Answer:
43.13Kg of melamine
Explanation:
The problem gives you the mass of urea and two balanced equations:[tex]CO(NH_{2})_{2}_{(l)}=HNCO_{(l)}+NH_{3}_{(g)}[/tex]
[tex]6HNO_{l}=C_{3}N_{3}(NH_{2})_{3}_{(l)}+3CO_{2}_{(g)}[/tex]
First we need to calculate the number of moles of urea that are used in the reaction, so:
molar mass of urea = [tex]60.06\frac{g}{mol}*\frac{1kg}{1000g}=0.06006\frac{Kg}{mol}[/tex]
The problem says that you have 161.2Kg of urea, so you take that mass of urea and find the moles of urea:
161.2Kg of urea[tex]*\frac{1molofurea}{0.06006Kgofurea}=[/tex]2684 moles of urea
Now from the stoichiometry you have:
2684 moles of urea[tex]*\frac{1molofHNCO}{1molurea}*\frac{1molofmelamine}{6molesofHNCO}[/tex] = 447 moles of melamine
The molar mass of the melamine is [tex]126.12\frac{g}{mol}[/tex] so we have:
[tex]447molesofmelamine*\frac{126.12g}{1molofmelamine}[/tex] = 5637.64 g of melamine
Converting that mass of melamine to Kg:
5637.64 g of melamine *[tex]\frac{1Kg}{1000g}[/tex] = 56.38 Kg of melamine, that is the theoretical yield of melamine.
Finally we need to calculate the mass of melamine with a yield of 76.5%, so we have:
%yield = 100*(Actual yield of melamine / Theoretical yield of melamine)
Actual yield of melamine = [tex]\frac{76.5}{100}*56.38Kg[/tex] = 43.13Kg of melamine
Using stoichiometry, the actual yield of Melamine from a reaction starting with 161.2 kg of urea with an overall yield of 76.5% can be found to be approximately 43.2 kg.
Explanation:The question asks about the actual yield of Melamine, C3N3(NH2)3, from a reaction starting with 161.2 kg of urea, CO(NH2)2, with an overall yield of 76.5%. To understand this prediction, we must use the concept of stoichiometry, which is a method used in chemistry to calculate the quantities of reactants or products in a chemical reaction.
First, calculate the mole of urea used: it is the mass of the urea divided by the molar mass of urea. The molar mass of urea (CO(NH2)2) is approximately 60 g/mole. So, you have 161.2*1000/60 = approximately 2686.67 moles of urea.
From the balanced chemical equation, we can see that one mole of urea produces one mole of HNCO, and six moles of HNCO produce one mole of Melamine. Therefore, in theory, 2686.67/6 = 447.78 moles of Melamine can be produced.
However, the reaction yield is only 76.5%, so the actual yield of the Melamine will be less. Using the percentage yield, we can calculate the actual yield: 447.78 * 0.765 = approximately 342.65 moles.
Finally, to convert from moles to mass, multiply by the molar mass of melamine, which is approximately 126 g/mole: 342.65*126 = approximately 43173.9g or 43.2 kg.
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Consider the formation of nitryl fluoride: 2NO2(g)+F2(g)⇌2NO2F(g) The reaction is first order in F2 and second order overall. What is the rate law? View Available Hint(s) Consider the formation of nitryl fluoride: The reaction is first order in and second order overall. What is the rate law? rate=k[NO2]2[F2]2 rate=k[NO2][F2]2 rate=k[F2] rate=k[NO2] rate=k[NO2][F2] rate=k[NO2]2[F2]
Answer:
Rate = [tex]k[NO_{2}][F_{2}][/tex]
Explanation:
Two reactants are present in this reaction which are [tex]NO_{2}and F_{2}[/tex]We know overall order of a reaction is summation of individual order with respect to reactants present in rate law equation.Here, overall order of reaction is 2 including first order with respect to [tex]F_{2}[/tex]So, rate of reaction should also be first order with respect to another reactant i.e. first order with respect to [tex]NO_{2}[/tex]So, rate law: rate = [tex]k[NO_{2}][F_{2}][/tex]You are given a protein solution with a concentration of 0.15 mg/ml.
v. Suppose that we want to prepare 100 microliters of 10 micrograms/microliters solution. How much of H2O and protein stock do we need to add to obtain the target concentration and volume? (THE CONCENTRATION IS NOT GREATER IN THE QUESTION, YOU NEED TO CONVERT IT TO MICROGRAMS/MICROLITERS.... THEN IT SHOULD MAKE SENSE)
Answer:
6,666.66 micro liter of water and protein stock we will need to add to obtain the target concentration and volume.
Explanation:
Concentration of given solution[tex]C_1 = 0.15 mg/mL[/tex]
1 mL = 1000 μL , 1 mg = 1000 μg
[tex]C_1=0.15 mg/mL=\frac{0.15\times 1000 \mu g}{1\times 1000 \mu L}=0.15 \mu g/\mu L[/tex]
The volume of the given solution =[tex]V_1= V[/tex]
Concentration of required solution = [tex]C_2=10 \mu g/\mu L[/tex]
Volume of required solution = [tex]V_2=100 \mu L[/tex]
[tex]C_1V_1=C_2V_2[/tex]
[tex]V=\frac{C_2V_2}{C_1}=\frac{10 \mu g/\mu L\times 100 \mu L}{0.15 \mu g/\mu L}=6,666.66 \mu L[/tex]
6,666.66 micro liter of water and protein stock we will need to add to obtain the target concentration and volume.
Identify the molecules with a dipole moment: (a) SF (b) CF (c) CCCB (d) CHCI (e) H.CO
Answer: An atom has either tendency of accepting or losing electron, on the basis of this virtue it is named as
Electronegative: An atom has tendency to attract the shared pair of electron towards itself is called electronegative.Electropositive: An atom that has tendency to give the shared pair of electron towards an electronegative atom is called electropositive.For the existence of dipole within a molecule there must be a difference in electronegativity of the atoms participating in it. All the molecules here have dipole moment because the atoms participating in them have a difference in electronegative.
A 0.4657 g sample of a pure soluble bromide compound is dissolved in water, and all of the bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate. The mass of the resulting AgBr is found to be 0.8878 g. What is the mass percentage of bromine in the original compound?
Answer:
The mass percentage of bromine in the original compound is 81,12%
Explanation:
Step 1: Calculate moles AgBr
moles AgBr = mass AgBr / molar mass AgBr
= 0.8878 g / 187.77 g/mol
= 0.00472812 moles AgBr
⇒
Since 1 mol AgBr contains 1 mol Br-
Then the amount of moles Br- in the original sample must also have been 0.00472812 moles
Step 2: Calculating mass Br-
mass Br- = molar mass Br x moles Br-
= 79.904 g/mol x 0.00472812 mol
= 0.377796 g Br-
⇒
There were 0.377796 g Br- in the original sample
Step 3: Calculating mass percentage Br-
⇒mass percentage = actual mass Br- / total mass x 100%
% mass Br = 0.377796 g / 0.4657 g x 100 %
= 81.12%
You have a sample of water that contains the organic compound
C7H12ON2 at a concentration of 50
mg/L. The compound can be oxidized by bacteria to form carbon
dioxide, water, and ammonia. How many mg/L of oxygen is needed to
biodegrade the compound? Note: Determine only carbonaceous
demand.
Answer:
131.4 mg/L of oxygen is needed to biodegrade the organic compound.
Explanation:
The chemical reaction will be written as:
[tex]2C_7H_{12}ON_2+23O_2\rightarrow 14CO_2+12H_2O+4NO_2[/tex]
Concentration of the organic compound = 50 mg/L
This means that 50 milligrams of organic compound in present in 1 L of the solution.
50 mg = 0.050 g
1 mg = 0.001 g
Moles of organic compound = [tex]\frac{0.050 g}{140 g/mol}=0.0003571 mol[/tex]
According to reaction, 2 moles of organic compound reacts with 23 moles of oxygen gas.
Then 0.0003571 moles of an organic compound will react with:
[tex]\frac{23}{2}\times 0.0003571 mol=0.004107 mol[/tex] oxygen gas.
Mass of 0.004107 moles of oxygen gas:
0.004107 mol × 32 g/mol = 0.1314 g = 131.4 mg
131.4 mg/L of oxygen is needed to biodegrade the organic compound.
Write 450.0000345 in Scientific Notation with 4 significant figures.
Answer:
[tex]4.500\times 10^{2}[/tex]
Explanation:
Scientific notation is the way of writing numbers which are either large or small. The number is written in the scientific notation when the number is between 1 and 10 and then multiplied by the power of 10.
The given number:
450.0000345 can be written as [tex]4.500000345\times 10^{2}[/tex]
Answer upto 4 significant digits = [tex]4.500\times 10^{2}[/tex]
Calculate the number of moles in 369 grams of CaoCl2? a) 1 b) 2 c) 3 d) 4
Answer:
First of all, take account the molar mass of the CaOCl2 (Calcium hypochlorite), which is 126.97 g/mol. If we have 126.97 g in a mol, 369g should be in aproximately 3 moles. Try to the think the rule of 3.
Explanation:
Balance the equation in the box. Click in the answer box to activate the palette. N2(g) + H2(g) → NH3(g)
Answer:
N2(g) + 3H2(g) → 2 NH3(g)
Explanation:
N2(g) + H2(g) → NH3(g)
We start equaling the number of N atoms in both sides multiplying by 2 the NH3.
N2(g) + H2(g) → 2 NH3(g)
So we equals the H atoms (there are six in products sites)
N2(g) + 3 H2(g) → 2 NH3(g)
How many degrees of freedom does each of the following systems have? (Answer as a number, i.e., 1, 2, 3, etc.)
1. Liquid water in equilibrium with its vapor?
2. Liquid water in equilibrium with a mixture of water vapor and nitrogen?
3. A liquid solution of alcohol in water in equilibrium with its vapor?
Answer:
Explanation:
Hello, since the Gibbs' phase rule states the following equation:
[tex]F=C-P+2[/tex]
Whereas C is the number of components and P the present phases, you answers are:
1. F=1-2+2=1.
2. F=2-2+2=2.
3. F=2-2+2=2.
Best regards.
If ine mole of pennies were divided amung 250 million peoplein
the US, how many dollars would each person recieve?
Answer:
[tex]2.4088\times 10^{13}[/tex] dollars each person will receive.
Explanation:
Number of people in which 1 mole of pennies is distributed = 250 million =
[tex]1 million = 10^6 [/tex]
250 million = [tex]2.5\times 10^8 [/tex] persons
Number of pennies in 1 mole = [tex]6.022\times 10^{23}[/tex]
Pennies per person:
[tex]\frac{6.022\times 10^{23} pennies}{2.5\times 10^8 persons}=2.4088\times 10^{15} pennies/person[/tex]
1 penny = 0.01 $
[tex]2.4088\times 10^{15} pennies/person=2.4088\times 10^{15}\times 0.01 \$/person=2.4088\times 10^{13} \$/person[/tex]
[tex]2.4088\times 10^{13}[/tex] dollars each person will receive.
Which statement below matches the correct response with the proper reasoning when comparing the volatility of CH2Cl2 with CH2Br2? a. CH2Cl2 is more volatile than CH2Br2 because the dipole-dipole interactions in CH2Cl2 are greater than the dipole-dipole interactions in CH2Br2. b. CH2Cl2 is more volatile than CH2Br2 because of the large dispersion forces in CH2Br2. c. CH2Br2 is more volatile than CH2Cl2because because the dipole-dipole interactions in CH2Br2 are greater than the dipole-dipole interactions in CH2Cl2d. CH2Br2 is more volatile than CH2Cl2 because of the large dispersion forces in CH2Br2
Answer:
b. CH₂Cl₂ is more volatile than CH₂Br₂ because of the large dispersion forces in CH₂Br₂
Explanation:
CH₂Cl₂ is more volatile than CH₂Br₂ (b.p of CH₂Cl₂ = 39,6 °C; b.p of CH₂Br₂ = 96,95°C). Thus, c. and d. are FALSE
Dipole-dipole interactions in CH₂Cl₂ are greater than the dipole-dipole interactions in CH₂Br₂ because Cl is more electronegative that Br (Cl = 3,16; Br = 2,96). But this mean CH₂Cl₂ is less volatile than CH₂Br₂ but it is false.
There are large dispersion forces in CH₂Br₂ because Br has more electrons and protons than Cl. Large disperson forces mean CH₂Br₂ is less volatile than CH₂Cl₂ and it is true.
I hope it helps!
A sample of nitric acid has a mass of 8.2g. It is dissolved in 1L of water. A 25mL aliquot of this acid is titrated with NaOH. The concentration of the NaOH is 0.18M. What titre volume was added to the aliquot to achieve neutralisation?
Answer:
18.075 mL of NaOH solution was added to achieve neutralization
Explanation:
First, let's formulate the chemical reaction between nitric acid and sodium hydroxide:
NaOH + HNO3 → NaNO3 + H2O
From this balanced equation we know that 1 mole of NaOH reacts with 1 mole of HNO3 to achieve neutralization. Let's calculate how many moles we have in the 25 mL aliquot to be titrated:
63.01 g of HNO3 ----- 1 mole
8.2 g of HNO3 ----- x = (8.2 g × 1 mole)/63.01 g = 0.13014 moles of HNO3
So far we added 8.2 grams of nitric acid (0.13014 moles) in 1 L of water.
1000 mL solution ---- 0.13014 moles of HNO3
25 mL (aliquot) ---- x = (25 mL× 0.13014 moles)/1000 mL = 0.0032535 moles
So, we now know that in the 25 mL aliquot to be titrated we have 0.0032535 moles of HNO3. As we stated before, 1 mole of NaOH will react with 1 mole of HNO3, hence 0.0032535 moles of HNO3 have to react with 0.0032535 moles of NaOH to achieve neutralization. Let's calculate then, in which volume of the given NaOH solution we have 0.0032535 moles:
0.18 moles of NaOH ----- 1000 mL Solution
0.0032535 moles---- x=(0.0032535moles×1000 mL)/0.18 moles = 18.075mL
As we can see, we need 18.075 mL of a 0.18 M NaOH solution to titrate a 25 mL aliquot of the prepared HNO3 solution.
Which is more stable, a trans-1,4-disubstitutedcyclohexane or
its cis isomer?
Answer:
If the substituents are the same, the cis distribution is more stable than trans distribution.
Explanation:
A cis cyclohexane is one in which both substituents are oriented towards the same face of the ring regardless of the conformation.
A trans cyclohexane has substituents on opposite sides of the ring.
In trans-1,4-disubstituted cyclohexane, the conformation with the two substituent groups in equatorial is more stable than the constitution that has the axial groups.
In cis-1,4-disubstituted cyclohexane, both conformations have the same stability as they have an axial and an equatorial substituent, so they have the same energy and there is no equilibrium shift.
What is the temperature (°C) of 1.75 g of O2 gas occupying 3.10 L at 1.00 atm? Enter your answer in the box provided.
Answer:
691.29 K or 418.14 °C
Explanation:
Hello, at first the moles of oxygen gas are required:
[tex]n_{O_2}=1.75 g * \frac{1mol O_2}{32 g O_2} =0.0547 mol[/tex]
Now, based on the ideal gas equation, we solve for the temperature:
[tex]PV=nRT\\T=\frac{PV}{nR}\\T=\frac{1atm * 3.10 L}{0.0547 mol*0.082 \frac{atm*L}{mol*K} }\\T=691.29 K[/tex]
Best regards.
How many U.S. gallons are there in a cubic mile? The total proven oil reserves of the U.S. are roughly 30 x 10°bbl. How many cubic miles is this?
Answer:
1 cubic mile = 1.101 * 10^12 US gallons
1 US bbl oil = 42 US gallons = 3.8143*10^ -11 cubic miles
Explanation:
The number of the exponent of the oil reserve is not very well shown in the question so, I provide you the conversion of bbl oil into cubic mile, the only thing you have to do is multiply the number of bbls of the reserve for the conversion in cubic miles and you'll have the answer.
On a website devoted to answering engineering questions, viewers were invited to determine how much power a 100-MW power plant generates annually. The answer declared to be best was submitted by a civil engineering student, who stated, "It produces 100 MW/hr so over the year that's 100*24*365.25 & do the math." a. Carry out the calculation, showing all the units. Answer b. What is wrong with the statement of the question? c. Why was the student wrong in saying that the plant produces 100 MW/hr?
Answer:
(a) The plant generates 3,153,600,000 MJ a year.
(b) The problem with the question is that "power" is not "generated". What is generated is Energy, and Power is the rate of generation of Energy.
(c) The student is wrong because the plants generates 100 MJ of energy a second. "MW/hr" is not a unit of energy or power, so it has no sense.
Maybe he get confused with MW-h, which is a unit of energy, used to measure electrical consumption.
Explanation:
(a) The power of the plant (100 MW) is the rate at which electrical energy is produced. It has units of [energy]/[time].
In this case, the plant produces 100 MJ/s. The energy produced can also be expressed in other units, like MJh.
To calculate the energy generated in one year, we have
[tex]Energy = Power * Time = \\\\Energy = 100 MW*1year\\\\Energy = (100 \frac{MJ}{s})*(1 year*\frac{365 days}{1year}* \frac{24hours}{1day}*\frac{3600s}{1hour})\\\\Energy=(100 \frac{MJ}{s})*(31,536,000s)=3,153,600,000MJ[/tex]
The plant generates 3,153,600,000 MJ a year.
(b) The problem with the question is that "power" is not "generated". What is generated is Energy, and Power is the rate of generation of Energy.
(c) The student is wrong because the plants generates 100 MJ of energy a second. "MW/hr" is not a unit of energy or power, so it has no sense.
Maybe he get confused with MW-h, which is a unit of energy, used to measure electrical consumption. That unit represents the amount of energy consumed or generated by a 1 MW unit in one hour. It is equivalent to 3600 MJ (1 MW-h = 3600 MJ).
In this unit, the 100 MW plant generates 876,000 MWh.
The area of a telescope lens is 6.676 x 10 mm. (a) What is the area in square feet (ft)? Enter your answer in scientific notation. x 10ft (b) If it takes a technician 51.7 s to polish 1.46 x 10mm', how long does it take her to polish the entire lens?
Answer:
a) A = 7.186 E-4 ft²
b) t = 236.403 s
Explanation:
A = 6.676 E1 mm²
a) A = 6.676 E1 mm² * ( ft / 304.8 mm )²
⇒ A = 7.186 E-4 ft²
b) t = 51.7 s → A = 1.46 E1 mm²
⇒ relation (r) = 6.676 E1 mm² / 1.46 E1 mm² = 4.573
⇒ t = 51.7 s * 4.573 = 236.403 s
Henry law is obeyed by a gas when gas has high
a) Pressure
b) Temperature
c) Solubility
d) Non of the above
Answer:
Option b
Explanation:
Henry law describes solubility of gases in liquids.
According to Henry's law, amount of gas dissolved in a liquid depends upon its partial pressure above the liquids.
Mathematically, Henry's law is represented as:
C = K × P
Where,
C = Solubility of gas or concentration of gas in liquids
K = Henry's constant
P = Partial pressure of the gas over the liquid
For, Henry's law to be valid, pressure should be not too high and temperature should not be too low. Henry's law is also valid in case of low dissolved gas concentrations.
So, among the given options, option b, temperature is correct.
Consider an atom with 6 protons, 6 neutrons, and 6 electrons. State the number of protons, neutrons, and electrons that would occur in a different isotope of the same element, and explain what changed by relating the change to the definition of an isotope.
Explanation:
An isotope is defined as the specie which contains same number of protons but different number of neutrons.
For example, [tex]^{12}_{6}C[/tex] and [tex]^{13}_{6}C[/tex] are isotopes.
In a neutral carbon atom, there are 6 protons and 6 neutrons.
As it is known that in a neutral atom the number of protons equal to the number of electrons.
This means that in a neutral carbon atom there are also 6 electrons.
Whereas [tex]^{13}_{6}C[/tex] is an isotope of carbon atom whose atomic number is 6 and atomic mass is 13.
Hence, calculate the number of neutrons in [tex]^{13}_{6}C[/tex] as follows.
Atomic mass = no. of protons + no. of neutrons
13 = 6 + no. of neutrons
no. of neutrons = 13 - 6
= 7
Hence, in a [tex]^{13}_{6}C[/tex] isotope of carbon atom there are 6 protons, 6 electrons and 7 neutrons.
This shows that change in number of neutrons take place according to the definition of an isotope.
Calculate the pH of: (a) 0.1M HCl; (b) 0.1M NaOH; (c) 3 X 10% M HNO3; (d) 5 X 10-10 M HCIO.; and (e) 2 x 10-8 M KOH.
Answer:
(a) [tex]pH = -Log (0.1M) = 1[/tex]
(b) [tex]pH = -Log (10^{-13}M) = 13[/tex]
(c) [tex]pH = -Log (3x10^{-3}M) = 2.5[/tex]
(d) [tex]pH = -Log (4.93x10^{-10}M) = 9.3[/tex]
(e) [tex]pH = -Log (5^{-7}M) = 6.3[/tex]
Explanation:
To calculate de pH of an acid solution the formula is:
[tex]pH = -Log ([H^{+}]) = 1[/tex]
were [H^{+}] is the concentration of protons of the solution. Therefore it is necessary to know the concentration of the protons for every solution in order to solve the problem.
(a) and (c) are strong acids so they dissociate completely in aqueous solution. Thus, the concentration of the acid is the same as the protons.
(b) and (e) are strong bases so they dissociate completely in aqueous solution too. Thus, the concentration of the base is the same as the oxydriles. But in this case it is necessary to consider the water autoionization to calculate the protons concentration:
[tex]K_{w} =[H^{+} ][OH^{-}]=10^{-14}[/tex]
clearing the [tex][H^{+} ][/tex]
[tex][H^{+} ]=\frac{10^{-14}}{[OH^{-}]}[/tex]
(d) is a weak base so it is necessary to solve the equilibrium first, knowing [tex]Ka=3.24x10^{-8}[/tex]
The reaction is [tex]HClO[/tex] → [tex]H^{+} + CO^{-}[/tex] so the equilibrium is
[tex]Ka=3.24x10^{-8}=\frac{x^{2}}{5x10^{-8}-x}[/tex]
clearing the x
[tex]{x^{2}={1.62x10^{-17}-3.24x10^{-8}x}[/tex]
[tex]x=[H^{+}]=4.93x10^{-10}[/tex]
How many significant figures are in the following? a) 0.1111 b) 2000 c) 35.6 d) 180,701
Answer:
a) 0.1111 : 4 significant figures
b) 2000 : 1 significant figure
c) 35.6 : 3 significant figures
d) 180,701 : 6 significant digits
Explanation:
Significant figures or digits of a number are the digits that carry meaning and contribute to the precision of the number.
a) 0.1111 : 4 significant figures, leading zeros are not significant digits.
b) 2000 : 1 significant figure, trailing zeros are not significant.
c) 35.6 : 3 significant figures, all digits are significant.
d) 180,701 : 6 significant digits, zeros between mom-zero digits are significant.
The number of significant figures in the given examples are:
a) 4 significant figures for 0.1111
b) 1 significant figure for 2000 (assuming no decimal)
c) 3 significant figures for 35.6
d) 6 significant figures for 180,701
To determine the number of significant figures in a number, certain rules are followed:
Non-zero digits are always significant.Any zeros between significant digits are significant.Leading zeros (zeros before non-zero digits) are not significant.Trailing zeros in a decimal number are significant.In a number without a decimal point, trailing zeros may or may not be significant, depending on whether a decimal point is specified or assumed.Applying these rules, we have:
0.1111 - All digits are significant because they are non-zero, giving 4 significant figures.2000 - Without additional context or a decimal point, we cannot be certain if the trailing zeros are significant, so this number has 1 significant figure.35.6 - All digits including the zero are significant, giving 3 significant figures.180,701 - All digits are significant, giving 6 significant figures.A 0.1153-gram sample of a pure hydrocarbon was burned in a C-H combustion train to produce 0.3986 gram of CO2and 0.0578 gram of H2O. Determine the masses of C and H in the sample and the percentages of these elements in this hydrocarbon.
Answer: The mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:
[tex]C_xH_y+O_2\rightarrow CO_2+H_2O[/tex]
where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.
We are given:
Mass of [tex]CO_2=0.3986g[/tex]
Mass of [tex]H_2O=0.0578g[/tex]
Mass of sample = 0.1153 g
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 0.3986 g of carbon dioxide, [tex]\frac{12}{44}\times 0.3986=0.1087g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:In 18g of water, 2 g of hydrogen is contained.
So, in 0.0578 g of water, [tex]\frac{2}{18}\times 0.0578=0.0066g[/tex] of hydrogen will be contained.
To calculate the percentage composition of a substance in sample, we use the equation:
[tex]\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}\times 100[/tex] ......(1)
For Carbon:Mass of sample = 0.1153 g
Mass of carbon = 0.1087 g
Putting values in equation 1, we get:
[tex]\%\text{ composition of carbon}=\frac{0.1087g}{0.1153g}\times 100=94.27\%[/tex]
For Hydrogen:Mass of sample = 0.1153 g
Mass of hydrogen = 0.0066 g
Putting values in equation 1, we get:
[tex]\%\text{ composition of hydrogen}=\frac{0.0066g}{0.1153g}\times 100=5.72\%[/tex]
Hence, the mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.
The masses of C and H in the sample are 0.1083 grams and 0.00324 grams, respectively. The percentages of C and H in the hydrocarbon are approximately 93.94% and 2.80%, respectively.
All the carbon in the hydrocarbon is converted to [tex]CO_2[/tex], we can use the mass of [tex]CO_2[/tex] to find the mass of carbon:
[tex]\[ \text{Mass of carbon} = \text{Mass of CO2} \times \frac{\text{Molar mass of carbon}}{\text{Molar mass of CO2}} \][/tex]
[tex]\[ \text{Mass of carbon} = 0.3986 \text{ g} \times \frac{12.01 \text{ g/mol}}{44.01 \text{ g/mol}} \][/tex]
[tex]\[ \text{Mass of carbon} = 0.3986 \text{ g} \times \frac{12}{44} \][/tex]
[tex]\[ \text{Mass of carbon} = 0.3986 \text{ g} \times 0.2727 \][/tex]
[tex]\[ \text{Mass of carbon} = 0.1083 \text{ g} \][/tex]
All the hydrogen in the hydrocarbon is converted to [tex]H_2O[/tex], we can use the mass of [tex]H_2O[/tex] to find the mass of hydrogen:
[tex]\[ \text{Mass of hydrogen} = \text{Mass of H2O} \times \frac{\text{Molar mass of hydrogen}}{\text{Molar mass of H2O}} \][/tex]
[tex]\[ \text{Mass of hydrogen} = 0.0578 \text{ g} \times \frac{1.008 \text{ g/mol}}{18.015 \text{ g/mol}} \][/tex]
[tex]\[ \text{Mass of hydrogen} = 0.0578 \text{ g} \times \frac{1}{18} \][/tex]
[tex]\[ \text{Mass of hydrogen} = 0.0578 \text{ g} \times 0.05597 \][/tex]
[tex]\[ \text{Mass of hydrogen} = 0.00324 \text{ g} \][/tex]
Now that we have the masses of carbon and hydrogen, we can calculate the percentages of these elements in the hydrocarbon:
[tex]\[ \text{Percentage of carbon} = \frac{\text{Mass of carbon}}{\text{Mass of hydrocarbon}} \times 100\% \][/tex]
[tex]\[ \text{Percentage of carbon} = \frac{0.1083 \text{ g}}{0.1153 \text{ g}} \times 100\% \][/tex]
[tex]\[ \text{Percentage of carbon} \approx 93.94\% \][/tex]
[tex]\[ \text{Percentage of hydrogen} = \frac{\text{Mass of hydrogen}}{\text{Mass of hydrocarbon}} \times 100\% \][/tex]
[tex]\[ \text{Percentage of hydrogen} = \frac{0.00324 \text{ g}}{0.1153 \text{ g}} \times 100\% \][/tex]
[tex]\[ \text{Percentage of hydrogen} \approx 2.80\% \][/tex]
The half-life of a pesticide determines its persistence in the environment. A common pesticide degrades in a first-order process with a half-life of 2 days. What fraction (in decimal notation) of the pesticide remains in the environment after 18 days? Enter to 4 decimal places.
Answer:
0.0020 fraction of the pesticide remains in the environment after 18 days
Explanation:
For a first order reaction, [tex]\frac{N}{N_{0}}=(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}[/tex]
where N is remaining mass after "t" time , [tex]N_{0}[/tex] is initial mass, [tex]\frac{N}{N_{0}}[/tex] represents fraction of mass remains after "t" time and [tex]t_{\frac{1}{2}}[/tex] is half-life
Here t is 18 days and [tex]t_{\frac{1}{2}}[/tex] is 2 days
So, [tex]\frac{N}{N_{0}}=(\frac{1}{2})^{\frac{18}{2}}=(\frac{1}{2})^{9}=0.0020[/tex]
Hence 0.0020 fraction of the pesticide remains in the environment after 18 days
The fraction (in decimal notation) of the pesticide that remains in the environment after 18 days is 0.0020
We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
Half-life (t½) = 2 days
Time (t) = 18 days
Number of half-lives (n) =?n = t / t½
n = 18 / 2
n = 9Thus, 9 half-lives has elapsed.
Finally, we shall determine the fraction of the pesticide that remains in the environment. This can be obtained as follow:Original amount (N₀) = 1
Number of half-lives (n) = 9
Fraction remaining (N / N₀) =?N = 1/2ⁿ × N₀
Divide both side by N₀
N / N₀ = 1/2ⁿ
N / N₀ = 1/2⁹
N / N₀ = 0.0020Thus, the fraction of the pesticide that remains in the environment is 0.0020
Learn more: https://brainly.com/question/20516965
Forces between similar molecules are said to be
---Select---cohesiveadhesiveconcaveconvex while those between
different types ofmolecule are said to be
---Select---cohesiveadhesiveconcaveconvex. Water 'beads' due to its
strong---Select---cohesiveadhesiveconcaveconvex forces. The
meniscus of water in a glasstube is
---Select---cohesiveadhesiveconcaveconvex because the
---Select---cohesiveadhesiveconcaveconvex forces are strong.
Answer:
Forces between similar molecules are said to be cohesive while those between different types of molecules are said to be adhesive.
Water 'beads' due to its strong cohesive forces. The meniscus of water in a glass tube is concave because the adhesive forces are strong.
Explanation:
The water in a tube has stronger adhesive forces between the water and glass molecules, so the cohesive forces between water molecules are weaker. That makes the water 'ascend' through the tube, giving a concave form of the meniscus. Another example is mercury, which is the opposite. In this case, the cohesive forces are stronger than the adhesive ones, thus the meniscus is convex.
How many ml of concentrated hydrochloric acid would be required to make 1 L of a 0.2 M solution? Assume that concentrated hydrochloric acid is 12.1 M
Answer: The volume of concentrated hydrochloric acid required is 16.53 mL
Explanation:
To calculate the volume of concentrated solution, we use the equation:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the concentrated solution
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted solution
We are given:
Conversion factor: 1 L = 1000 mL
[tex]M_1=12.1M\\V_1=?mL\\M_2=0.2M\\V_2=1L=1000mL[/tex]
Putting values in above equation, we get:
[tex]12.1\times V_1=0.2\times 1000\\\\V_1=16.53mL[/tex]
Hence, the volume of concentrated hydrochloric acid required is 16.53 mL
a 2.60 g sample of titanium metal chemically combines
withchlorine gas to form 10.31g of a titanium chloride.
a. What is the empirical formula of the titaniumchloride
b. What is the percent by mass of titanium and chlorine in
thesample?
Answer:
For a: The empirical formula for the given compound is [tex]TiCl_4[/tex]
For b: The percent by mass of titanium and chlorine in the sample is 25.55 % and 74.78 % respectively.
Explanation:
For a:We are given:
Mass of Titanium = 2.60 g
Mass of sample = 10.31 g
Mass of Chlorine = 10.31 - 2.60 = 7.71 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Moles of titanium =[tex]\frac{\text{Given mass of Titanium}}{\text{Molar mass of Titanium}}=\frac{2.60g}{47.867g/mole}=0.054moles[/tex]
Moles of Chlorine = [tex]\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{7.71g}{35.5g/mole}=0.217moles[/tex]
Step 2: Calculating the mole ratio of the given elements.For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.054 moles.
For Titanium = [tex]\frac{0.054}{0.054}=1[/tex]
For Chlorine = [tex]\frac{0.217}{0.054}=4.01\approx 4[/tex]
Step 3: Taking the mole ratio as their subscripts.
The ratio of Ti : Cl = 1 : 4
Hence, the empirical formula for the given compound is [tex]TiCl_4[/tex]
For b:To calculate the percentage by mass of substance in sample, we use the equation:
[tex]\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}\times 100[/tex] .......(1)
For Titanium:Mass of sample = 10.31 g
Mass of titanium = 2.60 g
Putting values in above equation, we get:
[tex]\%\text{ composition of titanium}=\frac{2.60g}{10.31g}\times 100=25.22\%[/tex]
For Chlorine:Mass of sample = 10.31 g
Mass of chlorine = 7.71 g
Putting values in above equation, we get:
[tex]\%\text{ composition of chlorine}=\frac{7.71g}{10.31g}\times 100=74.78\%[/tex]
Hence, the percent by mass of titanium and chlorine in the sample is 25.55 % and 74.78 % respectively.
(c) Draw the structure of any two aminoacids which bears a heterocyclic unit as all or part of the R side chain of the aminoacid, and highlight and name the particular heterocyclic unit present in the aminoacid.
Answer:
Here I show you tryptophan and tyrosine
Explanation:
of the 21 amino acids, there is five amino acid that have a heterocyclic group as part of the R side chain: Histidine, proline, Phenylalanine, tyrosine, and tryptophan. Each one has a unique nature and the heterocyclic group, mainly in tyrosine, and tryptophan allows to absorb the UV light (280 nm)
Nault 25000L 250 our = 25000L 15040 8. How many grams of CaCl2 are needed to make 150.0 mL of a 0.500 M CF solution? (Note: CaCl2 is a soluble salt. The molar mass of CaCl2 is 110.98 g/mol.)
Answer:
You need 8,324 g of CaCl₂ yo make this solution
Explanation:
Molarity is a way to express concentration in a solution, in units of moles of solute per liter of solution.
To know the grams of CaCl₂ it is necessary to know, first, the moles of this substance with the desired volume and concentration , thus:
0,1500 L × [tex]\frac{0,500 mol}{L}[/tex] = 0,075 CaCl₂ moles
Now, with the molar mass of CaCl₂ you will obtain the necessary grams, thus:
0,075 CaCl₂ moles × [tex]\frac{110,98 g}{mol}[/tex] = 8,324 g of CaCl₂
So, you need 8,324 g of CaCl₂ to make 150,0 mL of a 0,500M solution
I hope it helps!
Menthol (molar mass = 156.3 g/mol), the strong-smelling substance in many cough drops, is a compound of carbon, hydrogen and oxygen. When 0.1595g of menthol was burned in a combustion apparatus, 0.449g of CO2 and 0.184g of H2O formed. What is menthol’s molecular formula?
Answer: The molecular formula for the menthol is [tex]C_{10}H_{20}O[/tex]
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of [tex]CO_2=0.449g[/tex]
Mass of [tex]H_2O=0.184g[/tex]
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 0.449 g of carbon dioxide, [tex]\frac{12}{44}\times 0.449=0.122g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:In 18g of water, 2 g of hydrogen is contained.
So, in 0.184 g of water, [tex]\frac{2}{18}\times 0.184=0.0204g[/tex] of hydrogen will be contained.
Mass of oxygen in the compound = (0.1595) - (0.122 + 0.0204) = 0.0171 gTo formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.122g}{12g/mole}=0.0102moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0204g}{1g/mole}=0.0204moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0171g}{16g/mole}=0.00107moles[/tex]
Step 2: Calculating the mole ratio of the given elements.For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00107 moles.
For Carbon = [tex]\frac{0.0102}{0.00107}=9.53\approx 10[/tex]
For Hydrogen = [tex]\frac{0.0204}{0.00107}=19.54\approx 20[/tex]
For Oxygen = [tex]\frac{0.00107}{0.00107}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.The ratio of C : H : O = 10 : 20 : 1
Hence, the empirical formula for the given compound is [tex]C_{10}H_{20}O_1=C_{10}H_{20}O[/tex]
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is :
[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]
We are given:
Mass of molecular formula = 156.3 g/mol
Mass of empirical formula = 156 g/mol
Putting values in above equation, we get:
[tex]n=\frac{156.3g/mol}{156g/mol}=1[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O[/tex]
Thus, the molecular formula for the menthol is [tex]C_{10}H_{20}O[/tex]
In an evaporator 25 ton / h of a solution of 10% NaOH, 10% NaCl, and 80% water by weight. During evaporation, the water evaporates and the salt precipitates like crystals They are allowed to settle and are removed. The outgoing concentrated solution of the Evaporator contains 50% NaOH, 2% NaCl and 48% water. Based on This information is requested:
1. Draw the process flow diagram, indicating each of its streams and compositions (known and unknown).
2. Calculate the kilograms of precipitated salt and the kilograms of solution concentrated for every hour of work.
Answer:
1- Flow Diagram (file attached)
2- Kilograms of precipitated salt: 2400kg/h
Kilograms of solution concentrated per hour: 5000kg/h
Explanation:
We have an Evaporator with 25ton/h of a solution with: 10% NaOH, 10% NaCl, and 80% water. That means that we have an input which is a stream with the following composition fractions:
Stream 1:
NaOH= 0.1
NaCl= 0.1
Water= 0.8
Then we have 3 outputs, 3 streams that leave the evaporator, which are:
Stream 2:
Only contains water so its composition fraction of water is 1.
Stream 3:
Only contains NaCl so its composition fraction of NaCL is 1.
Stream 4:
NaOH= 0.5
NaCl= 0.02
Water= 0.48
We know that stream 1 is 25 ton/h and enter the evaporator but we do not know the flow rate of stream 4 that is the concentrated solution leaving the evaporator, so we will make a particular mass balance of the component NaOH that is present in both streams:
fraction of NaOH in stream 1 ×flow rate of stream 1= fraction of NaOH in stream 4× flow rate of stream 4
0.1×25 ton/h = 0.5× flow rate of stream 4
flow rate of stream 4= (0.1×25 ton/h)/0.5= 5ton/h= 5000kg/h
Now to know the kilograms of precipitated salt, which is the flow rate of stream 3 we make a particular mass balance of the component NaCl:
(fraction of NaCl in stream 1 ×flow rate of stream 1)- (fraction of NaCl in stream 3×flow rate of stream 3)=flow rate of stream 4× fraction of NaCl in stream 4
(0.1×25 tn/h)- (1×flow rate of stream 3)= 5tn/h × 0.02
flow rate of stream 3= 2.4 tn/h =2400 kg/h