Answer:
1m = 100cm
3* 1m= 3*100cm = 300cm
Hope it helps
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Organic compounds undergo a variety of different reactions, including substitution, addition, elimination, and rearrangement. An atom or a group of atoms in a molecule is replaced by another atom or a group of atoms in a substitution reaction. In an addition reaction, two molecules combine to yield a single molecule. Addition reactions occur at double or triple bonds. An elimination reaction can be thought of as the reverse of an addition reaction. It involves the removal of two atoms or groups from a molecule. A rearrangement reaction occurs when bonds in the molecule are broken and new bonds are formed, converting it to its isomer. Classify the following characteristics of the organic reactions according to the type of organic reaction.
a. Reactions involving the replacement of one atom or group of atoms.
b. Reactions involving removal of two atoms or groups from a molecule.
c. Products show increased bond order between two adjacent atoms.
d. Reactant requires presence of a π bond.
e. Product is the structural isomer of the reactant.
1. Substitution reaction
2. Addition reaction
3. Elimination reaction
4. Rearrangement reaction
Answer:
Reactions involving the replacement of one atom or group of atoms. - Substitution reaction
Reactions involving removal of two atoms or groups from a molecule - Elimination reaction
Products show increased bond order between two adjacent atoms - Elimination reaction
Reactant requires presence of a π bond - Addition reaction
Product is the structural isomer of the reactant - Rearrangement reaction
Explanation:
When an atom or a group of atoms is replaced by another in a reaction, then such is a substitution reaction. A typical example is the halogenation of alkanes.
A reaction involving the removal of two atoms or groups from a molecule resulting in increased bond order of products is called an elimination reaction. A typical example of such is dehydrohalogenation of alkyl halides.
Any reaction that involves a pi bond is an addition reaction because a molecule is added across the pi bond. A typical example is hydrogenation of alkenes.
Rearrangement reactions yield isomers of a molecule. Rearrangement may involve alkyl or hydride shifts in molecules.
Reactions involving the replacement of one atom or group of atoms is substitution reaction, reactions involving removal of two atoms or groups from a molecule and products show increased bond order between two adjacent atoms is elimination reaction, reactant requires presence of a π bond in addition reaction and product is the structural isomer of the reactant is rearrangement reaction.
What is chemical reaction?Chemical reactions are those reactions in which reactants undergoes through a variety of changes for the formation of new product.
Substitution reaction: In this reaction any atom or molecule of reactant is replaced by any outside atom or molecule.Addition reaction: In this reaction addition of any reagent takes place across the double or triple bond of any reactant for the formation of product.Elimination reaction: In this reaction any molecule or two atoms will eliminate from the reactant as a result of which we get a bond order increased product.Rearrangement reaction: In this reaction atoms or bonds of a reactant get rearranged for the formation of new product.Hence, classification of above points are done according to their characteristics.
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g Phosphorus -32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The half-life of phosphorus-32 is 14.3 days. What mass of phosphorus-32 is left from an original sample of 175 mg of Na332PO4 after 35.0 days
Answer:
6.88 mg
Explanation:
Step 1: Calculate the mass of ³²P in 175 mg of Na₃³²PO₄
The mass ratio of Na₃³²PO₄ to ³²P is 148.91:31.97.
175 mg g Na₃³²PO₄ × 31.97 g ³²P/148.91 g Na₃³²PO₄ = 37.6 mg ³²P
Step 2: Calculate the rate constant for the decay of ³²P
The half-life (t1/2) is 14.3 days. We can calculate k using the following expression.
k = ln2/ t1/2 = ln2 / 14.3 d = 0.0485 d⁻¹
Step 3: Calculate the amount of P, given the initial amount (P₀) is 37.6 mg and the time elapsed (t) is 35.0 days
For first-order kinetics, we will use the following expression.
ln P = ln P₀ - k × t
ln P = ln 37.6 mg - 0.0485 d⁻¹ × 35.0 d
P = 6.88 mg
Convert 400.0 ng/dL to cg/L
Answer:
.04 cg/L
Explanation:
Which diagram shows the correct direction of electron flow in an electrolytic cell?
1)
2)
3)
4)
Sorry I couldn't put the diagram in, Thanks.
Answer:
A
Explanation:
What are the lengths of the diagonals of the kite?
The answer ( 13 and 8 )
x²=5²+12²
x²=25+144
x²=169
x=13
x²=5²+6²
x²=25+36
x²=61
x=7.8
x=8
At a given temperature, K = 1.3x10^-2 for the reaction:
N2(g) + 3H2(g) ⇌ 2NH3(g)
Calculate values of K for the following reactions at this temperature.
a. 1/2N2 + 3/2H2(g) ⇌ NH3(g)
b. 2NH3(g) ⇌ N2(g) + 3H2(g)
c. NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
d. 2N2(g) + 6H2(g) ⇌ 4NH3(g)
Answer:
a) 0.11
b)76.9
c) 8.8
d) 1.7*10^-4
Explanation:
Step 1: Data given
K = 1.3 * 10^-2 for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g)
Step 2: Formula of K
aA(g) + bB(g) ⇌ cC(g) + dD(g)
K = [C]^c *[D]^d / [A]^a * [B]^b
K = 1.3 * 10^-2 = [NH3]² / [H2]³*[N2]
Step 3:
a) 1/2N2 + 3/2H2(g) ⇌ NH3(g)
N2(g) + 3H2(g) ⇌ 2NH3
1/2N2 + 3/2H2(g) ⇌ NH3(g) =>K' = [tex]\sqrt{K}[/tex]
K' = [tex]\sqrt{1.3*10^-2}[/tex] = 0.11
b. 2NH3(g) ⇌ N2(g) + 3H2(g)
N2(g) + 3H2(g) ⇌ 2NH3
2NH3(g) ⇌ N2(g) + 3H2(g) =>K' = 1/K
K' = 1/(1.3*10^-2) = 76.9
c. NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
N2(g) + 3H2(g) ⇌ 2NH3
NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
=>K' = [tex]\frac{1}{\sqrt{K} }[/tex]
K' = [tex]\frac{1}{\sqrt{1.3*10^-2} }[/tex]
K' = 8.8
d. 2N2(g) + 6H2(g) ⇌ 4NH3(g)
N2(g) + 3H2(g) ⇌ 2NH3
2N2(g) + 6H2(g) ⇌ 4NH3(g)
K' = K²
K' = (1.3*10^-2)²
K' = 1.7 *10 ^-4
Values of equilibrium constant at given temperature for the following reactions are 0.11, 76.9, 8.8 and 1.7 × 10⁻⁴ respectively.
How we calculate equilibrium constant?Equilibrium constant is define as the ration of the concentrations of product to the concentrations of reactant with respect to the exponent of their coefficients.
Given chemical reaction is:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Equilibrium constant for this reaction is:
K = [NH₃]² / [N₂][H₂]³
K = 1.3 × 10⁻² (given)
Equilibrium constant K₁ for below reaction will be written as:1/2N₂(g) + 3/2H₂(g) ⇌ NH₃(g)
K₁ = √K
Because concentration of all given species is 1/2 of the given reaction, so value of K₁ will be written as:
K₁ = √(1.3 × 10⁻²) = 0.11
2NH₃(g) ⇌ N₂(g) + 3H₂(g)
K₂ = 1/K
Because concentration of reactant and products are reciprocal from the concentration of original given reaction, so value of K₂ will be written as:
K₂ = 1/1.3 × 10⁻² = 76.9
NH₃(g) ⇌ 1/2N₂(g) + 3/2H₂(g)
K₃ = 1/√K
Because concentrations of given species is reciprocal as well as half of the given original reaction, so value of K₃ will be written as:
K₃ = 1/√(1.3 × 10⁻²) = 8.8
2N₂(g) + 6H₂(g) ⇌ 4NH₃(g)
K₄ = K²
Because concentrations of given species is double of the given original reaction, so value of K₄ will be written as:
K₄ = (1.3 × 10⁻²)² = 1.7 × 10⁻⁴
Hence, the value of K for given reactions are 0.11, 76.9, 8.8 and 1.7 × 10⁻⁴ respectively.
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What is the standard enthalpy change for the decomposition of one mole of SO3?
Answer:
see explanation
Explanation:
SO₃(g) + 395.77 Kj/mole => S°(s) + 3/2O₂(g)
The standard heat of formation for SO₃(g) is given by the following rxn:
S°(s) + 3/2O₂(g) => SO₃(g) + 395.77 kJ/mole. Reversing this reaction is the decomposition of SO₃(g) into its basic elements in their standard state (25°C, 1atm) and is endothermic with +295.77Kj/mole.
A 11.79 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 14.151 g . Add subscripts to correctly identify the empirical formula of the new oxide.
Answer:
MoO₃
Explanation:
To solve this question we must find the moles of molybdenum in Mo2O3. The moles of Mo remain constant in the new oxide. With the differences in masses we can find the mass of oxygen and its moles obtaining the empirical formula as follows:
Moles Mo2O3 -Molar mass: 239,878g/mol-
11.79g * (1mol / 239.878g) = 0.04915 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.09830 moles Mo
Mass Mo in the oxides:
0.09830 moles Mo * (95.95g/mol) = 9.432g Mo
Mass oxygen in the new oxide:
14.151g - 9.432g = 4.719g oxygen
Moles Oxygen:
4.719g oxygen * (1mol/16g) = 0.2949 moles O
The ratio of moles of O/Mo:
0.2949molO / 0.09830mol Mo = 3
That means there are 3 moles of oxygen per mole of Molybdenum and the empirical formula is:
MoO₃To standardize a hydrochloric acid solution, it was used as a titrant with a solid sample of sodium hydrogen carbonate, NaHCO3. The solid sample had a mass of 0.3967g, and 41.77 mL of acid was required to reach the equivalence point. Calculate the concentration of the standard solution.
Answer:
0.113 M
Explanation:
The reaction that takes place is:
NaHCO₃ + HCl →NaCl + CO₂ + H₂OFirst we convert 0.3967 g of NaHCO₃ into moles, using its molar mass:
0.3967 g ÷ 84 g/mol = 4.72x10⁻³ mol NaHCO₃As 1 mol of NaHCO₃ reacts with 1 mol of HCl, in 41.77 mL of the HCl solution there were 4.72x10⁻³ moles of HCl.
With the calculated number of moles and the given volume we calculate the concentration of the solution:
Converting 41.77 mL ⇒ 41.77 mL / 1000 = 0.04177 LConcentration = 4.72x10⁻³ mol / 0.04177 L = 0.113 MA company manufacturing KMnO4 wants to obtain the highest yield possible. Two of their research scientists are working on a technique to increase the yield.
Both scientists started with 50.0 g of manganese oxide.
What is the theoretical yield of potassium permanganate when starting with 50.0 g MnO2?
The equation for the production of potassium permanganate is as follows:
2 MnO2 + 4 KOH + O2 → 2 KMnO4 + 2 KOH + H2
Answer:
The theoretical yield potassium permanganate, KMnO₄ when starting with 50.0 g MnO₂ is 90.8 g
Explanation:
Molar mass of MnO₂ = (55 + 2 × 16) = 87.0 g/mol
Molar mass of KMnO₄ = (39 + 55 + 4 × 16) = 158 g/mol
Moles of MnO₂ in 50 g = reacting mass / molar mass
where reacting mass = 50 g
Moles of MnO₂ in 50 g = 50 g /87 g/mol = 0.575 moles
The equation for the production of potassium permanganate is as follows:
2 MnO2 + 4 KOH + O2 → 2 KMnO4 + 2 KOH + H2
From the equation of the reaction above, 2 moles of MnO₂ produces 2 moles of KmNO₄. The mole ratio of MnO₂ to KMnO₄ is 1 : 1
Therefore, 0.575 moles of MnO₂ will produce theoretically 0.575 moles of KMnO₄
Mass of 0.575 moles of KMnO₄ = number of moles × molar mass
Mass of 0.575 moles of KMnO₄ = 0.575 moles × 158 g/mol = 90.8 g of KMnO₄
Therefore, the theoretical yield potassium permanganate when starting with 50.0 g MnO₂ is 90.8 g
Which of the following is the most plausible explanation for the fact that the saponification of the triacylglycerol in the passage resulted in four different fatty acid salts?
a. The triacylglycerol molecule consisted of four different fatty acid units.
b. Glycerol was transformed into a fatty acid salt under the reaction conditions.
c. One of the fatty acid salts was unsaturated, and it completely isomerized under the reaction conditions.
d. One of the fatty acid salts was unsaturated, and a small percentage isomerized under the reaction conditions.
Answer: The correct option is C (One of the fatty acid salts was unsaturated, and it completely isomerized under the reaction conditions).
Explanation:
Fats and oils belongs to a general group of compounds known as lipids. Fatty acids are weak acid and are divided into two:
--> Saturated fatty acids: These have NO double bonds in their hydrocarbon chain, and
--> Unsaturated fatty acids: These have one or more double bonds in their hydrocarbon chain.
SAPONIFICATION is defined as the process by which fats and oil is hydrolyzed with caustic alkali to yield propane-1,2,3-triol and the corresponding sodium salt of the component fatty acids. During this process, One hydroxide ion is required to hydrolyze one ester linkage of a triacylglycerol molecule. Because there are three ester linkages in a triacylglycerol, three equivalents of sodium hydroxide will be needed to completely saponify the triacylglycerol. This explains the reason why saponification of the triacylglycerol iresulted in four different fatty acid salts.
Hydrofluoric acid is used in the preparation of numerous pharmaceuticals (e.g., Prozac) and industrial materials (e.g., Teflon). It can be produced by the reaction of hydrogen and fluorine gases. Starting with initial concentrations of 1.69 M for H2 and 1.69 M for F2, what would be the equilibrium concentration of HF? The equilibrium constant for the reaction Kc = 115.
a. 2.85 M.
b. 4.00 M.
c. 0.85 M.
d. 3.37 M.
e. 1.69 M.
Answer:
the equilibrium concentration of HF is 2.85 M
Option a) 2.85 M is the correct answer.
Explanation:
Given the data in the question;
H₂ + F₂ ⇄ 2HF
I 1.69 M 1.69 M 0
C -x -x +2x
E 1.69-x 1.69-x +2x
given that Kc = 115
Kc = [ HF ]² / [H₂][F₂]
we substitute
115 = [ 2x ]² / [ 1.69-x ][ 1.69-x ]
lets find the square root of both sides
10.7238 = 2x / [ 1.69-x ]
10.7238[ 1.69-x ] = 2x
18.123222 - 10.7238x = 2x
2x + 10.7238x = 18.123222
12.7238x = 18.123222
x = 18.123222 / 12.7238
x = 1.424356
Hence, equilibrium concentration of HF = 2x
that is;
HF = 2 × 1.424356
HF = 2.8487 ≈ 2.85 M
Therefore, the equilibrium concentration of HF is 2.85 M
Option a) 2.85 M is the correct answer.
A scientist collects a sample that has 2.00 × 1014 molecules of carbon dioxide gas.How many grams is this, given that the molar mass of CO2 is 44.01 g/mol?
Answer:
1.46 × 10⁻⁸ g
Explanation:
Step 1: Given data
Molecules of CO₂: 2.00 × 10¹⁴ molecules
Step 2: Convert molecules to moles
We need a conversion factor: Avogadro's number. There are 6.02 × 10²³ molecules in 1 mole of molecules.
2.00 × 10¹⁴ molecules × 1 mol/6.02 × 10²³ = 3.32 × 10⁻¹⁰ mol
Step 3: Convert moles to mass
We need a conversion factor: the molar mass. The molar mass of CO₂is 44.01 g/mol.
3.32 × 10⁻¹⁰ mol × 44.01 g/mol = 1.46 × 10⁻⁸ g
A metal X from two oxide A and B .3.oogm of A and B contain 0.72 and 1.16g of oxygen respectively.calculate the maases of metal in gm which combine with one gram of oxygen in each case
Answer:
Explanation:
Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.
Metal X can form 2 oxides (A and B).
A + B = 3g
The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.
The mass of metal X in the two oxides will be the same because it's the same metal.
Thus, we represent the mass of the metal in the two oxides as 2X.
2X + 0.72 + 1.16 = 3
2X + 1.88 = 3
2X = 3 - 1.88
2X = 1.12
X = 0.56
Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.
Thus, mass of metal (X) in 1g of oxygen in A is
0.56g ⇒ 0.72g
X ⇒ 1
X = 1 × 0.56/0.72
X = 0.78 g
Hence, 0.78g of the metal will combine with 1g of oxygen for A
Also, mass of metal (X) in 1g of oxygen in B is
0.56g ⇒ 1.16g
X ⇒ 1g
X = 1×0.56/1.16
X = 0.48 g
Thus, 0.48g of the metal will combine with 1g of oxygen for B
At 50.0 oC, a reinforced tank contains 675.5 grams of gaseous argon and 465.0 g of gaseous molecular chlorine with a total pressure of 4.00 atm. Calculate the following:
a. How many moles of Ar are in the tank?
b. How many moles of Cl, are in the tank?
c. Total moles of gas in the tank.
d. The mole fraction of Ar.
e. The mole fraction of Cl2.
f. The Partial Pressure of Ar.
g. The Partial Pressure of Cl2.
Answer:
For (a): The moles of Ar is 16.94 moles
For (b): The moles of [tex]Cl_2[/tex] is 16.94 moles
For (c): The total number of moles in a tank is 23.47 moles
For (d): The mole fraction of Ar is 0.722
For (e): The mole fraction of [tex]Cl_2[/tex] is 0.278
For (f): The partial pressure of Ar is 2.888 atm
For (g): The partial pressure of [tex]Cl_2[/tex] is 1.112 atm
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
For (a):Given mass of Ar = 675.5 g
Molar mass of Ar = 39.95 g/mol
Plugging values in equation 1:
[tex]\text{Moles of Ar}=\frac{675.5g}{39.95g/mol}=16.91 mol[/tex]
For (b):Given mass of [tex]Cl_2[/tex] = 465.0 g
Molar mass of [tex]Cl_2[/tex] = 70.9 g/mol
Plugging values in equation 1:
[tex]\text{Moles of }Cl_2=\frac{465.0g}{70.9g/mol}=6.56 mol[/tex]
For (c):Total moles of gas in the tank = [16.91 + 6.56] mol = 23.47 mol
Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:
[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex] .....(2)
where n is the number of moles
For (d):Moles of Ar = 16.94 moles
Total moles of gas in the tank = 23.47 mol
Putting values in equation 2, we get:
[tex]\chi_{Ar}=\frac{16.94}{23.47}\\\\\chi_{Ar}=0.722[/tex]
For (e):Total mole fraction of the system is always 1
Mole fraction of [tex]Cl_2[/tex] = [1 - 0.722] = 0.278
Raoult's law is the law used to calculate the partial pressure of the individual gases present in the mixture.
The equation for Raoult's law follows:
[tex]p_A=\chi_A\times p_T[/tex] .....(3)
where [tex]p_A[/tex] is the partial pressure of component A in the mixture and [tex]p_T[/tex] is the total partial pressure of the mixture
For (f):We are given:
[tex]\chi_{Ar}=0.722\\p_T=4.00atm[/tex]
Putting values in equation 3, we get:
[tex]p_{Ar}=0.722\times 4.00atm\\\\p_{Ar}=2.888atm[/tex]
For (g):We are given:
[tex]\chi_{Cl_2}=0.278\\p_T=4.00atm[/tex]
Putting values in equation 3, we get:
[tex]p_{Cl_2}=0.278\times 4.00atm\\\\p_{Cl_2}=1.112atm[/tex]
Liquid nitrogen becomes a gas when it is poured out of its container. The nitrogen is
Answer:
compressable
Explanation:
as liquid nitrogen came out from container the force exerted on it which changes it into liquid ends making it gas
liquid nitrogen boils at 77 K or -196° C
You have a sample of gold that contains 0.2 moles of gold (Au). How many gold atoms are present in the sample. HINT: Gold atoms represents the number of particles."
1.20 x 10^23 atoms
3.3 x 10^-25 atoms
1.01 x 10^-3 moles
BEST ANSWER IS
have a great summer
1.01 x 10^-3 moles
n = 0.20 mol
Required:N
Solution:N = n × 6.02 × 10²³ atoms/mol
N = 0.2 mol × 6.02 × 10²³ atoms/mol
N = 1.20 × 10²³ atoms
Therefore, there are 1.20 × 10²³ gold atoms in 0.2 mol of a gold sample.
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A beaker contains a 25 mL solution of an unknown monoprotic acid that reacts in a 1:1 stochiometric ratio with NaOH. Titrate the solution with NaOH to determine the concentration of the acid.Perform a titration by setting the concentration of the NaOH solution and adding it to the acid solution using the different Add Base buttons.The equivalence point of the titration is passed when the solution color changes.The unknown sample can be titrated multiple times by pressing the Retitrate button and starting over.Enter the concentration of the unknown acid solution.The base is 20.05 mL with 1.000 M
Answer:
0.80 M
Explanation:
Step 1: Write the generic neutralization reaction
HA + NaOH ⇒ NaA + H₂O
Step 2: Calculate the reacting moles of NaOH
20.05 mL of 1.000 M NaOH react.
0.02005 L × 1.000 mol/L = 0.02005 mol
Step 3: Calculate the reacting moles of HA
The molar ratio of NaOH to HA is 1:1. The reacting moles of HA is 1/1 × 0.02005 mol = 0.02005 mol.
Step 4: Calculate the concentration of HA
0.02005 moles of HA are in 25 mL.
[HA] = 0.02005 mol/0.025 L = 0.80 M
study the reaction given below in which excess magnesium ribbon (Mg)reacts with 50cm of a diluted sulphuric acid solution at room temperature
Questions
what Changes can be made to the following substance to increase the rate of reaction?
5.1.1 Magnesium
5.1.2 Sulphuric acid
Answer:
Magnesium reacts with dilute hydrochloric acid in a conical flask which is ... One student can add the magnesium ribbon to the acid and stopper the flask, ... 50 cm3 of 1M hydrochloric acid is a six-fold excess of acid.
The value of keq for the following reaction is 0.25
SO2(g) + NO2(g) _ SO3(g) + NO(g)
What is the value of at the same temperature if we multiply the reaction by 2
This question is concerned with the following oxides
• Sulfur dioxide
• Carbon monoxide
• Lithium oxide
• Aluminum (III) oxide
Which of the above oxides will not react with hydrochloric acid but will react with aqueous
sodium hydroxide?
Answer:
hi I used your code you got it
A student swings back and forth from position A to C, as shown.
Which of the following happens when the swing moves from Position C to Position B?
A. Both potential energy and kinetic energy of the student increase.
B. Both potential energy and kinetic energy of the student decrease.
C. Potential energy of the student decreases and kinetic energy of the student increases.
D. Kinetic energy of the student decreases and potential energy of the student increases.
Answer:
C
Explanation:
KE at B is max and PE is 0
KE at C is 0 and PE is max
so when student swings from C to B
its KE increases
and PE decreases
Anyone knows this? I don’t know this
QUESTION :WHICH OF THE FOLLOWING IS AN EXAMPLE OF A CONTROLLED EXPERIMENT TO TEST THIS?
ANSWER:
D. The temperatures of five breakers of 250 mL of water are varied, and 10 g of sugar is added to each breaker.
Groups on the periodic table also correspond with the number of ?
The question is incomplete, the complete question is;
Groups of the periodic table correspond to elements with a. the same color b. the same atomic number c. similar chemical properties d. similar numbers of neutrons
Answer:
similar chemical properties
Explanation:
In the periodic classification of elements, elements are divided into groups and periods. Elements in the same group of the periodic table have the same number of outermost electrons and share very similar chemical properties.
Elements in the same period have the same number of shells and the same maximum energy level of the outermost electron. Chemical properties carry markedly across a period.
The theoretical yield of zinc oxide in a reaction is 486 g. What is the percent
yield if 399 g is produced?
O A. 122%
O B. 4.93%
C. 82.1%
D. 29.6%
Answer:
the correct answer is c
Explanation:
becuase i had the same question
A certain first-order reaction is 45.0% complete in 65 s. What are the values of the rate constant and the half-life for this process
Answer:
0.01228s⁻¹ = rate constant
Half-life = 56.4s
Explanation:
The first order reaction follows the equation:
ln[A] = -kt + ln[A]₀
Where [A] is amount of reactant after time t = 45.0%, k is rate constante and [A]₀ initial amount of reactant = 100%
ln[45%] = -k*65s + ln[100%]
-0.7985 = -k*65s
0.01228s⁻¹ = rate constant
Half-life is:
Half-life = ln2 / k
Half-life = 56.4s
Ammonium sulfate (NH4)2SO4 is made by reacting 25.0 L of 3.0 mol/L H2SO4 with 3.1× 103 L of NH3 at a pressure of 0.68 atm and a temperature of 298 K according to the following reaction .
NH3(g) + H2SO4(aq) → (NH4)2SO4 (aq)
How many grams of ammonium sulfate are produced?
Answer: The mass of [tex](NH_4)_2SO_4[/tex] produced is 9910.5 g
Explanation:
For [tex]H_2SO_4[/tex]:Molarity is calculated by using the equation:
[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex] ......(1)
Molarity of [tex]H_2SO_4[/tex] = 3.0 M
Volume of solution = 25.0 L
Putting values in equation 1, we get:
[tex]\text{Moles of }H_2SO_4=(3.0mol/L\times 25.0L)=75mol[/tex]
For [tex]NH_3[/tex]:The ideal gas equation is given as:
[tex]PV=nRT[/tex] .......(2)
where,
P = pressure of the gas = 0.68 atm
V = volume of gas = [tex]3.1\times 10^3L[/tex]
n = number of moles of gas = ? moles
R = Gas constant = 0.0821 L.atm/mol.K
T = temperature of the gas = 298 K
Putting values in equation 2, we get:
[tex]0.68atm\times 3.1\times 10^3L=n\times 0.0821L.atm/mol.K\times 298K\\\\n=\frac{0.68\times 3.1\times 10^3}{0.0821\times 298}=86.16mol[/tex]
For the given chemical equation:
[tex]NH_3(g)+H_2SO_4(aq)\rightarrow (NH_4)_2SO_4(aq)[/tex]
By stoichiometry of the reaction:
If 1 mole of [tex]H_2SO_4[/tex] reacts with 1 mole of [tex]NH_3[/tex]
So, 75 moles of [tex]H_2SO_4[/tex] will react with = [tex]\frac{1}{1}\times 75=75mol[/tex] of [tex]NH_3[/tex]
As the given amount of [tex]NH_3[/tex] is more than the required amount. Thus, it is present in excess and is considered as an excess reagent
Thus, [tex]H_2SO_4[/tex] is considered a limiting reagent because it limits the formation of the product.
By the stoichiometry of the reaction:
If 1 mole of [tex]H_2SO_4[/tex] produces 1 mole of [tex](NH_4)_2SO_4[/tex]
So, 75 moles of [tex]H_2SO_4[/tex] will produce = [tex]\frac{1}{1}\times 75=75mol[/tex] of [tex](NH_4)_2SO_4[/tex]
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We know, molar mass of [tex](NH_4)_2SO_4[/tex] = 132.14 g/mol
Putting values in above equation, we get:
[tex]\text{Mass of }(NH_4)_2SO_4=(75mol\times 132.14g/mol)=9910.5g[/tex]
Hence, the mass of [tex](NH_4)_2SO_4[/tex] produced is 9910.5 g
what is the characteristics of tropical air mass
Answer:
Explanation:
Continental tropical air masses are extremely hot and dry. Arctic, Antarctic, and polar air masses are cold. The qualities of arctic air are developed over ice and snow-covered ground. Arctic air is deeply cold, colder than polar air masses.
HELP ME PLZ AND THANKS I WILL MARK YOU AS BRAINLIEST!!!
Answer:
See explanation.
Explanation:
Hello there!
In this case, since this problem is about gas laws, more specifically about the Gay-Lussac's one since the volume is said to be constant, we can use the following equation for its solution for the final pressure, P2:
[tex]\frac{P_2}{T_2} = \frac{P_1}{T_1}[/tex]
[tex]P_2= \frac{P_1T_2}{T_1}\\\\P_2 =\frac{12.0atm*450K}{300K}\\\\P_2= 18.0atm[/tex]
Thus, we fill in the table as follows:
Initial Final
Pressure 12.0 atm 18.0 atm
Volume 4.0 L 4.0 L
Temperature 300K 450K
Regards!
The main product of free radical bromination of methane is
A) ethane
B) chloromethane
C) bromonethane
D) bromine
Answer: C
Explanation: