water is known as the mixture
Answer:
iv. Iron
water is not a mixture
hydrogen is the simplest element
air is pure
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
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?
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.
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.
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:
Find the volume occupied by 128g of SO2.
The volume occupied by the given amount of sulfur dioxide will be 84 L.
please show working my dear citizen
15.27
The following equilibria were attained at 823 K:
COO(s) + H2() Co(s) + H2O(g) K = 67
COO(s) + CO(8) = Co(s) + CO2(8) K = 490
Based on these equilibria, calculate the equilibrium con-
stant for
H2(g) + CO2(g) = CO(g) + H2O(g) at 823 K.
The equilibrium constant for the reaction is K = 0.137
We obtain the equilibrium constant considering the following equilibria and their constants:
COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
COO(s) + CO(g) → Co(s) + CO₂(g) K₂ = 490
We write the first reaction in the forward direction because we need H₂(g) in the reactants side:
(1) COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
Then, we write the second reaction in the reverse direction because we need CO₂(g) in the reactants side. Thus, the equilibrium constant for the reaction in the reverse direction is the reciprocal of the constant for the reaction in the forward direction (K₂):
(2) Co(s) + CO₂(g) → COO(s) + CO(g) K₂ = 1/490
From the addition of (1) and (2), we obtain:
COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
+
Co(s) + CO₂(g) → COO(s) + CO(g) K₂ = 1/490
-------------------------------------------------
H₂(g) + CO₂(g) → CO(g) + H₂O(g)
Notice that Co(s) and COO(s) are removed that appear in the same amount at both sides of the chemical equation.
Now, the equilibrium constant K for the reaction that is the sum of other two reactions is calculated as the product of the equilibrium constants, as follows:
K = K₁ x K₂ = 67 x 1/490 = 67/490 = 0.137
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Given the following balanced reaction: 2Na(s) + F2(g) --> 2NaF(s)
a) How many moles of NaF will be made from 2.6 moles of F2?
b) How many moles of NaF will be made from 4.8 moles of Na?
Answer:
yes it is corrwect iyt is absolitle correct
Explanation:
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
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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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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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
What volume of each solution contains 0.14 mol of KCl? Express your answer using two significant figures.
1.8 M KCl
Answer:
Solution given:
1 mole of KCl[tex]\rightarrow [/tex]22.4l
1 mole of KCl[tex]\rightarrow [/tex]74.55g
we have
0.14 mole of KCl[tex]\rightarrow [/tex]74.55*0.14=10.347g
74.55g of KCl[tex]\rightarrow [/tex]22.4l
10.347 g of KCl[tex]\rightarrow [/tex]22.4/74.55*10.347=3.11litre
volume of each solution contains 0.14 mol of KCl contain 3.11litre.
[tex]\:[/tex]
1 mole of KCl → 22.4l
1 mole of KCl → 74.55g
we have
0.14 mole of KCl → 74.55*0.14=10.347g
74.55g of KCl → 22.4l
10.347 g of KCl → 22.4/74.55*10.347=3.11litre
volume of each solution contains 0.14 mol of KCl contain 3.11litre.
Consider the Fischer ester synthesis of methyl benzoate from benzoic acid and methanol in the presence of sulfuric acid as a catalyst. A reaction was performed in which 3.3 g of benzoic acid was reacted with excess methanol to make 1.7 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.
Answer:
46.2%
Explanation:
Number of moles benzoic acid reacts = 3.3g/122.12 g/mol = 0.027 moles
Since the reaction is 1:1, 0.027 moles of methyl benzoate is formed.
Hence;
Theoretical yield of methyl benzoate = 0.027 moles × 136.15 g/mol = 3.68 g
% yield = actual yield/theoretical yield × 100
% yield = 1.7 g/3.68 g × 100
% yield = 46.2%
A solution is made by dissolving 5.84 grams of NaCl in enough distilled water to give a final volume of 1.00 L. What is the molarity of the solution
Group of answer choices
0.0250 M
0.400 M
0.100 M
1.00 M
Answer:
Explanation:
1. A solution is made by dissolving 5.84g of NaCl is enough distilled water to a give a final volume of 1.00L. What is the molarity of the solution? a. 0.100 M b. 1.00 M c. 0.0250 M d. 0.400 M 2. A 0.9% NaCl (w/w) solution in water is a. is made by mixing 0.9 moles of NaCl in a 100 moles of water b. made and has the same final volume as 0.9% solution in ethyl alcohol c. a solution that boils at or above 100°C d. All the above (don't choose this one) 3. In an exergonic process, the system a. gains energy b. loses energy c. either gains or loses energy d. no energy change at all
Answer:
[tex]\boxed {\boxed {\sf 0.100 \ M }}[/tex]
Explanation:
Molarity is a measure of concentration in moles per liter.
[tex]molarity = \frac{moles \ of \ solute}{liters \ of \ solution}}[/tex]
The solution has 5.84 grams of sodium chloride or NaCl and a volume of 1.00 liters.
1. Moles of SoluteWe are given the mass of solute in grams, so we must convert to moles. This requires the molar mass, or the mass of 1 mole of a substance. These values are found on the Periodic Table as the atomic masses, but the units are grams per mole, not atomic mass units.
We have the compound sodium chloride, so look up the molar masses of the individual elements: sodium and chlorine.
Na: 22.9897693 g/mol Cl: 35.45 g/molThe chemical formula (NaCl) contains no subscripts, so there is 1 mole of each element in 1 mole of the compound. Add the 2 molar masses to find the compound's molar mass.
NaCl: 22.9897693 + 35.45 = 58.4397693 g/molThere are 58.4397693 grams of sodium chloride in 1 mole. We will use dimensional analysis and create a ratio using this information.
[tex]\frac {58.4397693 \ g\ \ NaCl} {1 \ mol \ NaCl}[/tex]
We are converting 5.84 grams to moles, so we multiply by that value.
[tex]5.84 \ g \ NaCl *\frac {58.4397693 \ g\ NaCl} {1 \ mol \ NaCl}[/tex]
Flip the ratio. It remains equivalent and the units of grams of sodium chloride cancel.
[tex]5.84 \ g \ NaCl *\frac {1 \ mol \ NaCl}{58.4397693 \ g\ NaCl}[/tex]
[tex]5.84 *\frac {1 \ mol \ NaCl}{58.4397693 }[/tex]
[tex]0.09993194823 \ mol \ NaCl[/tex]
2. MolarityWe can use the number of moles we just calculated to find the molarity. Remember there is 1 liter of solution.
[tex]molarity= \frac{moles \ of \ solute}{liters \ of \ solution}[/tex]
[tex]molarity= \frac{ 0.09993194823 \ mol \ NaCl}{1 \ L}[/tex]
[tex]molarity= 0.09993194823 \ mol \ NaCl/L[/tex]
3. Units and Significant FiguresThe original measurements of mass and volume have 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 9 in the ten-thousandths place tells us to round the 9 to a 0, but then we must also the next 9 to a 0, and the 0 to a 1.
[tex]molarity \approx 0.100 \ mol \ NaCl/L[/tex]
1 mole per liter is 1 molar or M. We can convert the units.
[tex]molarity \approx 0.100 \ M \ NaCl[/tex]
The molarity of the solution is 0.100 M.
which of the following measurements is equivalent to 5.461x10^-7m?
Answer:
B. 0.0000005461m
I used the method of moving the decimal.
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
How many neutrons does Carbon- 14 and Carbon -15 have? *
Answer: 8 for both
Explanation:
Consider the reaction: NaNO3(s) + H2SO4(l) NaHSO4(s) + HNO3(g) ΔH° = 21.2 kJ
How much heat must absorbed by the reaction system to convert 100g of NaNO3 into NaHSO4(s)?
Answer:
endet nach selam nw
4gh7
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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Heating water makes most solids in it
soluble, and it makes gases
soluble.
Increasing the pressure on a gas above water makes the gas
soluble. Compounds with comparatively stronger ionic bonds are
soluble.
Answer:
1. more
2. less
3. more
4. less
Explanation:
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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There are three isotopes of carbon. They have mass number of 12, 13 and 14. The average atomic mass of carbon is 12.0107 amu. What does this say about the relative abundances of the three isotopes?
Answer:
lots more of the carbon 12 than the others
havent calculated it percentage-wise but you can see its very close to 12 meaning it is of far greater abundance that carbon 13 and 14
Explanation:
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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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.
2. Write the chemical equation for the reaction NaOH Sodium Hydroxide AgNO3 Silver Nitrate
Answer:
AgNO3 + NaOH = AgOH + NaNO3.
Explanation:
Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.
Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.However, the equation balanced here is the initial reaction which produces AgOH and NaNO3.
What is the balanced form of the following equation?
Br2 + S2O32- + H2O → Br1- + SO42- + H+
Answer:
5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺
Explanation:
We will balance the redox reaction through the ion-electron method.
Step 1: Identify both half-reactions
Reduction: Br₂ ⇒ Br⁻
Oxidation: S₂O₃²⁻ ⇒ SO₄²⁻
Step 2: Perform the mass balance, adding H⁺ and H₂O where appropriate
Br₂ ⇒ 2 Br⁻
5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺
Step 3: Perform the charge balance, adding electrons where appropriate
2 e⁻ + Br₂ ⇒ 2 Br⁻
5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻
Step 4: Make the number of electrons gained and lost equal
5 × (2 e⁻ + Br₂ ⇒ 2 Br⁻)
1 × (5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻)
Step 5: Add both half-reactions
5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺
What volume of 1.50 mol/L stock solution is needed to make 125 mL of 0.60 mol/L solution?
Chemistry 11 Solutions
978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 8 Solutions and Their Properties • MHR | 85
Amount in moles, n, of the NaCl(s):
NaCl
2.5 g
m
n
M
58.44 g
2
4.2778 10 m l
ol
o
/m
u
Molar concentration, c, of the NaCl(aq):
–2 4.2778 × 10 mol
0.100
0.42778 mol/L
0.43 mol
L
/L
n
c
V
The molar concentration of the saline solution is 0.43 mol/L.
Check Your Solution
The units are correct and the answer correctly shows two significant digits. The
dilution of the original concentrated solution is correct and the change to mol/L
seems reasonable.
Section 8.4 Preparing Solutions in the Laboratory
Solutions for Practice Problems
Student Edition page 386
51. Practice Problem (page 386)
Suppose that you are given a stock solution of 1.50 mol/L ammonium sulfate,
(NH4)2SO4(aq).
What volume of the stock solution do you need to use to prepare each of the
following solutions?
a. 50.0 mL of 1.00 mol/L (NH4)2SO4(aq)
b. 2 × 102 mL of 0.800 mol/L (NH4)2SO4(aq)
c. 250 mL of 0.300 mol/L NH4
+
(aq)
What Is Required?
You need to calculate the initial volume, V1, of (NH4)2SO4(aq) stock solution
needed to prepare each given dilute solution.
The dilution gives the relationship between the molarity and the volume of the solution. The volume of stock solution with a molarity of 1.50 mol/L is 50 mL.
What is dilution?Dilution is said to be the addition of more volume to the concentrated solution to make it less in molar concentration. This tells about the inverse and indirect relationship between the volume and the molar concentration of the solution.
Given,
Initial volume = V₁
Initial molar concentration (M₁) = 1.50 mol/L
Final volume (V₂) = 125 mL = 0.125 L
Final molar concentration (M₂)= 0.60 mol/L
The dilution is calculated as:
M₁V₁ = M₂V₂
V₁ = M₂V₂ ÷ M₁
Substituting the values in the above formula as
V₁ = M₂V₂ ÷ M₁
V₁ = (0.60 mol/L × 0.125 L) ÷ 1.50 mol/ L
V₁ = 0.05 L
= 50 mL
Therefore, 50 mL of stock solution is needed to make a 0.60 mol/L solution.
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The density of toluene (C7H8) is 0.867 and that of thiophene (C4H4S) is 1.065 g/ml. A solution is made by dissolving 10.00g thiophene in 250.00ml of toluene. a)Calculate the molarity of the solution
b)Assuming the volume are addictive ,calculate the molarity of the solution
Answer:
Calcular la molaridad de una solución que se preparó disolviendo 14 g de KOH en suficiente
agua para obtener 250 mL de solución. (masa molar del KOH = 56 g/mol).
Resolución: de acuerdo a la definición de “molaridad” debemos calcular primero, el número de mol de soluto (KOH) que
se han disuelto en el volumen dado, es decir, “se transforma g de soluto a mol de soluto” por medio de la masa molar,
así:
56 g de KOH 14 g de KOH
----------------- = ------------------- X = 0,25 mol de KOH
1 mol X
Ahora, de acuerdo con la definición de molaridad, el número de mol debe estar contenido en 1000 mL (o 1 L) de
solución, que es el volumen estándar para esta unidad de concentración, lo que se determina con el siguiente planteamiento:
0,25 mol X
----------------------- = ------------------------- X = 1 mol de KOH
250 mL de solución 1000 mL de solución
Explanation:
It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high concentration, thus reducing the amount of by-product formed. What is the by-product
Answer:
Biphenyl
Explanation:
The reaction of bromo benzene with magnesium-ether solution yields a Grignard reagent.
The byproduct of this reaction is biphenyl. It is formed when two unreacted bromobenzene molecules are coupled together.
Hence, It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high concentration, thus reducing the amount of biphenyl by-product formed.
