metal = 143.557 g
Mass of metal = 143.557 g - 64.232g = 79.325 g
Mass of water, flask, stopper and metal = 226.196 g
Mass of water = 226.196 g - 143.557 g = 82.639 g
Since mass of water =volume of water
Volume occupied by metal = 89.385 cm^3 - 82.639 cm^3 = 6.746 cm^3
Density of metal = mass/volume = 79.325 g/6.746 cm^3
= 11.76 g/cm^3
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
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.