50 POINTS PLEASE HELP!
A gas diffuses 1/7 times faster than hydrogen gas (H2).
What is the molar mass of the gas?
100.10 g/mol
98.78 g/mol
86.68 g/mol
79.98 g/mol
Answer: The molar mass of the gas is 9.878 g/mol.
Explanation:
According to Graham's law, the rate of diffusion is inversely proportional to square root of molar mass of gas.
[tex]Rate = \frac{1}{\sqrt{M}}[/tex]
where,
M = molar mass of gas
As given gas diffuses 1/7 times faster than hydrogen gas. So, its molar mass is calculated as follows.
[tex]\frac{R_{1}}{R_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\\[/tex]
where,
[tex]M_{1}[/tex] = molar mass of hydrogen gas
[tex]M_{2}[/tex] = molar mass of another given gas
[tex]R_{1}[/tex] = rate of diffusion of hydrogen
[tex]R_{2}[/tex] = rate of diffusion of another given gas = [tex]\frac{1}{7}R_{1}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{R_{1}}{R_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\\\frac{R_{1}}{\frac{1}{7}R_{1}} = \sqrt{\frac{M_{2}}{2}}\\7 \times 1.414 = M_{2}\\M_{2} = 9.878 g/mol[/tex]
Thus, we can conclude that the molar mass of the gas is 9.878 g/mol.
Answer:
98.78 g/mol
Explanation:
Just took the test
Can you pls answer This question
Answer:
a) pentane
b) 1,3-dimethyl butane
c) 3,4-diethyl-5-methlyl heptane
d) 1,3,5-triethyl-2,4,6-trimethyl cyclohexane
Explanation:
The essence of IUPAC nomenclature is to provide a generally accepted nomenclature for chemical compounds which enables the structure of a compound to be drawn from its name without ambiguity.
In order to name straight chain organic compounds, we have to identify the parent chain and then give numbers to the substituents or locants such that they have the lowest possible numbers in alphabetical order.
The correct names of the organic compounds in the question are;
a) pentane
b) 1,3-dimethyl butane
c) 3,4-diethyl-5-methlyl heptane
d) 1,3,5-triethyl-2,4,6-trimethyl cyclohexane
When something is hot , its atom has more energy
True or False
Will mark as brainliest if the answer is correct
Select the mixture which is capable of tolerating addition of a strong acid without altering the pH? a.) 1.0 M HCl, 1.0 M NaCl b.) 1.0 M NH3,1.0 M NaCl c.) 1.0 M HF, 1.0 M NaCl d.) 1.0 M HNO2, 1.0 M NaNO2 e.) none
Answer:
d.) 1.0 M HNO2, 1.0 M NaNO2
Explanation:
A buffer is a mixture that is capable of tolerating addition of a strong acid or base without altering the pH substantially.
A buffer is the mixture of a WEAK ACID AND ITS CONJUGATE BASE or a WEAK BASE AND ITS CONJUGATE ACID
In the problems:
a) HCl is a strong acid and NaCl a salt. NO buffer
b) NH3 is a weak base and NaCL a salt. NO buffer
c) HF is a weak acid and NaCl a salt. NO buffer.
d) HNO2 is a weak acid and NaNO2 its conjugate base. A BUFFERGather data: experiment with a variety of initial concentrations of no2 and n2o4. for each set of initial concentrations, use the gizmo to determine the equilibrium concentrations of each substance. in the last column, find kc for that trial. run three trials for each set of initial conditions.
Solution :
[tex]$H_2 +I_2\rightleftharpoons 2HI$[/tex]
Initial concentration Equilibrium concentration [tex]$K_c=\frac{[HI]^2}{[H_2][I_2]}$[/tex]
(mol/L) (mol/L)
[tex]$[H_2]$[/tex] [tex]$[I_2]$[/tex] [tex]$[HI]$[/tex] [tex]$[H_2]$[/tex] [tex]$[I_2]$[/tex] [tex]$[HI]$[/tex]
[tex]$2.4 \times 10^{-2}$[/tex] [tex]$1.38 \times 10^{-2}$[/tex] 0 [tex]$1.14 \times10^{-2}$[/tex] [tex]$0.12 \times 10^{-2}$[/tex] [tex]$2.52 \times 10^{-2}$[/tex] 46.42
[tex]$2.4 \times 10^{-2}$[/tex] [tex]$1.68 \times 10^{-2}$[/tex] 0 [tex]$0.92 \times10^{-2}$[/tex] [tex]$0.20 \times 10^{-2}$[/tex] [tex]$2.96 \times 10^{-2}$[/tex] 47.61
[tex]$2.4 \times 10^{-2}$[/tex] [tex]$1.98 \times 10^{-2}$[/tex] 0 [tex]$0.77 \times10^{-2}$[/tex] [tex]$0.31 \times 10^{-2}$[/tex] [tex]$3.34 \times 10^{-2}$[/tex] 46.73
[tex]$2.4 \times 10^{-2}$[/tex] [tex]$1.76 \times 10^{-2}$[/tex] 0 [tex]$0.92 \times10^{-2}$[/tex] [tex]$0.22 \times 10^{-2}$[/tex] [tex]$3.08 \times 10^{-2}$[/tex] 46.86
[tex]$0$[/tex] [tex]$0$[/tex] [tex]$3.04 \times 10^{-2}$[/tex] [tex]$0.345 \times10^{-2}$[/tex] [tex]$0.345 \times 10^{-2}$[/tex] [tex]$2.35 \times 10^{-2}$[/tex] 46.39
[tex]$0$[/tex] [tex]$0$[/tex] [tex]$7.58 \times 10^{-2}$[/tex] [tex]$0.86 \times10^{-2}$[/tex] [tex]$0.86 \times 10^{-2}$[/tex] [tex]$5.86 \times 10^{-2}$[/tex] 46.42
Average [tex]$K_c=46.738$[/tex]
The diagram shows the water cycle. At point 4, condensation is occurring. How is water changing?
A) Water vapor (gas) is becoming liquid.
B) Water is being stored in the atmosphere.
C) Liquid water in the atmosphere is becoming a gas.
D) Liquid water from the earth's surface is becoming water vapor (gas)
Answer:
water vapor (gas) is becoming liquid
balance the following equations:
a. ____AgI + _____Na2S → _____Ag2S + _____NaI
b. ____Na3PO4 + ____HCl →_____NaCl + _____H3PO4
c. _____Ba3N2 + _____H2O →_____Ba(OH)2 + _____NH3
d. _____TiCl4 + _____H2O →_____TiO2 + _____HCl
e. _____CaCl2 + _____Na3PO4 →_____Ca3(PO4)2 + _____NaCl
Answer:
A. 2AgI + Na2S → Ag2S + 2NaI
B. Na3PO4 + 3HCl → 3NaCl + H3PO4
C. Ba3N2 + 6H2O → 3Ba(OH)2 + 2NH3
D. TiCl4 + 2H2O → TiO2 + 4HCl
E. 3CaCl2 + 2Na3PO4 → Ca3(PO4)2 + 6NaCl
Explanation:
What is the specific heat capacity of silver metal if 110.00 g of the metal
absorbs 396 J of heat and the temperature rises 15.0°C ?
Answer:
"0.24 J/g.k" is the appropriate solution.
Explanation:
Mass of substance,
m = 110.00 g
Heat,
q = 396 J
Change in temperature,
ΔT = 15.0°C
Now,
The specific heat will be:
⇒ [tex]q=ms \Delta T[/tex]
On putting the given values, we get
⇒ [tex]396=110.00\times s\times 15.0[/tex]
⇒ [tex]396=s\times 1650[/tex]
⇒ [tex]s=\frac{396}{1650}[/tex]
⇒ [tex]=0.24 \ J/g.k[/tex]
How many joules of heat is needed when 5 g of water is heated from 25°C to 75°C?
Answer:
1045 J
Explanation:
By using the formula Eh= mass×specific heat capacity × change in tempreture
With it being a known fact that the specific heat capacity of water is 4.16g/j/°c
balance
Al(OH)3 + H2SO4 → Al2(SO4)3 + H2O
Equation balance:
2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
2Al(OH)3+3H2SO4
H=12
SO4=3
O=6
Al=2
Al2(SO4)3+6H2O
H=12
SO4=3
O=6
Al=2
Name the layers of the atmosphere and one fact about each layer.
When we send electricity to pass through electrolysis jar we see gas bubble around both electrodes. Water electrolysis produces 30cm3 at cathode. a. What gas does it appear at each electrode? b. What would you do to identify the gas in the test tubes at each electrode? c. Calculate the volume of gas which appears at anode.
Answer:
a) anode - hydrogen gas
cathode - oxygen gas
b) hydrogen gas - The gas burns with a pop sound
Oxygen gas - The gas rekindles a glowing splint
c) The volume of gas produced at the anode is 60cm3
Explanation:
The electrolysis of water produces two volumes of hydrogen at the anode and one volume of oxygen at the cathode.
The hydrogen gas burns with a pop sound. It is easily identified by this test. Also, oxygen is known to rekindle a glowing splint and this is a commonly used test for oxygen gas.
Since the ratio of hydrogen produced at the anode to oxygen produced at the cathode is 2:1 by volume; then when 30cm3 of oxygen is produced at the cathode, 60cm3 of hydrogen is produced at the anode.
At 298 K, the rate constant for a reaction is 0.0346 s-1. What is the rate constant at 350K if the Ea = 50.2kJ/mol
Answer:
0.702 /s
Explanation:
Rate constant at [tex][298 \mathrm{~K}, \mathrm{~K}_{1}=3.46 \times 10^{-2} \mathrm{~s}^{-1}[/tex]
Rate constant at [tex]350 \mathrm{~K}, \mathrm{~K}_{2}=?[/tex]
[tex]T_{1}=298 \mathrm{~K}[/tex]
[tex]T_{2}=350 \mathrm{~K}[/tex]
Activation energy, [tex]\mathrm{Ea}=50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}[/tex]
Use the following equation to calculate [tex]K_{2}$ at $350 \mathrm{~K}[/tex]
Use the following equation to calculate [tex]K_{2}$ at $350 \mathrm{~K}[/tex]
[tex]\ln \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{Ea}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right][/tex]
Therefore,
[tex]\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{1}{298 \mathrm{~K}}-\frac{1}{350 \mathrm{~K}}\right][/tex]
[tex]\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{52 \mathrm{~K}}{298 \mathrm{~K} \times 350 \mathrm{~K}}\right][/tex]
[tex]\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}} &=\mathrm{e}^{3.01}[/tex]
[tex]\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}} &=20.3[/tex]
[tex]K_{2} &=20.3 \times 3.46 \times 10^{-2} \mathrm{~s}^{-1}[/tex]
[tex]&=0.702 \mathrm{~s}^{-1}[/tex]
hence, the rate constant at [tex]350 \mathrm{~K} [/tex] is 0.702[tex]\mathrm{~s}^{-1}[/tex]
A gas with a volume of 250 mL at a temperature of 293K is heated to 324K. What is the new volume of the gas?
V2 = 276 ml
Explanation:
Given:
V1 = 250 ml V2 = ?
T1 = 293K. T2 = 324K
Here, we use Charles's law:
V1/T1 = V2/T2
Solving for P2, we get
V2 = (T2/T1)V1
= (324K/293K)(250 ml)
= 276 ml
Select the correct choices to complete the sentence.
In the ionic solid sodium chloride (NaCl), there is a________
ratio of sodium ions to chlorine ions.
one and half to one
tow to one
one to two
one to one
Answer:
one to one
Explanation:
Answer:
In the ionic solid sodium chloride, there is one to one ratio of sodium ions to chloride ions
La rapidez de una reaccio es 0,450 mol/s ¡cuanto tiempo, en segundos, serán necesarios para que se consuman 0,323 mol de reactivos
Answer:
[tex]tiempo=0.72s[/tex]
Explanation:
¡Hola!
En este caso, como podemos observar, la relación entre la rapidez de reaction, las moles y el tiempo de reacción es:
[tex]rapidez=\frac{mol}{tiempo}[/tex]
De este modo, despejamos el tiempo en esta ecuación para obtener:
[tex]tiempo=\frac{mol}{rapidez}[/tex]
Así, reemplazamos la rapidez y moles dadas para obtener:
[tex]tiempo=\frac{0.323mol}{0.450mol/s}\\\\tiempo=0.72s[/tex]
¡Saludos!
Glucose is a compound made from three elements: carbon, oxygen, and hydrogen. For glucose to be broken down into these elements, it must undergo _____.
A. filtration or distillation
B. an irreversible physical change, such as tearing or slicing
C. a reversible physical change, such as boiling or melting
D. a chemical change
Answer:
D ) Chemical Change
Explanation:
chemical formula is given.
CO→C+O
What does the formula describe?
A.
It describes a mixture being separated by a physical process.
B
It describes an atom being separated by a chemical process.
C
It describes an element being broken down into the atoms from which it was formed.
D
It describes a compound being broken down into the elements from which it was formed.
Answer:
D . It describes a compound being broken down into the elements from which it was formed.
Explanation:
A compound consists of two or more elements chemically combined together in a chemical reaction. The compound carbon monoxide is represented by the chemical symbol, CO. This means that it contains 1 atom of the carbon element and 1 atom of the Oxygen element.
This compound is formed when carbon is directly oxidized in a limited supply of oxygen. In the above reaction, this compound is disintegrated into its individual components, namely, carbon and oxygen.
HELP ASAPPP!!! ON EXAM!
The arrangement of electrons within an atom is called it's...
a
Electron Flow
b
Electron Configuration
c
Atomic Number
d
Ionization Energy
Answer:
The arrangement of electrons within an atom is called electron configuration
If I help you, will you help other people?
Actually "this is the worst question i've ever seen use the internet bud" why don't you do that.
22.4 liters of CO2 at STP, 0◦C and 1 atm,
contains how many molecules? (Assume CO2
is an ideal gas.)
Answer:
There are [tex]6.01\times 10^{23}[/tex] number of molecules.
Explanation:
Given that,
Volume, V = 22.4 L
Temperature, T = 0°C = 273.15 K
Pressure, P = 1 atm = 1.013 bar
We need to find the number of molecules. Let there are n number of molecules. We know that, ideal gas law is as follows :
[tex]PV=nRT\\\\n=\dfrac{PV}{RT}\\\\n=\dfrac{1.013\times 22.4}{0.08 314\times 273.15}[/tex]
n = 0.999 moles
No. of molecules,
[tex]N=n\times N_A\\\\=0.999\times 6.023\times 10^{23}\\\\N=6.01\times 10^{23}[/tex]
So, there are [tex]6.01\times 10^{23}[/tex] number of molecules.
Which of the following is NOT one of the three main global climate zones?
A. Tropical
B. Latitude
C. Polar
D.Temprate
How many valence electrons does it have?
pls answer thank you
Answer:
One.
Explanation:
this is the structure of lithium which has valency 1 .Valency means no.electrons present in outermost shell or needed electrons to complete octet
help!!!!!!!!!!!!!!!!!!!!! i have less than 3 hours please help! image below!!!!
Answer:
friction
Explanation:
conduction doesn't really need much to start it. it basically has that reaction from the start so it's not conduction. Since heat is produced by friction in this situation it causes static to happen. so i think its friction
what were Lamarck's ideas about evolution and why were those ideas incorrect
What were Lamarck's ideas about evolution and Why were those ideas incorrect?
→ These and other examples show that Lamarck's theory does not explain how life formed and became the way it is. The other way that Lamarck's theory has been proven wrong is the study of genetics. Darwin knew that traits are passed on, but he never understood how they are passed on.
[tex]#CarryOnLearning[/tex]If HA, an acid, is 15 percent dissociated in a 2.00 molar solution, what is the Ka for the solution?
A.5.3 × 10-2
B.6.4 × 10-3
C.1.1 × 10-2
D.1.2 × 10-2
Answer:
A. 5.3 X 10^-2
Explanation:
I got 100 on PLATO
Help me plz plz bbbbb
Answer:
The respiratory system is the network of organs and tissues that help you breathe. basically your airways, lungs, and blood vessels so you quite literally need these to survive I mean your muscles need oxygen to function and your brain needs oxygen to function so without your respiratory system you wouldn't be able to move much and you would quickly pass out and die from the lack of oxygen to your brain there are other body processes that need oxygen as well but these are just two examples.
Explanation:
Answer:
Maybe it’s dark ??????????????
Explanation:
Hm
How many moles of water can be produced with 4.3 moles of H2 and 5.6 moles of O2? Which reactant is limiting? How many moles of the excess reactant will be left after the reaction? 2 H2 + O2 2 H2O
Answer:
Hydrogen H₂ will be the limiting reagent.
The excess reactant that will be left after the reaction is 3.45 moles.
4.3 moles of water can be produced.
Explanation:
The balanced reation is:
2 H₂ + O₂ → 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
H₂: 2 moles O₂: 1 mole H₂O: 2 molesTo determine the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 1 mole of O₂ reacts with 2 moles of H₂, how much moles of H₂ will be needed if 5.6 moles of O₂ react?
[tex]moles of H_{2} =\frac{5.6 moles of O_{2} *2 mole of H_{2} }{1 mole of O_{2}}[/tex]
moles of H₂= 11.2 moles
But 11.2 moles of H₂ are not available, 4.3 moles are available. Since you have less moles than you need to react with 5.6 moles of O₂, hydrogen H₂ will be the limiting reagent and oxygen O₂ will be the excess reagent.
Then you can apply the following rules of three:
If by reaction stoichiometry 2 moles of H₂ react with 1 mole of O₂, 4.3 moles of H₂ will react with how many moles of O₂?[tex]moles of O_{2} =\frac{1 mole of O_{2} *4.3 mole of H_{2} }{2 mole of O_{2}}[/tex]
moles of O₂= 2.15 moles
The excess reactant that will be left after the reaction can be calculated as:
5.6 moles - 2.15 moles= 3.45 moles
The excess reactant that will be left after the reaction is 3.45 moles.
If by reaction stoichiometry 2 moles of H₂ produce 2 moles of H₂O, 4.3 moles of H₂ produce how many moles of H₂O?[tex]moles of H_{2}O =\frac{2 moles of H_{2}O *4.3 mole of H_{2} }{2 mole of H_{2}}[/tex]
moles of H₂O= 4.3 moles
4.3 moles of water can be produced.
Un profesor desea preparar una disolución de agua oxigenada al 20%v/v y dispone solamente de 50 cm3 h2o2 ¿qué volumen d disolución deberá preparar? Con procedimiento
Answer:
250cm³ de disolución se deben preparar
Explanation:
El porcentaje volumen volumen (%v/v) se define como el volumen de soluto -En este caso, H2O2- en 100cm³ de disolución.
Se desea preparar una solución que contenga 20cm³ de H2O2 por cada 100cm³
Como solamente se cuenta con 50cm³ de H2O2, el volumen que se debe preparar es:
50cm³ H2O2 * (100cm³ Disolución / 20cm³ H2O2) =
250cm³ de disolución se deben prepararA chemist mixes 50.0mL of a 1.0M NaOH solution with 50.0mL of a 1.0M Ba(OH)2 solution. Assuming the two solutions are additive, what is the pH of the resulting solution
Answer:
[tex]pH=14.2[/tex]
Explanation:
Hello there!
In this case, according to the information in this problem, and considering these two bases are strong, it is necessary for us to calculate the total moles of OH ions as shown below:
[tex]n_{OH^-}^{from\ NaOH}=0.050L*1.0mol/L=0.050mol\\\\n_{OH^-}^{from\ Ba(OH)_2}=0.050L*1.0mol/L*2=0.10mol\\\\n_{OH^-}^{tot}=0.15mol[/tex]
Now, the as the solutions are additive, the total volume is then 0.100 L and the concentration:
[tex][OH^-]=\frac{0.15mol}{0.100L}=1.5[/tex]
And therefore, the pH is:
[tex]pH=14+log(1.5)\\\\pH=14.2[/tex]
Regards!
in the structure of an atom why are protons present in the centre and are not pulled outside by the electrons as both are positively charged with some unit of charge?
Answer:
Electrons and protons are not positively charged
Explanation:
Neutrons are the only ones positively charged in the structure of an atom.... Also like charges repel which means a -ive and a - ive will repel when together