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
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
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 .
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.
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
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
chemistry
Definition in your own words. I will check if you got it from online.
Word:
Malleable
(malleability)
How can you model the cycling of matter in the Earth system?
Answer:
The cycling of matter is important to many Earth processes and to the survival of organisms the existing matter must cycle continuously for this planet to support life Water, carbon, nitrogen, phosphorus, and even rocks move through cycles If these materials did not cycle, Earth could not support life.
Explanation:
Earth activities depend on matter cycling, and for organisms to survive, this planet's surface must cycle with the flow of matter.
What is Earth system?Rocks, as well as water, carbon, nitrogen, and phosphorus, go through cycles. The planet Earth could not support life if these materials did not cycle.
Subsystems exist within the Earth system. These subsystems include the exosphere, atmosphere, hydrosphere, lithosphere and geosphere, also referred to as the lithosphere, and the living environment (biosphere).
These systems are powered by energy that comes from both the Sun and the interior of the Earth. Through processes known as biogeochemical cycles, nutrients and elements also move through these systems along with energy.
Therefore, Earth activities depend on matter cycling, and for organisms to survive, this planet's surface must cycle with the flow of matter.
To learn more about Earth, refer to the link:
https://brainly.com/question/1204146
#SPJ6
16. Using the average atomic masses given in the inside front cover of this book, calculate the indicated quantities.
d. the number of moles of cobalt represented by 5.99 x 1021 cobalt atoms e. the mass of 4.23 mol of cobalt
f. the number of cobalt atoms in 4.23 mol of cobalt
g. the number of cobalt atoms in 4.23 g of cobalt
Answer:
d. 9.95 × 10⁻³ mol
e. 249 g
f. 2.55 × 10²⁴ atoms
g. 4.32 × 10²² atoms
Explanation:
d. the number of moles of cobalt represented by 5.99 x 10²¹ cobalt atoms
We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.
5.99 x 10²¹ atoms × 1 mol/6.02 × 10²³ atoms = 9.95 × 10⁻³ mol
e. the mass of 4.23 mol of cobalt
The molar mass of cobalt is 58.93 g/mol.
4.23 mol × 58.93 g/mol = 249 g
f. the number of cobalt atoms in 4.23 mol of cobalt
We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.
4.23 mol × 6.02 × 10²³ atoms/1 mol = 2.55 × 10²⁴ atoms
g. the number of cobalt atoms in 4.23 g of cobalt
First, we will calculate the moles of cobalt using the molar mass of cobalt.
4.23 g × 1 mol/58.93 g = 0.0718 mol
Then, we will calculate the number of cobalt atoms using Avogadro's number.
0.0718 mol × 6.02 × 10²³ atoms/1 mol = 4.32 × 10²² atoms
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
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
According to the Michaelis-Menten equation, when an enzyme is combined with a substrate of concentration s (in millimolars), the reaction rate (in micromolars/min) is
Answer:
The answer is "A"
Explanation:
Please find the complete question in the attachment file.
[tex]\to R(s)= \frac{As}{K+s}[/tex]
when the s in the approach, that is infinity R(s) tends
[tex]\to \frac{A}{\frac{K}{s}+1} \\\\ \to\frac{A}{0+1} \\\\ \to\frac{A}{1} \\\\ \to A[/tex]
1. Each substance written to the right of the arrow in a chemical equation is a
(1 point)
O catalyst
O reactant
O precipitate
O product
Answer: product
Explanation:
Each substance written to the right of the arrow in a chemical equation is referred to as a product.
When writing a chemical equation, the substance that's written to the left of arrow in the equation is the reactants.
On the other hand which is the right side is the product.
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 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
Learn more about serial dilution: https://brainly.com/question/2167827
As the food burned,
energy was
nergy. Thus, a
thermal
transformed intos
chemical
form of Select
nuclear
$ converted to a
form of
Select
energy.
Check
Answer:
I don't get it is it even a question?
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:
If snails and crayfish die, what happens to the other species?
Answer:
Other species will slowly die, because the species that eat snails and crayfish will not have anything to eat, then the thing that eats them will not have anything. This process will go on and on. So long story short populations will decrease, then off.
Magnesium reacts with a silver nitrate solution
The reaction is represented by the ionic equation given
State why this reaction is considered a redox reaction
Answer: Mg oxidised to Mg++ and Ag+ reduced to Ag
Explanation:No ionic equation given!!
assume equation is Mg + 2Ag+ + 2NO3- —> Mg++ + 2Ag + 2NO3-
redox reaction: Ag gains an electron = reduced
Mg loseselectrons = oxidised
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
A compound is made of 6.00 grams of oxygen, 7.00 grams of nitrogen, and 20.00grams of hydrogen. Find the percent composition of the compound.
A O-18.18%, N-21.21%, H-60.60%
B O-11.18%, N-22.21%, H-69.60%
C O-20%, N-30%, H-50%
D O-60.60%, N-21.21%, H-18.18%
The percent composition of the compound.
A O-18.18%, N-21.21%, H-60.60%
Further explanationGiven
6.00 grams of oxygen,
7.00 grams of nitrogen,
20.00 grams of hydrogen.
Required
The percent composition
Solution
Total mass :
= mass of O + mass of N + mass of H
= 6 + 7 + 20
= 33 g
% O = 6/33 x 100%= 18.18%
% N = 7/33 x 100%=21.21%
% H = 20/33 x 100% = 60.6 %
Suppose a 500.mL flask is filled with 0.40mol of N2 and 1.0mol of NO. The following reaction becomes possible:
N2g+O2g ->2NOg
The equilibrium constant K for this reaction is 5.93 at the temperature of the flask. Calculate the equilibrium molarity of N2. Round your answer to two decimal places.
Answer:
[N₂] = 1.1M
Explanation:
Based on the chemical reaction:
N₂(g) + O₂(g) ⇄ 2 NO(g)
Equilibrium constant, K, is defined as:
K = 5.93 = [NO]² / [N₂] [O₂]
Where [] are equilibrium concentrations of each specie
As initial concentrations are:
N₂ = 0.40mol / 0.500L = 0.8M
NO = 1mol / 0.500L = 2M
The equilbrium concentrations are:
[NO] = 2M - 2X
[N₂] = 0.8M +X
[O₂] = X
Replacing:
5.93 = [2 - 2X]² / [0.8+X] [X]
5.93 = 4 - 8X + 4X² / 0.8X + X²
4.744X + 5.93X² = 4 - 8X + 4X²
1.93X² + 12.744X - 4 = 0
Solving for X:
X = -6.9M → False solution. There are no negative concentrations
X = 0.3M. Real solution.
[N₂] in equilibrium is:
[N₂] = 0.8M +0.3M
[N₂] = 1.1M
is C5H10 ionic or covalent?
How many cm 3 are in 0.014 in 3? (1 in = 2.54 cm)
Answer:
0.229 cm³.
Explanation:
The following data were obtained from the question:
Volume (in in³) = 0.014 in³
Volume (in cm³) =?
1 in = 2.54 cm
Next, we shall determine a conversion scale to convert from in³ to cm³. This can be obtained as follow:
1 in = 2.54 cm
Therefore,
1 in³ = 2.54³ cm³
1 in³ = 16.387 cm³
Finally, we shall convert 0.014 in³ to cm³. This can be obtained as follow:
1 in³ = 16.387 cm³
Therefore,
0.014 in³ = 0.014 in³ × 16.387 cm³ / 1 in³
0.014 in³ = 0.229 cm³
Thus, 0.014 in³ is equivalent to 0.229 cm³.
When preparing for work in the fume hood, be sure to gather all necessary tools, glassware, and chemicals _________ to minimize the number of times the hood sash is raised and lowered. Work as much as possible in the _________ of the work surface to keep the area tidy and promote air flow. If you need to step away from the experiment to obtain another item, _________ the sash during this time.
Answer:
In advance
middle
lower
Explanation:
These are the safety precautions needed when carrying out duties in the fume hood.
When planning and preparing to work in a fume hood (a locally designed area to reduce exposure to hazardous fumes). It is advisable to make all equipment readily available at your disposal in advance to reduce and minimize the raising and lowering of the hood sash at intervals.
It is also pertinent to understand that working in the middle of the work surface helps to promote the movement of air and keeps the area neat and tidy.
However, if any case where there is a need to get a new tool or equipment during the process of working in a fume hood, it is advisable to lower the sash at that point in time.
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 < 0.00001Man 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
what is the formula for H-H
Answer:
H-H equation is written as follows:
pH=pK + log
{HCO3-}(base)
{H2CO3}(acid)
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.
definition of solubility
(science)
Answer:
th relative ability of a solute to devolve into a solvent
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%
