The empirical formula : C. C₁₆H₁₅N₂
Further explanationGiven
12.5 g of aniline
7.1 grams of H2O
1.48 grams of N2
Required
The empirical formula
Solution
Reaction :
mass H in H₂O :
= 2.1/18 x 7.1 g
= 0.79
mass N = 1.48
mass C :
= 12.5 g-(mass H+mass N)
= 12.5 - (0.79+1.48)
= 10.23
Mol ratio C : H : N =
= 10.23/12 : 0.79/1 : 1.48/14
= 0.853 : 0.79 : 0.106
= 8 : 7.5 : 1
= 16 : 15 : 2
Joseph Priestly is frequently credited with the discovery of oxygen, and was reported to have produced molecular oxygen from the decomposition reaction of mercury(II) oxide, which is the reverse of the synthesis of HgO depicted in the following equation. 4 Hg(l) + 2 O2(g) LaTeX: \rightarrow → 4 HgO(s) Determine the value of LaTeX: \Delta ΔH°rxn for the synthesis, given that
Joseph Priestly is frequently credited with the discovery of oxygen, and was reported to have produced molecular oxygen from the decomposition reaction of mercury(II) oxide, which is the reverse of the synthesis of HgO depicted in the following equation. [tex]4Hg(l)+2O_2(g)\rightarrow 4 HgO(s) [/tex]Determine the value of [tex]\Delta ΔH°rxn[/tex] for the synthesis, given that [tex]\Delta H_f^0[/tex] for HgO is -90.7 kJ/mol.
Answer: The enthalpy change for this reaction is, -362.8 kJ
Explanation:
The balanced chemical reaction is,
[tex]4Hg(l)+2O_2(g)\rightarrow 4HgO(s)[/tex]
The expression for enthalpy change is,
[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
[tex]\Delta H=[(n_{HgO}\times \Delta H_{HgO})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{Hg}\times \Delta H_{Hg})][/tex]
where,
n = number of moles
[tex]\Delta H_{O_2}=0[/tex] (as heat of formation of substances in their standard state is zero
[tex]\Delta H_{Hg}=0[/tex] (as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
[tex]\Delta H=[(4\times -90.7)]-[(2\times 0)+(4\times 0)][/tex]
[tex]\Delta H=-362.8kJ[/tex]
Therefore, the enthalpy change for this reaction is, -362.8 kJ
Explain the differences between an ideal gas and a real gas.
Answer:
Ideal Gas
The ideal gas is extremely small and the mass is almost zero and no volume Ideal gas is also considered as a point mass.
Real Gas
The molecules of real gas occupy space though they are small particles and also have volume.
anation:
The differences between an ideal gas and a real gas are that the ideal gas follows the gas laws perfectly under all conditions. Whereas a real gas deviates from ideal gas behaviors.
The ideal gas law, also known as the general gas equation, is a fundamental principle in thermodynamics and relates the pressure, volume, temperature, and number of moles of an ideal gas.
An ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions of temperature and pressure. It is assumed to have no volume, no intermolecular forces, and elastic collisions between its particles. An ideal gas also obeys the ideal gas law.
On the other hand, a real gas is a gas that does not follow the gas laws perfectly under all conditions of temperature and pressure. Real gases have volume and intermolecular forces that affect their behavior. These forces cause deviations from ideal gas behavior, especially at high pressures and low temperatures.
In summary, while an ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions, a real gas is a gas that deviates from ideal gas behavior due to its volume, intermolecular forces, and non-elastic collisions between its particles.
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is C5H10 ionic or covalent?
The density of a sample of gasoline is 0.70 g/cm3. What is the mass of 1 liter of this gasoline?
Group of answer choices
0.7 g
70 g
700 g
1,429 g
Answer:
700g
Explanation:
Given parameters:
Density of gasoline = 0.7g/cm³
Volume of gasoline = 1L = 1000cm³
Unknown:
Mass of the gasoline = ?
Solution:
Density is the mass per unit volume of a substance. It can be expressed as;
Density = [tex]\frac{mass}{volume}[/tex]
So;
Mass = density x volume
Mass = 0.7 x 1000 = 700g
An ionic compound has a generic formula of QR2.
Which elements could the Q and R represent?
Once you choose an answer, check it by plugging those elements into the QR2 formula to see if it looks right.
Q= Sodium R= Oxygen
Q= Magnesium R= Chlorine
Q= Oxygen R= Sodium
Q= Chlorine R= Magnesium
Answer:
Q= Magnesium R= Chlorine
Explanation:
The element Q should be magnesium and R is chlorine.
An ionic compound is a compound that is formed by the combination of a metal and non-metal. Such bonds forms when there is a transfer of electrons from the metals to the non-metals. This leaves a net positive charge on the metal and a negative charge on the non-metal.
The electrostatic attraction leads to the formation of the bond.
To solve this problem, the hypothetical compound is QR₂
Mg Cl
2 8 2 2 8 7
So, Mg transfers 2 electrons to two atoms of chlorine.
This leads to the formation of the compound MgCl₂
PREDICT How do you think the atoms in metal elements are different from those in
nonmetals or metalloids? How might the atoms of different metals vary from one another?
Answer:
See explanation
Explanation:
The atoms of metals have fewer valence electrons than the atoms of metals and metalloids.
Atoms of metals have only very few valence electrons in their outermost shells hence they donate electrons during bonding. However, atoms of nonmetals have more electrons in their outermost shells and rather accept electrons during bonding. The atoms of metalloids just have a number of valence electrons that are intermediate between those of metals and nonmetals and mostly share electrons in covalent bonds.
Similarly, atoms of metallic elements differ from each other in the number of valence electrons present in the valence shell of the atom of each element. For instance, sodium has one electron in the valence shell of its atom while aluminium has three electrons in the valence shell of its atom.
The atoms of metallic elements are different from the atoms of non metals or metalloids base on the outer electron/ valency electrons and the its bonding pattern.
The atoms of different metals varies in it ability to bond quickly.
The atoms of metallic elements are different from the atoms of non metals or metalloids base on the outer electron/ valency electrons and how it bonds.
Metallic atoms have very few electrons in the outermost shell. The valency electrons of this metallic atoms are few and are easily lost during bonding. They have the ability to release there valency electrons easily. Example of this metals are sodium, potassium , calcium etc.
On the other hand non metallic elements have numerous electron in the outermost shell and easily receive electron during bonding. Example are chlorine, fluorine, oxygen etc.
The metalloid atoms like silicon and germanium have an average number of electron in their outermost shell. They are in between.
The atoms of different metals varies in it ability to bond quickly. For example the group 1 metals are very reactive than the group 2 metals. This simply means the group 1 metals(alkali metals) goes into bonding more easily than the group 2 metals(alkali earth metals).
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3.4 x 10-25 kg = ? microounces
Answer: 1.2 x 10^-17 microounces
Explanation:
Ounce = 28.5G microounce = 28.5*10^-6g
3.4*10^-25 kg = 3.4*10^-22 g = (3.4/2.85)*10^(-22+5) = 1.2*10-17
identify which element is oxidized and which element is reduced.
PLZ HELPP...
How many moles of water can be formed from 0.57 moles of hydrogen gas?
Answer:
0.57 water
Explanation:
To solve this problem, we need to write the reaction expression first.
The reactants are oxygen gas and hydrogen gas.
They react to give a product of water
2H₂ + O₂ → 2 H₂O
Given that;
Number of moles of hydrogen gas = 0.57moles
From the balanced reaction expression;
2 moles of hydrogen gas produces 2 moles of water
So;
0.57mole of hydrogen gas will also produce 0.57 water
How many moles of hydrogen gas are present in 65.0 liters at STP?
1456 moles
1.45 moles
3.00 moles
2.90 moles
Answer:
2.9moles of hydrogen gas
Explanation:
convert liters to dm³
since 1liter= 1dm³
thus, 65.0liters = 65.0dm³
number of moles = volume given/22.4dm³
= 65.0/22.4
=2.9moles
A series of dilute NaCl solutions are prepared starting with an initial stock solution of 0.100 M NaCl. Solution A is prepared by pipeting 10 mL of the stock solution into a 250-mL volumetric flask and diluting to volume. Solution B is prepared by pipeting 25 mL of solution A into a 100-mL volumetric flask and diluting to volume. Solution C is prepared by pipeting 20 mL of solution B into a 500-mL volumetric flask and diluting to volume. What is the molar concentration of NaCl in solutions A, B and C
Answer:
Solution A: 0.00400M
Solution B: 0.00400M
Solution C: 4.00x10⁻⁵M
Explanation:
Solution A is diluting the 0.100M NaCl from 10mL to 250mL. That is:
250mL / 10mL = 25 times.
That means molar concentration of sln A is:
0.100M / 25 = 0.00400M
Solution B is obtained diluting 25mL to 100mL:
100mL / 25mL = 4 times
0.00400M / 4 times = 0.00100M
And solution C is obtained diluting the solution C from 20mL to 500mL:
500mL / 20mL = 25 times
Solution C:
0.00100M / 25 times = 4.00x10⁻⁵M
The formula for serial dilution can be used to obtain the molarity of solution A, B , C.
For solution AM1V1 = M2V2
M2 = 0.100 M × 10 mL/250-mL
M2 = 0.004 M
For solution BM1V1 = M2V2
M2 = 0.004 M × 25 mL/100-mL
M2 = 0.001 M
For solution CM1V1 = M2V2
M2 = 0.001 M × 20 mL/500-mL
M2 = 0.00004 M
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Molecule A undergoes isomerization to molecule B in acetone. Using curved arrows, showing key intermediates and any formal charges, propose a detailed mechanism for this isomerization. Provide a brief explanation why this isomerization occurs.
Answer:
hello your question is incomplete attached below is the complete question
answer :
we use Isomerization because conjugated allylic carbocation is more stable when compared to a Non-conjugated Allylic carbocation
Explanation:
Reason for the mechanism
we use Isomerization because conjugated allylic carbocation is more stable when compared to a Non-conjugated Allylic carbocation
attached below is the detailed mechanism
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Answer:
I don't get it is it even a question?
Identify the term that matches each definition.
The front vent of a fume hood, which helps maintain proper air circulation____.
The horizontal, flat area of a fume hood upon which experiments are carried out____.
A characteristic that describes substances that evaporate readily, producing large amounts of vapors____.
The glass panel in front of the fume hood that shields the user from fumes and other hazard_____.
A. Airfoil.
B. Sash.
C. Work surface.
D. Volatile.
Answer:
A,
C.
D.
B.
Explanation:
The front vent of a fume hood that assists and maintain proper air circulation is Airfoil
The horizontal flat surface area of the fume hood where experiments are being carried out is Work Surface.
The main characteristics which demonstrate and describes how substances evaporate rapidly and readily into the thin air while producing a huge amount of vapor is known as Volatile
In front of the fume hood, lies the glass panel whose main purpose is to shield the user from the hazardous substance. This glass panel is known as the Sash.
A sample of saturated clay was placed in a container and weighed. The weight was 6N. The clay in its container was placed in an oven for 24 hours at 105° C. The weight reduced to a constant value of 5N. The weight of the container is 1N. If G-2.7, determine the:
(a) water content;
(b) void ratio;
(c) bulk unit weight;
(d) dry unit weight;
(e) effective unit weight.
Answer is given below
Explanation:
given data
weight = 6N
temp = 105° C
weight reduced = 5N
solution
weight of container is 1N
SO W = (6-1) = 5
And Wd = 5 - 1 = 4
so
moisture content is
moisture content = [tex]\frac{W-Wd}{Wd} \times 100[/tex] .......1
moisture content = [tex]\frac{5-4}{4} \times 100[/tex]
moisture content = 25%
and
as we know density of soil soild = 2700 kg/m³
density of water = 1000 kg/m³
and sp gravity of soil = [tex]\frac{2700}{1000}[/tex] = 2.7
so
now we get here bulk unit weight
bulk unit wt = [tex]Yw \times [\frac{G+e}{1+e}][/tex] ..........2
bulk unit wt = [tex]9.01 \times [\frac{2.7 + 0.675}{1+0.675}][/tex]
bulk unit wt = 19.766 KN/m³
and
so dry unit wt will be
dry unit wt = [tex]\frac{Ysat}{1+w}[/tex] ..............3
dry unit wt = [tex]\frac{19.766}{1+0.25}[/tex]
dry unit wt = 15.813 kN/m³
How many significant figures are in 3.20x10^2 g?
Answer:
3
Explanation:
For numbers with decimals, count the number after the decimal.
An atom has 81 electrons, 84 neutrons, and 82 protons. What element is this atom?
Answer:
Lead
Explanation:
The subatomic particles within an atom can be used to know the atom or element given.
Of particular interest is the number of protons within the atom.
The periodic table is based on the atomic number of atoms. This atomic number is the number of protons within an atomic space.
So; If we know the number of protons within an atom, we can know the element.
The number of protons given is 82, the element is therefore lead.
Answer:
The atomic number of polonium is 84. The atomic number lead is 82.
Explanation:
Calculate the number of oxygen atoms in a 50.0g sample of scheelite CaWO4
Answer:
0.696 atoms of oxygen
Explanation:
We'll begin by calculating the number of mole in 50 g of scheelite CaWO₄. This can be obtained as follow:
Mass of CaWO₄ = 50 g
Molar mass of CaWO₄ = 40 + 184 + (4×16)
= 40 + 184 + 64
= 288 g/mol
Mole of CaWO₄ =?
Mole = mass / Molar mass
Mole of CaWO₄ = 50 / 288
Mole of CaWO₄ = 0.174 mole
Finally, we shall determine the number of oxygen atom in 50 g (i.e 0.174 mole) of CaWO₄. This can be obtained as follow:
1 mole of CaWO₄ contains 4 atoms of oxygen.
Therefore, 0.174 mole of CaWO₄ will contain = 0.696 atoms of oxygen.
Thus, 50 g (i.e 0.174 mole) of CaWO₄ contains 0.696 atoms of oxygen.
balance the following equation by oxidation reduction method FeSO4
+
KMnO4+ H2SO4 → Fe2 (SO4)3+ k2SO4+MnSO4+H2O
Answer:
[tex]10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O[/tex].
Explanation:
Identify the elements with oxidation state changes:
Oxidation states of iron, [tex]\rm Fe[/tex]:
[tex]+2[/tex] in [tex]\rm FeSO_4[/tex] among the reactants.[tex]+3[/tex] in [tex]\rm Fe_2(SO_4)_3[/tex] among the products.Change to the oxidation state: [tex]+1[/tex] (oxidation) for each [tex]\rm Fe[/tex] atom.Oxidation state of manganese, [tex]\rm Mn[/tex]:
[tex]+7[/tex] in [tex]\rm KMnO_4[/tex] among the reactants.[tex]+2[/tex] in [tex]\rm MnSO_4[/tex] among the products.Change to the oxidation state: [tex](-5)[/tex] (reduction) for each [tex]\rm Mn[/tex] atom.The change in the oxidation state of [tex]\rm Mn[/tex] is five times the opposite of the change to the oxidation state of [tex]\rm Fe[/tex]. If there are one mole of [tex]\rm Mn\![/tex] atoms in each mole of this reaction, there would be five times as many [tex]\rm Fe\![/tex] atoms per mole reaction. In other words:
[tex]\displaystyle 5\, \overset{+2}{\rm Fe}\rm SO_4 + 1\, \rm K \overset{+7}{Mn} O_4 + ?\, H_2SO_4\\ \to \frac{5}{2}\, \overset{+3}{Fe} (SO_4)_3 + ?\, K_2SO_4 + 1\, \overset{+2}{Mn}SO_4 + ?\, H_2O[/tex].
(Notice that each mole of this reaction would include five times as many [tex]\rm Fe[/tex] atoms as [tex]\rm Mn[/tex] atoms.)
Multiply the coefficients by [tex]2[/tex] to eliminate the fraction:
[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + ?\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].
Find the unknown coefficients using the conservation of atoms.
Reactants:
[tex]2[/tex] potassium [tex]\rm K[/tex] atoms in two [tex]\rm K_2SO_4[/tex] formula units.Therefore, among the products:
[tex]2[/tex] potassium [tex]\rm K[/tex] atoms in one [tex]\rm K_2SO_4[/tex] formula unit.[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].
Products:
[tex]5 \times 3 + 2 + 1 = 18[/tex] sulfur [tex]\rm S[/tex] atoms in five [tex]\rm Fe_2(SO_4)_3[/tex] formula units, two [tex]\rm K_2 SO_4[/tex] formula units, and one [tex]\rm MnSO_4[/tex] formula unit.Reactants:
There are already ten [tex]\rm S[/tex] atoms in that ten [tex]\rm Fe(SO_4)_2[/tex] formula units. The other [tex]18 - 10 = 8[/tex] formula units would correspond to eight [tex]\rm H_2SO_4[/tex] molecules among the reactants of this reaction.[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].
Products:
There are [tex]8 \times 2 = 16[/tex] hydrogen [tex]\rm H[/tex] atoms in that eight [tex]\rm H_2SO_4[/tex] molecules.Therefore, among the products:
There would be [tex]16 / 2 = 8[/tex] molecules of [tex]\rm H_2O[/tex], with two [tex]\rm H[/tex] atoms in each [tex]\rm H_2O\![/tex] molecule.[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O[/tex].
chemistry
Definition in your own words. I will check if you got it from online.
Word:
Malleable
(malleability)
You want to clean a 500-ml flask that has been used to store a 0.9M solution. Each time the flask is emptied, 1.00 ml of solution adheres to the walls, and thus remains in the flask. For each rinse cycle, you pour 9.00 ml of solvent into the flask (to make 10.00 ml total), swirl to mix uniformly, and then empty it. What is the minimum number of such rinses necessary to reduce the residual concentration of 0.00001 M or below
Answer:
In the 5th cycle rinse, the residual concentration of the solution is < 0.00001M
Explanation:
In each rinse cycle, the dilution that you are doing of the solution is from 1.00mL to 10.00mL, that is a dilution of 10
In the first rinse the concentration must be of 0.9M 10 = 0.09M
2nd = 0.009M
3rd = 0.0009M
4th = 0.00009M
5th = 0.000009M →
In the 5th cycle rinse, the residual concentration of the solution is < 0.00001MA particular term in an atom in which LS coupling is a good approximation splits into three levels, each having the same L and same S but different J. If the relative spacings between the levels are in the proportion 5:3, find L and S.
Answer:
Explanation:
From the information given;
Consider using Lande's Interval rule which can be expressed as:
[tex]\Delta E = E_{j+1} - E_jj \ = \alpha (j+1)[/tex]
here;
[tex]j+1[/tex] = highest level of j
and
[tex]\dfrac{\Delta E_1}{\Delta E_2} = \dfrac{(j+2)}{(j+1)}[/tex]
[tex]\dfrac{5}{3} = \dfrac{(j+2)}{(j+1)}[/tex]
[tex]5(j+1) = 3(j+2)[/tex]
[tex]5j+5 = 3j+6[/tex]
[tex]2j = 1\\ \\ j = \dfrac{1}{2}[/tex]
recall that:
[tex]j = |S-L| \ \to \ |S+L |[/tex]
So;
[tex]S-L = \dfrac{1}{2} --- (1)[/tex]; &
[tex]S+L = \dfrac{5}{2} --- (1)[/tex]
Using the elimination method, we have:
[tex]2S = \dfrac{6}{2}[/tex]
[tex]S = \dfrac{3}{2}[/tex]
Since [tex]S = \dfrac{3}{2}[/tex]; then from (1)
[tex]\dfrac{3}{2} -L = \dfrac{1}{2}[/tex]
[tex]L = \dfrac{2}{2}[/tex]
[tex]L = 1[/tex]
Calculate the enthalpy change with the help of Hess’s law for the decomposition of hydrogen peroxide, (2H O (l) ------- 2H O(l) + O (g)), if the enthalpy of formation of water (2H (g) +O (g)) is –512KJ and vaporization of hydrogen peroxide (H O (l) = H (g) + O (g)) is 376 kJ.
Answer:
Explanation:
For answer see attached file .
A species of desert plant produces flowers that only bloom at night. How does this enhance the survival of the species?
A. This allows the plants to conserve water and not bloom during the heat of the day.
B. This species relies on nocturnal animals like moths for pollination and reproduction
C. This species relies on moonlight for photosynthesis.
D.This allows flowers to stay closed durng the day when herbivores are more likely to eat them.
Od
Answer:
A. This allows the plants to conserve water and not bloom during the heat of the day
Explanation:
Most desert plants only bloom at night because they take advantage of animals like moths and insects that fly at night for pollination and reproduction.
Because these plants are in the desert and do not get enough water except from short occasional rainfalls, they conserve water and not bloom during the heat of the day. They bloom at night when the temperature is low and this enhances their water conservation and survival.
an unknown substance has a mass of 57.4 g and occupies a volume of 34.3 ml. what is the density in g/ml?
Answer:
1.6734 g\ml..hope it helps
water is a unique material in that the density of the solid is lower than the density of the liquid (which is why ice forms at the top of a pond and why ice floats in our drinks). if the density for ice at 0C is .917g/mL and the density for water at 0C is .999 g/mL, what is the calculated free space (as %) for each of these materials. you will need to estimate the volume of water as the sum of 2 H atoms and 1 O atom with radii 37 and 66 pm respectively. note that you will also have to assume a quantity of water to perform this exercise
Answer:
% Free space in water = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%
% Free space in ice = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%
Explanation:
As given ,
Density for ice at 0⁰C = 0.917 g/ml
Density for water at 0⁰C = 0.999 g/ml
Radii of H atoms = 37 pm
Radii of O atoms = 66 pm
Now,
Consider 1 ml of water = 1 cm²
As , we know that mass of water in 1 cm² = 0.999 g
Moles of water = [tex]\frac{0.999}{18} = 0.056[/tex]
Volume of H₂O = 1.624×[tex]10^{-31}[/tex] m²
Now,
Volume occupied by water = 0.056×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²
= 5.48×[tex]10^{-9}[/tex] m²
⇒Volume occupied by water = 5.48×[tex]10^{-9}[/tex] m²
Now,
Free space = 1×[tex]10^{-6}[/tex] - 5.48×[tex]10^{-9}[/tex] = 9.95×[tex]10^{-7}[/tex] m²
% Free space = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%
Now,
Consider 1 ml of ice = 1 cm²
S.I unit of ice = 1×[tex]10^{-6}[/tex] m²
As , we know that mass of water in 1×[tex]10^{-6}[/tex] m² = 0.917 g
Moles of ice = [tex]\frac{0.917}{18} = 0.012[/tex]
Volume of H₂O = 6.022×[tex]10^{23}[/tex] ×0.012
Volume of ice unit = [tex]\frac{4}{3} \pi (37*10^{-12})^{3} *2 + \frac{4}{3} \pi (66*10^{-12})^{3} = 1.624*10^{-31}m^{3}[/tex]
Now,
Volume occupied by water = 0.012×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²
= 1.17×[tex]10^{-9}[/tex] m²
⇒Volume occupied by water = 1.17×[tex]10^{-9}[/tex] m²
Now,
Free space = 1×[tex]10^{-6}[/tex] - 1.17×[tex]10^{-9}[/tex] = 9.98×[tex]10^{-7}[/tex] m²
% Free space = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%
Two volumes of nitric oxide react with one volume of oxygen gas to form two volumes of a reddish-brown gas. Deduce the formula of this gas and sketch particle representations of its molecules.
Answer:
Explanation:
Nitric oxide is the gas NO, it reacts with oxygen as shown below;
2NO(g) + O2(g) -----> 2NO2(g)
Now the gas formed is the gas NO2 which is known to be reddish brown in colour.
A diagrammatic representation of this reaction is shown in the image attached to this answer.
Image credit: Chemlibretext
What produces the magnetic force of an electromagnet?
O magnetic fields passing through the device
O static charged particles on the wire
O movement of charged particles through the wire
O positive and negative charges repelling each other
Answer:
movement of charged particles through the wire .
Explanation:
When electricity is passed through the wire of electromagnet , moving electrons of the wire produces magnetic field . This magnetic field in increased due to high permeability of soft iron of the electromagnet . It is this magnetic field which creates magnetic force .
In each row, checkbox under the compound that can reasonably be expected to be more acidic in aqueous solution, e.g have the larger
Ka
H₂ SO₃ H₃ SO ₄
H₃ PO₄ H₃ PO₃
HCH₃ SO₂ HCH₃CO₂
Explanation:
H2SO3 is more acid than H2TeO3. Since S is more electronegative than Te is. In H2SO3, thus, dissociation of H+ would be smoother.
So, H2SO3's got high Ka.
HCH3SO2 is more acid than HCH3CO2. Since S is more electronegative than C. So, HCH3SO2 is a high Ka.
HClO2 is more acid than HClO. Since in HClO2, after the donation of H+ ion, the negative charge is set by two oxygen atoms, while in HClO, only one oxygen atom stabilizes the negative charge.
So, HClO2 is a high Ka
A student dissolves of aniline in of a solvent with a density of . The student notices that the volume of the solvent does not change when the aniline dissolves in it. Calculate the molarity and molality of the student's solution. Be sure each of your answer entries has the correct number of significant digits.
Answer:
Molarity: 0.21M
Molality: 0.20m
Explanation:
...dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL...
To solve this question, we need to find the moles of aniline in 3.9g using its molar mass. Then, we need to find the kg and Liters of solution in order to find molarity (Moles/L solution) and molality (Moles/kg of solvent):
Moles aniline:
Molar mass:
6C: 6* 12.01g/mol = 72.06g/mol
7H: 7*1.008g/mol = 7.056g/mol
N: 1*14.007g/mol = 14.007g/mol
72.06g/mol+7.056g/mol+14.007g/mol = 93.123g/mol
Moles of 3.9g: 3.9g * (1mol / 93.123g) = 0.04188moles
Liters solution:
200mL * (1L / 1000mL) = 0.200L
kg solvent:
200mL * (1.05g/mL) * (1kg/1000g) = 0.210L
Molarity:
0.04188mol / 0.200L = 0.21M
Molality:
0.04188mol / 0.210L =0.20m
