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 is the electron configuration for Na?
1s22s22p63s1. That's the ans
How many significant figures are in 3.20x10^2 g?
Answer:
3
Explanation:
For numbers with decimals, count the number after the decimal.
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
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
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
identify which element is oxidized and which element is reduced.
PLZ HELPP...
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
is C5H10 ionic or covalent?
Which of the following is the Arrhenius equation?
Answer:
A is the answer the reason why is because I got it right when I was doing my test
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
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 .
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 < 0.00001MAs the food burned,
energy was
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Answer:
I don't get it is it even a question?
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
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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Classify the following compounds as binary ionic, ternary ionic, or molecular.
Answer:bro u already have an answer why are you asking?
Explanation:
Every morning, Jeremiah uses a blender to make a smoothie for breakfast. Which of the
following shows the energy transformation that the blender demonstrates?
А
Electrical - light
B
Chemical → mechanical
с
Mechanical → chemical
D
Electrical mechanical
Answer:
D
Explanation:
Answer:
D: Electrical -> mechanical
Explanation:
Hope this helps!
a titanium bicycle frame displays 0.250 L of water and has a mass of 1.21kg. what is the density of the titanium on g/cm3?
Answer:
$4.49 g/cm3
Explanation:
Density=Mass/Volume
plzz put brainiest
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
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.
chemistry
Definition in your own words. I will check if you got it from online.
Word:
Malleable
(malleability)
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%
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 .
Suppose that you are given a mixture of proteins with their properties provided in the following chart.
Isoelectric point PI Molecular weight
Protein A 4.1 80
Protein B 9.0 81
Protein C 8.8 37
Protein D 3.9 172
Choose one combination of techniques that can be used to isolate Protein B from Proteins A, C, and D.
a. gel filtration chromatography and ultracentrifugation
b. dialysis and ultracentrifugation
c. dialysis and ion exchange chromatography
d. ion exchange chromatography and gel filtration chromatography
Answer:
The correct option is d. ion exchange chromatography and gel filtration chromatography
Explanation:
The mixture contains proteins with different isoelectric points and molecular weights. Ion exchange chromatography separates the proteins with different electric charges, which depend on the isoelectric point. At certain pH of the aqueous medium, the proteins will have a certain electric charge and they will be retained differently in the stationary phase. Proteins A and D have similar isoelectric points (around 4.0) and maybe they cannot be efficiently separated by this method; the same occurs with protein B and C (with an isoelectric point around 9.0). To separate proteins A from D and protein B from C, it could be used gel filtration chromatography. This separation method is based on differences in molecular weights. So, at the end of the process, we will obtain the proteins A, B, C, D separated one from each other.
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
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.
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
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
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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