Answer:
The molar mass of gas Q is 43.923 g/mol
Explanation:
The given volume of ethane gas that diffuses through a porous plug in 100 seconds = 100 cm³
Therefore;
The rate of diffusion of ethane gas through the porous plug, [tex]v_{ethane}[/tex], is given as follows;
[tex]v_{ethane}[/tex] = (100 cm³/100 s) = 1 cm³/s
The molar mass of ethane, C₂H₆ = 2×12 g/mol + 6×1 g/mol = 30 g/mol
The given volume of gas, Q, that diffuses through a porous plug in 121 seconds = 100 cm³
∴ The rate of diffusion of the gas, Q, [tex]v_Q[/tex] = 100/121 cm³/s
Graham's Law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of the molecular mass of the gas
Mathematically, we have;
[tex]\dfrac{v_A}{v_B} =\sqrt{\dfrac{m_B}{m_A} }[/tex]
Where;
[tex]v_A[/tex] = The rate of diffusion of gas A
[tex]v_B[/tex] = The rate of diffusion of gas B
[tex]m_A[/tex] = The molar mass of the gas A
[tex]m_B[/tex] = The molar mass of the gas B
Therefore, for ethane and gas Q, measured under the same condition, we have;
[tex]\dfrac{v_{ethane}}{v_Q} =\sqrt{\dfrac{m_Q}{m_{ethane}} }[/tex]
[tex]\dfrac{1 \ cm^3/s}{\dfrac{100}{121} \ cm^3/s} =\sqrt{\dfrac{m_Q}{30 \ g/mol} }[/tex]
[tex]m_Q = \left ({\dfrac{121}{100} } \right) ^2 \times 30 \ g/mol = 43.923 \ g/mol[/tex]
The molar mass of gas Q, [tex]m_Q[/tex] = 43.923 g/mol.
describe how thin layer chromatography is used in the isolation and extraction of lipids
Thin layer chromatography(TLC) works with the principle of separation through adsorption.
It is used in the isolation and extraction of lipids through the following steps:
apply the lipid samples spots in the bottom of the plate.also apply sample solution to the marked spotpour the mobile phase into the TLC chamber and use a moist filter paper to cover it. this is done to maintain equal humidity.then place the plate in the TLC chamber and close it with a lid.the plate is immersed into the solvent (mobile phase) for its development. this is done, keeping in mind that the sample spot should be above the solvent.once the sample spots are developed, they are removed and dried.this is later viewed using the UV light chamber to see the isolation of the lipid sample.Learn more here:
https://brainly.com/question/3137660
What are 5 uses of nitrogen?
Answer:
nitrogen is used in the production of 1) fertilisers 2) nitric acid 3) nylon 4) dyes and 5) explosives
please give me brainliest if you can :))
5+64
this is TOTALLY SOOOO hard so help me
^thats what she said
Answer:
69
Explanation:
I thought that I answer is correct
Answer:
The answer is 69
Explanation:
Because if you add one its 66, and so on, hope this helps! but if you add five it will be 69, if you need help with easy equations let me know
A student was given a 2.850-g sample of a mixture of potassium nitrate and potassium bromide and was asked to find the percentage of each compound in the mixture. She dissolved the sample and added a solution that contained an excess of silver nitrate, AgNO3. The silver ion precipitated all of the bromide ion in the mixture as AgBr. It was filtered, dried, and weighed. Its mass was 1.740 g. What was the percentage of each compound in the mixture
Answer:
See explanation
Explanation:
The reaction occurs as follows;
KBr(aq) + AgNO3(aq) ----> AgBr(s) + KNO3(aq)
Number of moles of AgBr formed = mass /molar mass =1.740 g/187.77 g/mol = 0.0093 moles
From the reaction equation;
1 mole of KBr yields 1 mole of AgBr
Hence the number of moles of KBr reacted = 0.0093 moles
Mass of KBr present = 0.0093 moles × 119g/mol = 1.11 g
Mass of KNO3 = 2.850 g - 1.11 g = 1.74 g
Percentage of KBr = 1.11 g/2.850 g × 100 = 38.9%
Percentage of KNO3 = 1.74 g/2.850 g × 100 = 61.1%
I need help!!!! please
6.94 is much more closet to 7.016, so Li 7 is more abundant
When 120 g of carbon reacts completely with 320 g of oxygen the mass of carbon dioxide formed will be?
Liquids which do not wer the glass surface form a convex meniscus.True or false
Answer:
true is your answer
Explanation:
hope it helps
A compound was found to contain 90.6% lead (Pb) and 9.4% oxygen. What is the empirical formula for this compound?
Answer:
the answer is 47.9 and ik because I just had that question
The empirical formula of the compound is O₄Pb₃.
What is the empirical formula?
An Empirical system is the chemical system of a compound that offers the proportions of the elements gifted within the compound however not the real numbers or arrangement of atoms. This would be the lowest complete variety ratio of the elements within the compound.
Amount of lead (Pb) = 90.6%
⇒and amount of oxygen = 9.4%
taking the whole number ratio
o = 4
Pb = 3
∴ ⇒O: Pb=4:3
O4Pb3 answer.
Learn more about empirical formula here:-https://brainly.com/question/1603500
#SPJ2
two easy uses of mixture
Explanation:
it helps to make juices.
It helps to make concentrated acid into dilute acid.
Answer:
Explanation:
Here are a few more examples:
1. Smoke and fog (Smog)
2. Dirt and water (Mud)
Which of the following is NOT a physical property?
Viscosity
Reactivity
Malleability
Density
Answer:
reactivity is NOT a physical property
Define force and speed
Force = The external energy that changes or tends to change the state of any body or object is called force.
Speed = The rate of distance is called speed.
Answer:
Force is a push or pull which changes or tend to change the position of a body.
The rate of change its position with time or magnitude is called speed.
PLEASE HELP ASAP
Barney was a very inventive scientist and tried to perform the same experiments as those by famous scientists from long ago. One of the most challenging experiments was to roll out a very thin film of gold and put X-ray film in a circle around it. The difficult part was to get ahold of some radioactivity that he could shoot at the thin film for gold. One day, he ran into The Great Gazoo and told him his goal. The Great Gazoo said he had the perfect substance for him that was radioactive and should work. The element was named after him and had the symbol Gz. Barney quickly went back to his lab to run the experiment. What Barney noticed is that all the radioactive particles went through the gold film and none of them ricocheted back, but all went straight through the gold. Being puzzled, he quickly went to the lab to analyze this unknown radioactive substance. He found that the atomic number was 119, the mass number was 305. After a few more tests, Barney realized what was wrong with Gz. Your task is to figure out what was wrong with Gz by answering the following questions:
Whose experiment was Barney trying to imitate?
Where would Gz be located on Earth’s periodic table, column and row?
Give as many details on Gz that you could predict based on its location on the periodic table: type of chemical (metal, metalloid, or nonmetal), reactivity, ionization energy, electronegativity, and size of element.
What would the elements electron configuration end in?
Why did all the radioactive particles go straight through the gold film?
What would the daughter product be for Gz? Give all details for the atomic number, mass number, number of protons, neutrons and electrons.
If Gz had worked correctly, like Barney intended, what type of decay should Gz have had?
If Gz had worked correctly, what would the daughter product be? Give all details for the atomic number, mass number, number of protons, neutrons and electrons.
If Gz had worked correctly, what element would it become?
Answer:
a) Barney wants to repeat the rutheford experiment
b) the element will be in row 8 column 1 being an alkali metal
c) it should be a radioactive element, with a lot of mass
Explanation:
n this exercise they indicate that the particle Gz has atomic number 119 and atomic mass 305 amu, when reviewing the element periodicity table with this atomic number it has not yet been discovered, it should be in row 8 column 1 therefore it should be an alkali metal .
Therefore, it has only one electron in its last orbit.
a) Barney wants to repeat the rutheford experiment
b) the element will be in row 8 column 1 being an alkali metal
c) it should be a radioactive element, with a lot of mass
Hydrogen and oxygen combine in
the ratio of 1:8 by mass to form
water. What mass of oxygen gas
would be required to react
completely with 3 g of hydrogen
gas?
Answer:
hhshshhxujejushwhbwhsjs s hs
Write the empirical formula for at least four ionic compounds that could be formed from the following ions: NO3, Pb^4+, NH4, SO4
Explanation:
here's the answer to your question
double arrow just mean that it's a reversible process, and the reaction can go back and forth.
hii pls help me to balance the equation and state the symbols thanksss
Ca(s)+ H2O (l)----->H2 +Ca(oH)2
Ca(s)+2H2O(l)----->H2(g)+Ca(oH)2
I hope this helps
sorry if it's wrong
The reactivity of elements can be predicted based on the position of an element in the periodic table. Based on the periodic table, which of the following elements would be expected to be the most reactive?
Iodine (I)
Iodine (I)
Chlorine (Cl)
Chlorine (Cl)
Fluorine (F)
Fluorine (F)
Bromine (Br)
Answer:
Chlorine (Cl) is most reactive out the the listed elements.
Answer:
iodine
Explanation:
It is the most reactive on the periodic table.
16. The valency of sodium is +1 and that of chlorine is -1, why?
Answer:
It because ,sodium is a metal and chlorine is a non metal
A piece of wood displaces 12.5 mL of water and has a density of 0.97 g/ml. what is the mass of the piece of wood? what will happen if the wood is placed in water?
HELP PLEASE! ASAP YOUR DA BEST! 15 POINTS YALL
Density=0.97g/mL
Volume=12.5mL
[tex]\\ \sf \longmapsto Density=\dfrac{Mass}{Volume}[/tex]
[tex]\\ \rm\longmapsto Mass=Density(Volume)[/tex]
[tex]\\ \sf \longmapsto Mass=0.97(12.5)[/tex]
[tex]\\ \sf \longmapsto Mass=12.125g[/tex]
A piece of sodium metal can be described as?
which chloride is a coloured solid rtp
Answer:
sodium chloride
hope that helped :)
Answer:
Chlorine is a greenish yellow gas at room temperature and atmospheric pressure. ... yielding chlorine water, and from this solution a solid hydrate of ideal ...
If the temperature of a reversible reaction in dynamic equilibrium decreases, how will the equilibrium change?
A. It will shift towards the endothermic reaction.
B. It will not change.
C. It will shift towards the products.
D. It will shift towards the exothermic reaction.
Answer:
D. It will shift towards the exothermic reaction.
Explanation:
water and development are substitute of one another.
explanation with it, please
9. During an experiment the students prepared three mixtures A)Starch in water B) Sodium chloride solution C) Tincture of Iodine. i) Students observed a visible beam of light through mixture A. Why? ii) Tincture of lodTe did not show Tyndall effect. Explain reason. ill) How can you relate particle size to Tyndall effect?
Answer:
See explanation
Explanation:
Tyndall effect refers to the scattering of light in a solution. Tyndall effect occurs when the size of particles in the solution exceeds 1 nm in diameter. Such solutions are actually called false solutions.
In tincture of iodine, the size of particles in solution is less than 1 nm in diameter hence the solution does not exhibit Tyndall effect. Hence, tincture of iodine is a true solution.
Therefore, if the size of particles in solution exceeded 1nm in diameter, Tyndall effect is observed.
A chemist sets up a chemical reaction but finds that none of the reactant molecules have the required activation energy. What is the result?
A.
Products will form with less energy input.
B.
No products will be formed.
C.
The products will form too quickly.
D.
Products will convert to reactants.
Answer:
B.
No products will be formed.
At the beginning of the reaction: N2 + 3H2 -> 2NH there is 1 mol of N. and 3 moles of H, and at the end of it is present a mixture formed by 2.5 moles in total, what is the performance for the reaction?
5.- Al comenzar la reacción: N2+ 3H2 -> 2NH existe 1 mol de N. y 3 moles de H, y al finalizarla está presente una mezcla formada por 2,5 moles en total, ¿cuál es el rendimiento para la reacción?
Answer:
75%
Explanation:
call realistic moles of N2 in the reaction is a
moles of H2 :3a
moles of NH3: 2a
the moles of residual N2 after reaction: 1-a
the moles of residual H2 after react: 3- 3a
the moles of mixture: 2a+1-a+3-3a=2.5
-> a=0.75
H=0.75/1=75%
A chemist determines by measurements that 0.0550 moles of nitrogen gas participate in a chemical reaction. Calculate the mass of nitrogen gas that participates.
Be sure your answer has the correct number of significant digits.
The mass of nitrogen gas that participated in the chemical reaction is 1.54g
HOW TO CALCULATE MASS OF AN ELEMENT:
Mass of a substance can be calculated by multiplying the number of moles in mol of the substance by its molecular mass in g/mol. That is;mass (M) = molar mass (MM) × number of moles (n)According to this question, a chemist determines by measurements that 0.0550 moles of nitrogen gas (N2) participate in a chemical reaction.
The molecular mass of nitrogen gas (N2) = 14.01(2)= 28.02g/mol
Hence, the mass of the nitrogen gas that participated in the chemical reaction is calculated as follows:
Mass (g) = 0.0550 mol × 28.02 g/molMass = 1.5411Therefore, the mass of nitrogen gas that participated in the chemical reaction is 1.54g
Learn more: https://brainly.com/question/18269198
Describe the formation of oxygen molecule
Answer:
oxygen molecule has two oxygen atom . Each O atom share 2 electrons to form two covalent bonds out of which one is sigma bond and other is pi bond . sigma bond is formed by axial overlap 2p atomic orbitals of oxygen and pi bond is formed of lateral overlap of 2p atomic orbitals of oxygen .
Which statements are true in regard to the VSEPR theory?
Select all that apply.
Molecules acquire a shape that results in the greatest charge.
Molecules acquire a shape that minimizes the repulsions of electron groups.
Molecules acquire a shape that maximizes the distance of electron groups.
Molecules acquire a shape that results in the lowest possible energy state.
The correct option is D. 2) Anti-bonding electrons or lone pairs. These lone pairs, and bonds helps to form the shape which keeps these electrons separate as possible.
The gases in a hair spray can are at a temperature of 27oC and a pressure of 30 lbs/in2. If the
gases in the can reach a pressure of 90 lbs/in2, the can will explode. To what temperature must
the gases be raised in order for the can to explode? Assume constant volume. Show your work.
Using Gay Lussac's Law
[tex]\boxed{\sf \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}}[/tex]
[tex]\\ \sf\longmapsto \dfrac{30}{27}=\dfrac{90}{T_1}[/tex]
[tex]\\ \sf\longmapsto T_1=\dfrac{90\times 27}{30}[/tex]
[tex]\\ \sf\longmapsto T_1=81°C[/tex]
Key Notes:-Gay Lussac's Law:-
It states that at constant volume (V),The pressure(P) of fixed amount of gas caries directly with its absolute temperature(T).
[tex]\\ \sf\longmapsto P\propto T[/tex]
[tex]\\ \sf\longmapsto \dfrac{P}{T}=Constant[/tex]
[tex]\\ \sf\longmapsto \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}[/tex]
[tex]\\ \sf\longmapsto P_1T_2=P_2T_1[/tex]
The final temperature of the gases with 90 lbs/in² is equal to 627°C assuming the volume is constant.
What is Gay Lussac's law?Gay-Lussac's law can be described as if the volume (V) of the gas remains constant then the pressure (P) is directly proportional to the absolute temperature (T) of the gas.
Gay Lussca's law in mathematical expression can be represented as:
P/T = k
It can be also represented as the pressure (P) being directly proportional to the temperature (T).
P ∝ T ( V of gas is constant)
or, P₁/T₁ = P₂/T₂
Given, the initial temperature of the gas, T₁ = 27°C = 27 + 273 = 300 K
The initial pressure of the gas, P₁ = 30 lbs/in²
The final pressure of the gas, P₂ = 90 lbs/in²
The final temperature of the gases in a hair spray can be determined as:
[tex]{\displaystyle \frac{P_1}{T_1} =\frac{P_2}{T_2}[/tex]
[tex]{\displaystyle T_2 =\frac{P_2\times T_1}{P_1}[/tex]
[tex]{\displaystyle T_2 =\frac{90\times 300}{30}[/tex]
T₂ = 900 K = 900 - 273
T₂ = 627°C
Learn more about Gay Lussac's law, here:
brainly.com/question/11387781
#SPJ2
Oxide of nitrogen that is acidic
Answer:
Nitrogen oxides are used in the production of nitric acid, lacquers, dyes, and other chemicals. Nitrogen oxides are used in rocket fuels, in the nitrification of organic chemicals, and in the manufacture of explosives.
Explanation: