Answer:
.032 M .
Explanation:
Molecular weight of ammonium nitrate is 80 .
3.2 g = 3.2 / 80 moles
= .04 moles
volume = 100 mL = 0.1 L
Molarity of 100 mL solution = .04 moles / 0.1 L
= 0.4 M solution.
Now 2 mL solution of 0.4 M is diluted to a volume of 25 mL .
Using the formula S₁ V₁ = S₂V₂
0.4 M x 2 mL = S₂ x 25 mL
S₂ = .4 x 2 / 25
= .032 M
Hence required concentration is .032 M .
In practice, the second law of thermodynamics means that:
a. Systems move from ordered behavior to more random behavior.
b. Systems move from random behavior to more ordered behavior.
c. Systems move between ordered and random behavior patterns based on temperature.
d. Systems are constantly striving to reach equilibrium.
Answer:
Systems move from ordered behavior to more random behavior.
Explanation:
Entropy refers to the degree of disorderliness in a system. The second law of thermodynamics can be restated in terms of entropy as follows; “any spontaneous process in any isolated system always results in an increase in the entropy of that system.''(science direct)
According to this law, systems tend towards a more disorderly behaviour (increase in entropy) hence the answer given above.
A nuclease enzyme breaks the covalent bond originally connecting the phosphate to the 5' carbon in a nucleic acid. After allowing this enzyme to completely digest the nucleic acid down to monomers, you perform tests to determine where the phosphate is attached to each monomer. Where do you expect to find this phosphate
Answer:
The phosphate will remain attached to the 5' carbon of the deoxy or the ribose sugar in the nucleic acid monomers.
Explanation:
The structure of nucleic acid polymers is built up from monomers of nucleotides.
A nucleotide consists of a sugar backbone which is either a ribose or deoxyribose sugar, a nitogenous base which is either a purine or pyrimidine, and a phosphate group. The nitrogenous base is attached to the carbon number 1 or C-1 of the sugar backbone by a covalent bond. The phosphate group on the other hand is covalently attached to the carbon number 5 or 5' carbon of the sugar backbone.
When polymers of nucleic acids are formed, the phosphate at the 5' carbon of the sugar backbone is covalently linked in a phosphodiester bond to the 3' carbon of the sugar backbone in another nucleotide molecule, thus extending the strands of the nucleic acid molecule.
Nucleases are enzymes that break down the phosphodiseter bonds in nucleic acids resulting in nucleotide monomers. After complete digestion ofmthe nucleic acid polymer by nucleases, the phosphate will remain attached to the 5' carbon of the deoxy or the ribose sugar in the nucleic acid monomers.
The freezing point of a substance is -20°C. Its boiling point is 120°C.
a. At 80°C the substance is in the state
b. At -50°C the substance is in the state.
C. At 140°C the substance is in the state.
Answer:
a. liquid
b. solid
c. gas, (should be at it's boiling point)
Explanation: If the normal melting point of a substance is below room temperature, the substance is a liquid at room temperature. Benzene melts at 6°C and boils at 80°C; it is a liquid at room temperature. If both the normal melting point and the normal boiling point are above room temperature, the substance is a solid.
if you need an explanation to each lmk
Consider the reaction below. How much heat is absorbed if 5.00 moles of nitrogen react
with excess oxygen?
2 N2 (8) + O2(g) → 2 N20 (8) AHrxn- +163.2 kJ
Explanation:
The given chemical reaction is:
[tex]2 N_2 (g) + O_2(g) -> 2 N_20 (g) delta Hrxn= +163.2 kJ[/tex]
When two moles of nitrogen reacts with oxygen, it requires 163.2kJ of energy.
When 5.00 mol of nitrogen requires how much energy?
[tex]5.00 mol x \frac{163.2 kJ }{2 mol} \\=408 kJ[/tex]
Hence, the answer is 408 kJ of heat energy is required.
Consider the reaction: A(aq) + 2B (aq) === C (aq). Initially 1.00 mol A and 1.80 mol B
were placed in a 5.00-liter container. The mole of B at equilibrium was determined to
be 1.00 mol. Calculate K value.
0.060
5.1
25
17
Ugh
Answer:
17
Explanation:
Step 1: Calculate the needed concentrations
[A]i = 1.00 mol/5.00 L = 0.200 M
[B]i = 1.80 mol/5.00 L = 0.360 M
[B]e = 1.00 mol/5.00 L = 0.200 M
Step 2: Make an ICE chart
A(aq) + 2 B(aq) ⇄ C(aq)
I 0.200 0.360 0
C -x -2x +x
E 0.200-x 0.360-2x x
Then,
[B]e = 0.360-2x = 0.200
x = 0.0800
The concentrations at equilibrium are:
[A]e = 0.200-0.0800 = 0.120 M
[B]e = 0.200 M
[C]e = 0.0800 M
Step 3: Calculate the concentration equilibrium constant (K)
K = [C] / [A] × [B]²
K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17
Forcus on the yellow highlighted texts, your help is appreciated.
[tex]{ \sf{ \red{no \: pranks}}}[/tex]
Answer:
Transition temperature is the temperature at which a substance changes from one state to another.
Allotropy is the existence of an element in many forms.
How many protons does Tin have?
A. 50
B. 68
C. 118
Hello There!
Tin has 50 protons.Hope that helps you!
~Just a felicitous girlie
#HaveASplendidDay
[tex]SilentNature[/tex]
Write balanced equations for the reaction of each of the following carboxylic acids with NaOH. Part A formic acid Express your answer as a chemical equation. A chemical reaction does not occur for this question. Request Answer Part B 3-chloropropanoic acid Express your answer as a chemical equation. nothing A chemical reaction does not occur for this question.
Answer:
Part A
HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)
Part B
ClCH2CH2CO2H(aq) + NaOH(aq) ------> ClCH2CH2CO2Na(aq) + H2O(l)
Explanation:
The reaction between an alkanoic acid and a base is a neutralization reaction. The reaction occurs as follows;
RCOOH + NaOH ----> RCOONa + H2O
We have to note the fact that the net ionic reaction still remains;
H^+(aq) + OH^-(aq) ---> H2O(l)
In both cases, the reaction can occur and they actually do occur as written.
A sample of gas is held at constant volume. If the number of moles of this sample of gas is doubled and the pressure of this sample of gas is halved, what happens to the absolute temperature of the gas?
Select one
a. The absolute temperature is doubled.
b. The absolute temperature is halved.
c. The absolute temperature is quadrupled.
d. The absolute temperature is quartered.
e. The absolute temperature stays the same.
Answer:
number of moles of gas increases the volume also increases.
HELP ASAS 15 POINTS
When using the process of evaporation to separate a mixture, what is left behind in the evaporating dish?
A. None of these.
B. The liquid evaporates and the solid is left in the dish.
C. The mixture does not separate, and the entire mixture evaporates.
D. The mixture does not separate, and the entire mixture remains in the dish.
Answer:
liquid will be evaporated while solid remains
does anyone know how to solve this and what the answer would be?
Dynamic equilibrium is showed at the point at which solid liquid and gas intersect.
At the point at which solid liquid and gas intersect represents a system that shows dynamic equilibrium. There is equal amount of reactants and products at the point of dynamic equilibrium because the transition of substances occur between the reactants and products at equal rates, means that there is no net change. Reactants and products are formed at the rate that no change occur in their concentration.
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which of the following is is a chemical property of pure water
Answer:
Pure water has an acidity of about 7 on the pH scale. -is a chemical property of pure water. Pure water has an acidity of about 7 on the pH scale
Answer: không màu , không mùi không vị
Explanation:
Determine the number of moles of aluminum in 2.154 x 10-1 kg of Al. Group of answer choices 5816 mol 7.984 mol 6.02 X 1023 mol 4.801 mol 8.783
Answer:
Avogadro's number is 1 mol = 6.02 * 10^23 elements
It means that 1 mol of atoms is 6.02 * 10^23 atoms
1 mol of atoms = 6.02 * 10^23 atoms
From there, if you divide both sides by 1 mol of atoms, you get
1 = 6.02 * 10^23 atoms / 1 mol of atoms.
That means, that to pass from a number of moles of atoms to number of atoms you have to multipby by the conversion factor
6.02*10^23 atoms Al/ 1 mol Al
That is the second option of the list.
Explanation:
6) Hydrogen gas can be generated from the reaction between aluminum metal and hydrochloric acid:
2 Al(s) + 6 HCl(aq) + 2 AICI3, (aq) + 3 H2(g)
a. Suppose that 3.00 grams of Al are mixed with excess acid. If the hydrogen gas produced is directly collected
into a 850 mL glass flask at 24.0 °C, what is the pressure inside the flask (in atm)?
b. This hydrogen gas is then completely transferred from the flask to a balloon. To what volume (in L) will the
balloon inflate under STP conditions?
c. Suppose the balloon is released and rises up to an altitude where the temperature is 11.2 °C and the pressure is
438 mm Hg. What is the new volume of the balloon (in L)?
Stoichiometry refers to the relationship between the moles of reactants and products.
This question must be solved using both stoichiometry and the gas laws
The reaction equation is;
2 Al(s) + 6 HCl(aq) --------> 2 AICI3, (aq) + 3 H2(g)
Using stoichiometryNumber of moles of Al = 3g/27g/mol = 0.11 moles
According to the reaction equation;
2 moles of Al yields 3 moles of H2
0.11 moles of Al yields 0.11 * 3/2 = 0.165 moles
Using the gas lawsFrom the ideal gas equation;
PV=nRT
P = ?
n= 0.165 moles
V = 0.85 L
T = 297 K
R = 0.082 atmLK-1mol-1
P= nRT/V
P = 0.165 * 0.082 * 297/0.85
P= 4.73 atm
Under STP conditions;P1 = 4.73 atm
T1 = 297 K
V1 = 0.85 L
P2 = 1 atm
T2 =273 K
V2 =?
From the general gas equation;P1V1/T1 = P2V2/T2
P1V1T2 = P2V2T1
V2 = P1V1T2/P2T1
V2 = 4.73 * 0.85 * 273/1 * 297
V2 = 3.69 L
P1 = 760 mmHg
T1 = 273 K
V1 = 3.69
P2 = 438 mm Hg
T2 = 284.2 K
V2 =?
P1V1/T1 = P2V2/T2
P1V1T2 = P2V2T1
V2 = P1V1T2/P2T1
V2 = 760 * 3.69 * 284.2/438 *273
V2 = 797010.48/119574
V2= 6.67 L
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a. You have a stock solution of 14.8 M NH3. How many milliliters of this solution should you dilute to make 1000.0 mL of 0.250 M NH3?
b. If you take a 10.0 mL portion of the stock solution and dilute it to a total volume of 0.500 L, what will be the concentration of the final solution?
Answer:A) V = 16.892 ml
Explanation:
M1 * V1 = M2 * V2
14.8 M * V1 =0.250 M * 1000 ml
V1 = 16.892 ml
a. The volume of 16.89 milliliters of the stock solution of 14.8 M should be diluted to make 1000.0 mL of 0.250 M.
b. The concentration of the final solution is 0.296 M.
What is the dilution law?The concentration or the volume of the concentrated or dilute solution can be calculated by using the equation:
M₁V₁ = M₂V₂
where M₁ and V₁ are the concentration and volume of the concentrated solution respectively and M₂ and V₂ are the concentration and volume of the dilute solution.
A stock solution is a solution that has a high concentration and that will be diluted to a low concentration by the addition of water in it.
Given, a stock solution of concentration, M₁ = 14.8 M
The concentration of the diluted solution, M₂ = 0.250 M
The volume of diluted solution, V₂ = 1000ml
Substitute the value of the molarity and volume in equation (1):
(14.8)× (V₁) = (1000) × (0.250)
V₁ = 16.89 ml
Similarly, for part (b): M₁ = 14.8 M, V₁ = 10 ml and V₂ = 0.5L = 500 ml
(14.8)× (10) = (500) × (M₂)
M₂ = 0.296 M
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What is Bose Einstein state of matter and their examples
Answer:
A BEC ( Bose - Einstein condensate ) is a state of matter of a dilute gas of bosons cooled to temperatures very close to absolute zero is called BEC.
Examples - Superconductors and superfluids are the two examples of BEC.
Explanation:
How many grams of H₂SO₄ are contained in 2.00 L of 6.0 M H₂SO₄?
Please explain and show work.
Answer:
1176 grams
Explanation:
nH2SO4 =2*6=12 mol
mH2SO4=12*98=1176 grams
Answer:
solution given:
molarity of H₂SO₄=6 M
volume=2L
no of mole =6M*2=12mole
we have
mass =mole* actual mass=12*98=1176g
the mass is 1176g.
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
Hãy cho biết giá trị và ý nghĩa của số lượng tử n, l, m, ms khi mô tả trạng thái của electron trong nguyên tử?
I need to know what is the median of the data
Answer:
The median is also the number that is halfway into the set. To find the median, the data should be arranged in order from least to greatest. If there is an even number of items in the data set, then the median is found by taking the mean (average) of the two middlemost numbers.
I hope it helps
Tech A says that hydrocarbons are a result of complete combustion. Tech B says that a catalytic converter creates a chemical reaction, changing carbon monoxide and hydrocarbons to water and carbon dioxide. Who is correct
Answer:
Neither Tech A nor B is correct
Explanation:
Combustion is a chemical reaction that occurs when a chemical molecule(s) interacts quickly with oxygen and produces heat.
When hydrocarbon undergoes a complete combustion reaction, they produce water and CO2.
Tech B is also incorrect because the main purpose of a catalytic converter is to accelerate and speed up the chemical reaction rates, Hence, they are not involved in chemical reaction formation. Catalytic converters are utilized as a control device in exhaust emission to lessen the effect of toxic gas fumes.
At what velocity (m/s) must a 20.0g object be moving in order to possess a kinetic energy of 1.00J
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.
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Draw a formula for Thr-Gly-Ala (T-G-A) in its predominant ionic form at pH 7.3. You may assume for the purposes of this question that the pKa values of the acidic groups of amino acid residues in the peptide are the same as in the amino acid itself.
Answer:
gggggggggg
Explanation:
gggggggg
The tripeptide formed from threonine, glycine and alanine is neutral at the pH of 7.3. The carboxylic end is negative charged by donating its proton to form the NH₃⁺ group.
What is peptide?Peptides are protein units formed from two or more amino acids bonded through peptide bonds. There are essential and non-essential amino acids. Essential amino acids have to be uptake from food and non-essential amino acids are synthesized inside the body.
Threonine is an essential amino acid with a CH₃CHOH side group. Glycine has the simplest side group hydrogen and alanine has CH₃ side chain. Both glycine and alanine are non-essential amino acids.
Each amino acids are represented with a three letter code or one letter symbol. Thus threonine is T, G for glycine and A for alanine. At a pH of 7.3 the peptide formed from these amino-acids contains a negatively charged carboxylic end.
A positively charged amino end made by protonation from the acid group make the overall charge zero. The structure of the peptide is given in the uploaded image.
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At 445oC, Kc for the following reaction is 0.020. 2 HI(g) <--> H2 (g) + I2 (g) A mixture of H2, I2, and HI in a vessel at 445oC has the following concentrations: [HI] = 1.5 M, [H2] = 2.50 M and [I2] = 0.05 M. Which one of the following statements concerning the reaction quotient, Qc, is TRUE for the above system?
a. Qc = Kc; the system is at equilibrium.
b. Qc is less than Kc; more H2 and I2 will be produced.
c. Qc is less than Kc; more HI will be produced.
d. Qc is greater than Kc; more HI will be produced.
Explanation:
The given balanced chemical equation is:
[tex]2 HI(g) <--> H_2 (g) + I_2 (g)[/tex]
The value of Kc at 445oC is 0.020.
[HI]=1.5M
[H2]=2.50M
[I2]=0.05M
The value of Qc(reaction quotient ) is calculated as shown below:
Qc has the same expression as the equilibrium constant.
[tex]Qc=\frac{[H_2][I_2]}{[HI]^2} \\Qc=(2.50Mx0.05M)/(1.5M)^2\\Qc=0.055[/tex]
Qc>Kc,
Hence, the backward reaction is favored and the formation of Hi is favored.
Among the given options, the correct answer is option d. Qc is greater than Kc; more HI will be produced.
what characterizes a homogeneous mixture?
Answer:
a mixture that doesn't really show the ingredients or things put into the material or food.
If 50.0 g of sulfuric acid and 40.0 grams of barium chloride are mixed, how many grams of sulfuric acid and how many grams of barium chloride remain after the double replacement reaction is complete?
After the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.
First, we will write the balanced equation for the reaction
H₂SO₄ + BaCl₂ → BaSO₄ + 2HCl
This means 1 mole of BaCl₂ is needed to react completely with 1 mole of H₂SO₄ to give 1 mole of BaSO₄ and 2 moles of HCl
From the question, 50.0g of sulfuric acid is mixed with 40.0 grams of barium chloride. To determine the quantity of each substance remaining after the complete reaction, we will first determine the number of moles present in each of the reactant.
For H₂SO₄
mass = 50.0g
Molar mass = 98.079 g/mol
From the formula
Number of moles = Mass / Molar mass
∴ Number of moles of H₂SO₄ = 50.0g / 98.079 g/mol
Number of moles of H₂SO₄ = 0.5098 mol
For BaCl₂
mass = 40.0 g
Molar mass = 208.23 g/mol
∴ Number of moles of BaCl₂ = 40.0g / 208.23 g/mol
Number of moles of BaCl₂ = 0.1921 mol
Since the number of moles of H₂SO₄ is more than that of BaCl₂, then H₂SO₄ is the excess reagent and BaCl₂ is the limiting reagent (that is, it will be used up completely during the reaction)
From the equation, 1 mole of H₂SO₄ is needed to completely react with 1 mole of BaCl₂
∴ 0.1921 mol of H₂SO₄ will be needed to completely react with 0.1921 mol of BaCl₂.
Therefore, after the reaction is complete, 0 mole (i.e 0 grams) of BaCl₂ will remain and (0.5098 mole - 0.1921 mole) of H₂SO₄ will remain.
Number of moles H₂SO₄ that will remain = 0.5098 mole - 0.1921 mole = 0.3177 moles
Now, we will convert this to grams
From the formula
Mass = Number of moles × Molar mass
Mass of H₂SO₄ that will remain = 0.3177 moles × 98.079 g/mol
Mass of H₂SO₄ that will remain = 31.1597 g
Mass of H₂SO₄ that will remain ≅ 31.16 g
Hence, after the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.
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The shape of a molecule is determined by:
A. All of these
B. The number of electron clouds around the atom.
C. The number of bonds.
D. Mutual repulsion between electrons.
Hydrogengasand oxygengas react to form water vapor. Suppose you have of and of in a reactor. Calculate the largest amount of that could be produced. Round your answer to the nearest .
The question is incomplete. The complete question is :
Hydrogen [tex](H_2)[/tex] gas and oxygen [tex](O_2)[/tex] gas react to form water vapor [tex](H_2O)[/tex]. Suppose you have 11.0 mol of [tex]H_2[/tex] and 13.0 mol of [tex]O_2[/tex] in a reactor. Calculate the largest amount of [tex]H_2O[/tex] that could be produced. Round your answer to the nearest 0.1 mol .
Solution :
The balanced reaction for reaction is :
[tex]$2H_2(g) \ \ \ \ + \ \ \ \ \ O_2(g)\ \ \ \rightarrow \ \ \ \ 2H_2O(g)$[/tex]
11.0 13.0
11/2 13/1 (dividing by the co-efficient)
6.5 mol 13 mol (minimum is limiting reagent as it is completely consumed during the reaction)
Therefore, [tex]H_2[/tex] is limiting reagent. It's stoichiometry decides the product formation amount from equation above it is clear that number of moles for [tex]H_2O[/tex] will be produced = number of moles of [tex]H_2[/tex]
= 11.0 mol
Which of the following are examples of single replacement reactions? Select all that apply.
Answer:
Na2S(aq)+Cd(No3)2(aq)=CdS(s)+2NaNo3(aq)
Answer: it’s checkbox 2&3
How many neutrons does Carbon- 14 and Carbon -15 have? *
Answer: 8 for both
Explanation:
