Answer:
5.37% w/w is the mass percent of vinegar assuming a molarity of 0.8935mol/L
Explanation:
Assuming the molarity of vinegar is 0.8935mol/L:
Mass percent is defined as 100 times the ratio between mass of solute (In this case, acetic acid), and the mass of the solution
To solve this question we need to find the mass of acetic acid from the moles using the molar mass and the mass of the solution from the volume in liters using the density:
Mass Acetic acid -Molar mass: 60.052g/mol-
0.8935mol * (60.052g / mol) = 53.656g Acetic Acid
Mass Solution:
1L = 1000mL * (1.00g/mL) = 1000g Solution
Mass Percent:
53.656g Acetic Acid / 1000g Solution * 100 =
5.37% w/w is the mass percent of vinegar assuming a molarity of 0.8935mol/LThe mass percent of acetic acid in the original sample of vinegar of molarity 0.8935mol/L is 5.37% w/w.
How do we calculate the mass percent?Mass percent of any solute present in any solution will be calculated as the:
Mass % of solute = (mass of solute / mass of solution) × 100
Let the molarity of vinegar = 0.8935mol/L
Means 0.8935 moles of vinegar present in the 1 liter of the solution.
Now we calculate mass from moles as:
n = W/M, where
W = required mass
M = molar mass = 60.052g /mol
W = (0.8935mol)(60.052g/mol) = 53.656g
Mass of solution = 1L = 1000mL×(1.00g/mL) = 1000g Solution
Then the mass % of acetic acid:
Mass % = (53.656g / 1000g) × 100 = 5.37% w/w
Hence the required % mass is 5.37% w/w.
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What is Avogadro's number?
O A. 6.02 x 10-23
O B. 6.0223
C. 6.02 x 10
D. 6.02 x 1023
Answer:
E.6.02 10-23Explanation:
Answer:
6.02×10^23 I hope it helps you
Explanation:
I hope it helps you
Determine the mmol of both starting materials (factoring in that formic acid is not pure, but rather 88% weight/volume, or 88g/100 ml), showing your work. Determine the limiting reagent in this synthesis. Lastly, calculate the theoretical yield of benzimidazole that you could expect to form.
Solution :
Molecular Molar Mass Volume Density Mass Moles nmoles
formula (g/mol) (mL) (g/mL) (g)
[tex]$C_6H_8N_2$[/tex] 108.14 0.108 0.001 1
HCOOH 46.02 0.064 1.22 0.07808 0.0017 1.7
mmoles of o-phenylenediamine = 1 mmoles
mmoles of formic acid = 1.7 [tex]\approx[/tex] 2 mmoles
From the reaction of o-phenylenediamine and formic acid, we see,
1 mmole of o-phenylenediamine reacts with 1 mmole of formic acid.
But here, 2 mmoles of the formic acid , this means that the formic acid is an excess reagent and the o-phenylenediamine is the limiting reagent here.
The amount of product depends on the limiting reagent that is o-phenylenediamine. So, 1mmole of o-phenylenediamine will give 1mmole of product.
molar mass of Benzimidazole = [tex]118.14[/tex] g/mol
mmoles of Benzimidazole formed = [tex]1[/tex] mmol
Mass of benzimidazole formed = molar mass x [tex]\frac{nmoles}{1000}[/tex]
[tex]$=\frac{118.14 \times 1}{1000}$[/tex]
= 0.11814 g
So the theoretical yield of Benzimidazole is = 0.118 g = 118mg
Identify “A” in the following reaction: CH3¬COOH + Na2CO3 → A + CO2 + H2O
Consider an equilibrium (K1) that is established after 10 mL of compound A and 10 mL of compound B are mixed. Now, imagine the equilibrium (K2) where 1 mL of compound A is added to 100 mL of compound B. How are K1 and K2 related algebraically (read this question VERY carefully, at least one more time)
The equilibrium constant K₁ = Equilbrium constant K₂.
The equilibrium constant, K, of a reaction, is defined as:
"The ratio between concentration of products powered to their reaction quotient and concentration of reactants powered to thier reaction quotient".
For the reaction:
aA + bB ⇄ cC + dD
The equilibrium constant, K, is:
[tex]K = \frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]
Now, assuming the reaction of the problem is 1:1:
A + B ⇄ C + D
[tex]K = \frac{[C][D]}{[A][B]}[/tex]
The concentrations of the reactants are directly proportional to the volume added. Thus, we can assume that concentration = Volume. Replacing for K₁ and K₂:
[tex]K_1 = \frac{[C][D]}{[10mL][10mL]} = K_1 = \frac{[C][D]}{100mL^2}[/tex]
In the same way:
[tex]K_2 = \frac{[C][D]}{[1mL][100mL]} = K_2 = \frac{[C][D]}{100mL^2}[/tex]
Thus, we can say:
K₁ = K₂Learn more about chemical equilibrium in:
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what is the difference between Absorption and adsorption
Considering a fish breeder decided to breed small fishes which needs a pH between 6,0 to 7,0 to stay alive. He needs to adjust the water's pH that is 5,0 to a value of 6.5, having available only calcium carbonate. The mass in mg added to 5L of water is about:
A)2,5
B)5,5
C)6,5
D)7,5
E)9,5
A solution is made by mixing 34.5 g of sugar with 75.0 g of water. What is the mass percent of sugar in this solution?
Please explain and show work.
the mass percent of sugar in this solution is 46%.
Answer:
Solution given:
mass of solute=34.5g
mass of solvent=75g
mass percent=[tex]\frac{mass\:of\:solute}{mass\:of \:solvent}*100\%[/tex]
=[tex]\bold{\frac{34.5}{75.0}*100\%=46\%}[/tex]
We know that,
[tex]mass \: solute = \frac{mass \: of \: solute}{mass \: of \: solvent} [/tex]×100
[tex] \frac{34.5}{75.0} = 0.46[/tex]
hope it helps..
Given its formula and Avogadro's Number (6.02 x 10^23 molecules/mol), deduce how many molecules are present in 3 x 10^-16 grams of TCDD. Type in only a number without using scientific notation.
Answer:
5 × 10⁵ molecules (500,000 molecules)
Explanation:
Step 1: Convert 3 × 10⁻¹⁶ g to moles
We will use the molar mass of TCDD (321.97 g/mol).
3 × 10⁻¹⁶ g × 1 mol/321.97 g = 9 × 10⁻¹⁹ mol
Step 2: Convert 9 × 10⁻¹⁹ mol to molecules
The required conversion factor is Avogadro's number (6.02 × 10²³ molecules/mol).
9 × 10⁻¹⁹ mol × 6.02 × 10²³ molecules/1 mol = 5 × 10⁵ molecules
A decomposition of a sample of diphosphorus trioxide forms 1.29 g phosphorus to every 1.00 g oxygen. The decomposition of a sample of diphosphorus pentoxide forms 0.775 g phosphorus to every 1.00 g oxygen.
Required:
How many grams of P205 are formed when 5.89 g of P react with excess oxgen?
Answer:
There is 13.48 grams of P2O5 formed
Explanation:
Step 1: Data given
A decomposition of a sample of diphosphorus trioxide forms 1.29 g phosphorus to every 1.00 g oxygen.
Mass of P = 5.89 grams
Molar mass of O2 = 32.0 g/mol
atomic mass of P = 30.97 g/mol
molar mass of P2O5 = 141.94 g/mol
Step 2: The balanced equation
4P(s)+5O2(g)⇔ 2P2O5(s)
Step 3: Calculate moles of P
Moles P = Mass P / atomic mass P
Moles P = 5.89 grams / 30.97 g/mol
Moles P = 0.190 moles
Step 4: Calculate moles of P2O5
For 4 moles P we need 5 moles O2 to produce 2 moles P2O5
For 0.190 moles of P we'll have 0.190/2 = 0.095 moles P2O5
Step 5: Calculate mass of P2O5
Mass P2O5 = moles P2O5 * molar mass P2O5
Mass P2O5 = 0.095 moles * 141.94 g/mol
Mass P2O5 = 13.48 grams
There is 13.48 grams of P2O5 formed
A sample of gas in a balloon has an initial temperature of 18 ∘C and a volume of 33 L. If the temperature changes to 47 ∘C, and there is no change of pressure or amount of gas, what is the new volume, V2, of the gas?
Answer:
The final volume of the sample of gas is 36.287 liters.
Explanation:
Let suppose that sample of gas is a closed system, that is, a system with no mass interactions with surroundings, and gas is represented by the equation of state for ideal gases, which is described below:
[tex]P\cdot V = n\cdot R_{u}\cdot T[/tex] (1)
Where:
[tex]P[/tex] - Pressure, in atmospheres.
[tex]V[/tex] - Volume, in liters.
[tex]n[/tex] - Molar quantity, in moles.
[tex]T[/tex] - Temperature, in Kelvin.
[tex]R_{u}[/tex] - Ideal gas constant, in atmosphere-liters per mole-Kelvin.
As we know that sample of gas experiments an isobaric process, we can determine the final volume by the following relationship:
[tex]\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}[/tex] (2)
Where:
[tex]V_{1}[/tex] - Initial volume, in liters.
[tex]V_{2}[/tex] - Final volume, in liters.
[tex]T_{1}[/tex] - Initial temperature, in Kelvin.
[tex]T_{2}[/tex] - Final temperature, in Kelvin.
If we know that [tex]V_{1} = 33\,L[/tex], [tex]T_{1} = 291.15\,K[/tex] and [tex]T_{2} = 320.15\,K[/tex], then the final volume of the gas is:
[tex]V_{2} = V_{1}\cdot \left(\frac{T_{2}}{T_{1}} \right)[/tex]
[tex]V_{2} = 33\,L \times \frac{320.15\,K}{291.15\,K}[/tex]
[tex]V_{2} = 36.287\,L[/tex]
The final volume of the sample of gas is 36.287 liters.
PLEASE HELP ASAP
Use the equation below to answer the following questions.
2Al(s) + 3Cu(NO3)2(aq) 3Cu(s) + 2Al(NO3)3(aq)
Determine the oxidation state of the atoms in the equation's reactants and products: (6 points)
Oxidation state of Al in reactant:
in product:
Oxidation state of Cu in reactant:
in product:
Oxidation state of N in reactant:
in product:
Oxidation state of O in reactant:
in product:
Explain why this is a redox reaction.
Thank you!
Answer:
hlo.......................,
Complete the balanced dissociation equation for the compound below in aqueous solution. If the compound does not dissociate, write NR after the reaction arrow.
HI (aq) -->
Answer:
[tex]{ \bf{HI _{(aq)} \: → \: H {}^{ + } _{(aq)} \: + \: \: I {}^{ - } _{(aq)} }}[/tex]
What is the relationship between temperature and kinetic energy?
Answer:
"[Temperature is a measurement of the average kinetic energy of the molecules in an object or a system. Kinetic energy is the energy that an object has because of its motion. The molecules in a substance have a range of kinetic energies because they don't all move at the same speed.]"
Answer:
Temperature is directly proportional to the average translational kinetic energy of molecules in an ideal gas
Explanation:
The sample concentration was measured at 50mg/ml. The loading concentration needs to be 10mg/ml. The final volume needs to be 25ul. What is the volume of sample needed and the amount of buffer needed to reach 25ul
Answer:
a) [tex]V_1=5ul[/tex]
b) [tex]v=20ul[/tex]
Explanation:
From the question we are told that:
initial Concentration [tex]C_1=50mg/ml[/tex]
Final Concentration [tex]C_2=10mg/ml[/tex]
Final volume needs [tex]V_2 =25ul[/tex]
Generally the equation for Volume is mathematically given by
[tex]C_1V_1=C_2V_2[/tex]
[tex]V_1=\frac{C_1V_1}{C_2}[/tex]
[tex]V_1=\frac{10*25}{50}[/tex]
[tex]V_1=5ul[/tex]
Therefore
The volume of buffer needed is
[tex]v=V_2-V_1\\\\v=25-5[/tex]
[tex]v=20ul[/tex]
What is the molarity of a solution that contains 17g of NH₃ in 0.50L of solution?
Please explain as well!
Explanation:
Because molarity is mol/L, we'll have to convert 17g to mol.
After obtaining the mol, we'll divide that by the volume to obtain Molarity.
[tex]\\ \sf\longmapsto NH_3[/tex]
[tex]\\ \sf\longmapsto 14u+3(1u)[/tex]
[tex]\\ \sf\longmapsto 14u+3u[/tex]
[tex]\\ \sf\longmapsto 17u[/tex]
[tex]\\ \sf\longmapsto 17g/mol[/tex]
Moles of Ammonia:-
[tex]\boxed{\sf No\:of\:moles =\dfrac{Given\:mass}{Molar\:mass}}[/tex]
[tex]\\ \sf\longmapsto No\:of\:moles=\dfrac{17}{17}[/tex]
[tex]\\ \sf\longmapsto No\;of\:moles=1mol[/tex]
Volume of solution=0.50LWe know
[tex]\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Volume\:of\: Solution\:in\:L}}[/tex]
[tex]\\ \sf\longmapsto Molarity=\dfrac{1}{0.50}[/tex]
[tex]\\ \sf\longmapsto Molarity=2M[/tex]
How are radioactive isotopes used to diagnose thyroid problems?
Answers
A.
Production of radioactive iodine by the thyroid is measured.
B.
Uptake of radioactive iodine by the thyroid gland is measured.
C.
Destruction of damaged cells by radioisotopes is measured.
D.
Absorption of gamma rays from ingested radioisotopes is measured.
Question
Answer:
A
Explanation:
uptakw of raduoactive iofine by the thyroid gland js measurd
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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. Which of the following statement is not related to a chemical reaction ? A. New substances are formed B. Atoms of the elements transform into atoms of other elements C. The properties of the new substances will be different D. There will be bond breaking and bond forming
Answer:
the answer should be B because elements do not tranform into other elements in a chemical reaction
am I right please?
Which of the following releases hormones into your bloodstream?
A. Endocrine system
B. Sympathetic nervous system
C. Lobal system
a
D. Autonomic nervous system
Answer:
answer is A. Endocrine system
Endocrine glands secrete hormones straight into the bloodstream. Hormones help to control many body functions, such as growth, repair and reproduction.
Answer:
A endocrine system
this is the answer
Which of the following is used in EBRT?
O Silver tube
O Gold tube
O Copper tube
O Iron tube
Copper tube is used in EBRT.
What is meant by EBRT?External Beam Radiation. Therapy (EBRT) is a type of radiation therapy that directs a beam of radiation from outside the body, toward cancerous tissues inside the body.External beam radiation therapy (EBRT) is the most common type of radiation therapy. It directs high-energy radiation beams at the cancer.Copper tube is used in EBRT.
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A new member of your team is visually impaired. He's having a hard time reading his computer screen. How can you show that you value diversity? a) Suggest the organization may not be a great fit. O b) Tell him to apply for a different job within the organization that does not b) require computer usage. 12 Od Tell him that he might want to schedule an eye exam. c) d) Provide access to technology that can magnify or read what is on the computer screen.
Answer:
C
Explanation:
If he can't read his computer screen. Means his eyes have problem
A new member of your team is visually impaired. He's having a hard time reading his computer screen. value diversity require computer usage. 12 Tell him that he might want to schedule an eye exam. Therefore, option C is correct.
What is value diversity ?Institutions and community members who value diversity do so in order to recognize the advantages of both their differences and commonalities. They make a deliberate effort to forge enduring connections between individuals and organizations with a broad membership.
When people's unique qualities, abilities, interests, and viewpoints are recognized and supported, they develop a stronger sense of self and health and achieve better results in their academic and professional endeavors.
Although diversity and inclusion are related ideas, they are not the same thing. Diversity has to do with representation or how something is put together. The degree to which the contributions, presence, and viewpoints of other groups of individuals are appreciated and incorporated into a setting is referred to as inclusion.
Thus, option C is correct.
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Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product of the instantaneous amounts of A and B not converted to chemical C. Initially, there are 100 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 25 grams of C is formed in 9 minutes. How much is formed in 36 minutes
Answer:
In 36 minutes, 100 grams of Chemical C is formed.
Explanation:
Combination of chemicals A and B = chemical C
Chemical A available = 100 grams
Chemical B available = 50 grams
Proportion of A mixed with C = A2
Proportion of B mixed with C = B1
Therefore, Chemical C = A2 + B1
If 25 grams of C is formed in 9 minutes
In one minute 25/9 grams of C will be formed
Therefore, in 36 minutes, 25/9 * 36 = 100 grams
What is the molecule shown below?
A. Pentane
B. Trimethylethane
C. 2,2-dimethylpropane
D. 3-dipropane
Q2
Answer:
C
Explanation:
if we were to followw the IUPAC
If the concentration of [CO32−] were doubled (say, by adding a highly soluble carbonate salt such as Cs2CO3), what would be the new concentration of [Li+] in the saturated solution?
Answer:
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A sample of oxygen occupied 621 mL when the pressure increased to 1095.93mm Hg. At constant temperature, what volume did the gas initially occupy when the pressure was 774.29mm Hg?
a.) 879.0
b.) 438.7
c.) 890.2
d.) 1366
Answer:
for you ta bangla de Mayo for bangla ki you would know it has changed an anyone come by with home home but is in an going forward you as as on Saturday to what our current meeting office next weekend too with we before too
Explanation:
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Draw the organic product(s) of the following reaction.
NaNH2/ NH3(l)
CH2CH2CH2-Câ¡C -C-H â
Answer:
H-C = C-H NaNH2 [tex]\ \to \0}[/tex] H-C = CNa
H-C = C - CH2 CH2 CH2 CH3
Explanation:
NaNH2 acts as base in this reaction. The organic products released after the reaction of carbon hydrogen atom with sodium amide. These products released after the chemical reaction when carbon and hydrogen atom reacts and NaNH2 acts as base then substitution nucleophilic reaction takes place.
9. Discuss the general trend in Chemical Properties of the Representative Elements
Answer:
Elements in the same period show trends in atomic radius, ionization energy, electron affinity, and electronegativity.
How many grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride?
Answer:
it is 11.55 and ik because I just had that question
18.0 grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride.
Let's consider the following balanced equation.
4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g)
The molar ratio of C₈H₄N₂ to Cu(C₃₂H₁₆N₈) is 4:1. The moles of Cu(C₃₂H₁₆N₈) produced from 0.125 moles of C₈H₄N₂ are:
[tex]0.125 mol C_8H_4N_2 \times \frac{1molCu(C_{32}H_{16}N_8)}{4mol C_8H_4N_2} = 0.0313 molCu(C_{32}H_{16}N_8)[/tex]
The molar mass of Cu(C₃₂H₁₆N₈) is 576.07 g/mol. The mass corresponding to 0.0313 moles of Cu(C₃₂H₁₆N₈) is:
[tex]0.0313 moles \times \frac{576.07g}{mol} = 18.0 g[/tex]
18.0 grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride.
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how many moles of lithium atoms are contained in 5.2 g of lithium
Answer:
[tex]\boxed {\boxed {\sf 0.75 \ mol \ Li}}[/tex]
Explanation:
We are asked to convert 5.2 grams of lithium to moles of lithium.
1. Molar MassTo convert from grams to moles, we need the molar mass. This is the measurement of the mass in 1 mole of a substance. It can be found on the Periodic Table because it is the same value as the atomic mass, but the units are grams per mole instead of atomic mass units.
Look up the molar mass of lithium.
Li: 6.94 g/mol 2. Convert Grams to MolesCreate a ratio using the molar mass of lithium.
[tex]\frac { 6.94 \ g \ Li}{ 1 \ mol \ Li}[/tex]
Multiply by the value we are converting: 5.2 grams of lithium.
[tex]5.2 \ g \ Li *\frac { 6.94 \ g \ Li}{ 1 \ mol \ Li}[/tex]
Flip the ratio so the units of grams of lithium cancel.
[tex]5.2 \ g \ Li *\frac{ 1 \ mol \ Li} { 6.94 \ g \ Li}[/tex]
[tex]5.2 *\frac{ 1 \ mol \ Li} { 6.94 }[/tex]
[tex]\frac{5.2} { 6.94 } \ mol \ LI[/tex]
[tex]0.749279538905 \ mol \ Li[/tex]
3. RoundThe original measurement of grams (5.2) has 2 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 9 in the thousandths place to the right tells us to round the 4 up to a 5.
[tex]0.75 \ mol \ Li[/tex]
5.2 grams of lithium is equal to 0.75 moles of lithium atoms.
