Answer:
Approximately [tex]92.9\; \rm mL[/tex].
Explanation:
Look up relevant relative atomic mass data on a modern periodic table:
[tex]\rm Al[/tex]: [tex]26.982[/tex].[tex]\rm S[/tex]: [tex]32.06[/tex].[tex]\rm O[/tex]: [tex]15.999[/tex].Calculate the formula mass of [tex]\rm Al_2(SO_4)_3[/tex]:
[tex]M(\mathrm{Al_2(SO_4)_3}) = 2 \times 26.982 + 3 \times (32.06 + 4 \times 15.999) = 342.132\; \rm g \cdot mol^{-1}[/tex].
Calculate the number of moles of [tex]\rm Al_2(SO_4)_3[/tex] formula units that corresponds to [tex]110\; \rm g[/tex]:
[tex]\displaystyle n(\mathrm{Al_2(SO_4)_3}) = \frac{m}{M} = \frac{110\; \rm g}{342.132\; \rm g \cdot mol^{-1}} \approx 0.322\; \rm mol[/tex].
Calculate the volume of a [tex]3.25\; \rm M[/tex] [tex]\rm Al_2(SO_4)_3[/tex] solution that would contain [tex]0.322\; \rm mol[/tex] of [tex]\rm Al_2(SO_4)_3\![/tex] formula units. Note, that the unit of concentration "[tex]\rm M[/tex]" is equivalent to [tex]\rm mol \cdot L^{-1}[/tex].
[tex]\displaystyle V = \frac{n}{c} = \frac{0.322\; \rm mol}{3.25\; \rm mol \cdot L^{-1}} \approx 0.0989\; \rm L[/tex].
Convert that to milliliters as requested:
[tex]\displaystyle V \approx 0.0989\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 98.9\; \rm mL[/tex].
the isotopes of an elements have the same masse
Answer:
True
Explanation:
But have different neutrons
What is the density of water if you have 50.0 grams of water and a volume of 50.0 millimeters
Answer:
[tex]\boxed {\tt 1.0 \ g/mL}[/tex]
Explanation:
Density can be found by dividing the mass by the volume.
[tex]d=\frac{m}{v}[/tex]
The mass of the water is 50.0 grams.
The volume of the water is 50.0 milliliters.
[tex]m= 50.0\ g \\v=50.0 \ mL[/tex]
Substitute the values into the formula.
[tex]d=\frac{50.0 \ g}{50.0 \ mL}[/tex]
Divide.
[tex]d= 1.0 \ g/mL[/tex]
The density of the water is 1.0 grams per milliliter. Also, remember that the density of pure water is always 1.0 g/mL or g/cm³
1. Which statement is true about oxygen,
sulfur, and selenium? scap.8.0
^ They belong to the same group.
© They belong to the same period.
The mass number doubles between each
element.
0 The atomic number increases by eight
between each clement,
Answer:
they belong to same grop
What is the relationship between the specific rotation of (2S ,3S)-dibromopentane and (2R, 3R)-dibromopentane?
A) It is different in magnitude and opposite in sign
B) it is equal in magnitude and have the same sign
C) it is zero
D) it is equal in magnitude but opposite in sign
E) It is not possible to predict the specific rotation
Answer:
D) it is equal in magnitude but opposite in sign
Explanation:
The compounds (2S ,3S)-dibromopentane and (2R, 3R)-dibromopentane are a pair of enantiomers.
Enantiomers are stereoisomers that that are mirror images of each other and rotate plane polarized light. The magnitude of rotation of plane polarized light is known as specific rotation of the compound measured in degrees.
Enantiomers have the same magnitude of specific rotation but opposite in sign. This is because one enantiomer rotates plane polarized light to the left while the other rotates plane polarized light to the right.
Does Sodium Chloride, have the same properties as sodium and chlorine
A negatively charged rod is placed near two neutral metal spheres, as shown below. Which statement describes the charging method shown? The spheres are charged through friction. The spheres develop the same charge. Electrons move from the rod to Sphere A. Electrons move from Sphere A to Sphere B.
Answer:
D. Electrons move from Sphere A to Sphere B.
Explanation:
The statement that best describes the charging method is "Electrons move from Sphere A to Sphere B". This is further explained below.
What is Charge?A charge is simply defined as the property of a unit of matter that expresses how many more or fewer electrons it has than protons.
In conclusion, For a negatively charged rod placed near two neutral metal spheres, electrons move from Sphere A to Sphere B.
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Liz and Michelle are testing which ball will bounce the highest. They test 4 different balls. Each ball is dropped five times from the same height. The height of the bounce is measured. At the end of the experiment, the data for each ball was averaged to obtain an average bounce height.
Is this a well designed experiment?
A) No, it does not give any data.
B) No, they should have used more types of balls.
C) Yes, they tested more than one thing at a time.
D) Yes, they averaged the measurements and had multiple trials.
Answer:
D
Explanation:,
In designing an experiment there must be multiple trials. Multiple trials eliminate errors when the results are averaged.
This gives better results than when single measurements are used to determine results of experiments. The result of this experiment can be regarded as being reliable because it was well designed.
When aqueous solutions of AgNO3 and KI are mixed, AgI precipitates. The balanced net ionic equation is ________. When aqueous solutions of AgNO3 and KI are mixed, AgI precipitates. The balanced net ionic equation is ________. AgNO3 (aq) + KI (aq) → AgI (s) + KNO3 (aq) Ag+ (aq) + I- (aq) → AgI (s) AgNO3 (aq) + KI (aq) → AgI (aq) + KNO3 (s) Ag+ (aq) + NO3 - (aq) → AgNO3 (aq) Ag+ (aq) + NO3 - (aq) → AgNO3 (s)
Answer:
Ag⁺ (aq) + I¯ (aq) —> AgI (s)
Explanation:
We'll begin by writing the dissociation equation for aqueous AgNO₃ and KI.
Aqueous AgNO₃ and KI will dissociate in solution as follow:
AgNO₃ (aq) —> Ag⁺(aq) + NO₃¯ (aq)
KI (aq) —> K⁺(aq) + I¯(aq)
Aqueous AgNO₃ and KI will react as follow:
AgNO₃ (aq) + KI (aq) —>
Ag⁺(aq) + NO₃¯ (aq) + K⁺ (aq) + I¯(aq) —> AgI (s) + K⁺ (aq) + NO₃¯ (aq)
Cancel out the spectator ions (i.e ions that appears on both sides of the equation) to obtain the net ionic equation. The spectator ions are K⁺ and NO₃¯.
Thus, the net ionic equation is:
Ag⁺ (aq) + I¯ (aq) —> AgI (s)
The net ionic equation of aqueous solutions of [tex]AgNO_3[/tex] and [tex]KI[/tex] to form [tex]AgI[/tex] precipitates is: B. [tex]Ag^{+}_{(aq)} + I^{-}_{(aq)}[/tex] -----> [tex]AgI_{(s)}[/tex]
A balanced chemical equation can be defined as a chemical equation wherein the number of atoms on the reactant (left) side is equal to the number of atoms on the product (right) side.
This ultimately implies that, both the charge on each atom and sum of the masses of the chemical compounds or elements in a chemical equation are properly balanced.
An ion can be defined as an atom or molecules (group of atoms) that has lost or gained one or more of its valence electrons, thereby, making it have a net positive or negative electrical charge.
First of all, we would write the dissociation equation for aqueous solutions of [tex]AgNO_3[/tex] and [tex]KI[/tex]:
For [tex]AgNO_3[/tex]:
[tex]AgNO_3_{(aq)}[/tex] -----> [tex]Ag^{+}_{(aq)} + NO_{3}^{-}_{(aq)}[/tex]
For [tex]KI[/tex]:
[tex]KI_{(aq)}[/tex] -----> [tex]K^{+}_{(aq)} + I^{-}_{(aq)}[/tex]
Next, we would write a chemical equation for the reaction of aqueous solutions of [tex]AgNO_3[/tex] and [tex]KI[/tex]:
[tex]AgNO_3_{(aq)} + KI_{(aq)}[/tex] -----> [tex]Ag^{+}_{(aq)} + NO_{3}^{-}_{(aq)}[/tex] [tex]+ \;K^{+}_{(aq)} + I^{-}_{(aq)}[/tex] ----->[tex]AgI_{(s)} + K^{+}_{(aq)} + NO_{3}^{-}_{(aq)}[/tex]
Note: Spectator ions refers to the ions that exist as a reactant and a product in a chemical equation because they are unchanged by the chemical reaction.
In this chemical reaction, the spectator ions are:
[tex]K^+[/tex][tex]NO_{3}^{-}[/tex]Finally, in order to obtain the net ionic equation, we would cancel out the two (2) spectator ions:
[tex]Ag^{+}_{(aq)} + I^{-}_{(aq)}[/tex] -----> [tex]AgI_{(s)}[/tex]
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How many liters of H2(g) at STP are produced per gram of Al(s) consumed in the following reaction? 2Al(s)+6HCl(aq)→2AlCl3(aq)+3H2(g)
Answer:
1.24 L of H₂ at STP .
Explanation:
2Al(s) +6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)
2 moles 3 x 22.4 L
2 x 27 g of Al reacts to give 3 x 22.4 L of H₂ at STP .
1 g of Al will react to give 3 x 22.4 / ( 2 x 27 ) L of H₂ at STP .
= 1.24 L of H₂ at STP .
The volume of hydrogen produced by 1 grams of Al has been 1.24 L.
The balanced chemical reaction has been given as:
[tex]\rm 2\;Al\;+\;6\;HCl\;\rightarrow\;2\;AlCl_3\;+\;3\;H_2[/tex]
From the equation, 2 moles of Aluminum gives 3 moles of Hydrogen
The mass of the compound from moles can be given as:
Mass = moles × molecular mass
Mass of 2 moles Al = 2 × 27 g
Mass of 2 moles Al = 54 g
Mass of 3 moles hydrogen = 3 × 2 g
Mass of 3 moles hydrogen = 6 g
From the equation,
54 g aluminum gives = 6 grams hydrogen
[tex]\rm 1\;gram\;aluminum\;=\;\dfrac{6}{54}\;\times\;1[/tex]
1 gram Aluminum = 0.11 grams hydrogen
The mass of hydrogen produced by 1 gram Al has been 0.11g. The moles equivalent to 0.11g hydrogen has been given as:
Mass = moles × molecular mass
0.11 g = moles × 2 g/mol
Mole of hydrogen = 0.055 mol
The moles of hydrogen produced by 1 gram of Al has been 0.055 mol.
According to the ideal gas equation, any gas at STP has 1 mole equivalent to 22.4 L. So,
1 mol = 22.4 L
0.055 mol = 0.055 × 22.4 L
0.055 mol = 1.244 L.
The volume of hydrogen produced by 1 grams of Al has been 1.24 L.
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How many lbs are in 5 kilograms
Answer:
There are approximately 11.0231 pounds in a kilogram
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a 50.0L tank with 14. mol of sulfur dioxide gas and 2.6 mol of oxygen gas, and when the mixture has come to equilibrium measures the amount of sulfur trioxide gas to be 1.6 mol. Calculate the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.
Answer:
Explanation:
From the given information:
The equation for the reaction can be represented as:
[tex]2SO_2 + O_2 \to 2SO_3[/tex]
The I.C.E table can be represented as:
2SO₂ O₂ 2SO₃
Initial: 14 2.6 0
Change: -2x -x +2x
Equilibrium: 14 - 2x 2.6 - x 2x
However, Since the amount of sulfur trioxide gas to be 1.6 mol.
SO₃ = 2x,
then x = 1.6/2
x = 0.8 mol
For 2SO₂; we have 14 - 2x
= 14 - 2(0.8)
= 14 - 1.6
= 12.4 mol
For O₂; we have 2.6 - x
= 2.6 - 1.6
= 1.0 mol
Thus;
[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,
[O₂] = 1/50 = 0.02 M ,
[SO₃] = 1.6/50 = 0.032 M
Kc = [SO₃]² / [SO₂]² [O₂]
= ( 0.032²) / ( 0.248² x 0.02)
= 0.8325
Recall that; the equilibrium constant for the reaction [tex]2SO_2 + O_2 \to 2SO_3[/tex] = 0.8325;
If we want to find:
[tex]SO_2 + \dfrac{1}{2}O_2 \to SO_3[/tex]
Then:
[tex]K_c = (0.8325)^{1/2}[/tex]
[tex]\mathbf{K_c = 0.912}[/tex]
Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.
how should the following italicized statement be taken??? elements can be broken further into simpler substances.
a. true
b. false.
Answer:
false
Explanation:
Elements cannot be broken down into a simpler substance.
Which element forms a cation?
A. gold
B. phosphorus
C. fluorine
D. nitrogen
Answer:
gold
Explanation:
Cations are positively charged, so the element that loses electrons to have the same amount of electrons as a noble gas atom is gold.
pentane or 2,2,3- trimethylhexane has the higher boiling point. why?
Answer:
2,2,3- trimethylhexane because it has more carbon atoms than pentane.
Explanation:
Hello.
In this case, since the physical properties of organic compounds are intensified as the number of carbon atoms start increasing on it, the larger the amount of carbon atom, the higher the boiling point since more energy is required to allow the liquid-phase molecules to transcend to the vapor-phase.
In such a way, since pentane has five carbon atoms and 2,2,3- trimethylhexane has nine carbon atoms, 2,2,3- trimethylhexane has the highest boiling point.
Best regards.
How many valence electrons must a lithium atom lose to obtain a complete valence shell?
A. one
B. two
C. three
D. four
Answer:
C. three
Explanation:
HELP FAST PLZ!!!!! Which phase change allows a substance to transform from a liquid to a
gas?
melting
Ofreezing
O ionization
condensation
deionization
O
evaporation
sublimation
Answer:evaporation
Explanation:
Answer:
Pretty sure its evaporation, if its not I'm very sorry.
Calculate the fractional saturation for hemoglobin when the partial pressure of oxygen is 40 mm Hg. Assume hemoglobin is 50%% saturated with oxygen at a partial pressure of 22 mm Hg and that the Hill coefficient is 3.
Answer:
The fractional saturation for hemoglobin is 0.86
Explanation:
The fractional saturation for hemoglobin can be calculated using the formula
[tex]Y_{O_{2} } = \frac{(P_{O_{2} })^{h} } {(P_{50})^{h} + (P_{O_{2} })^{h} }[/tex]
Where [tex]Y_{O_{2} } \\[/tex] is the fractional oxygen saturation
[tex]{P_{O_{2} }[/tex] is the partial pressure of oxygen
[tex]P_{50}[/tex] is the partial pressure when 50% hemoglobin is saturated with oxygen
and h is the Hill coefficient
From the question,
[tex]{P_{O_{2} }[/tex] = 40 mm Hg
[tex]P_{50}[/tex] = 22 mm Hg
h = 3
Putting these values into the equation, we get
[tex]Y_{O_{2} } = \frac{(P_{O_{2} })^{h} } {(P_{50})^{h} + (P_{O_{2} })^{h} }[/tex]
[tex]Y_{O_{2} } = \frac{40^{3} }{22^{3} + 40^{3} }[/tex]
[tex]Y_{O_{2} } = \frac{64000 }{10648 + 64000 }[/tex]
[tex]Y_{O_{2} } = \frac{64000 }{74648 }[/tex]
[tex]Y_{O_{2} } = 0.86[/tex]
Hence, the fractional saturation for hemoglobin is 0.86.
Quickly please! Which is a group of tissues that work together to carry out a common function?
cell
organ
organelle
organ system
Answer:
organ system
Explanation:
Answer:
B) organ
Explanation:
5. What would be the temperature change if 12.5 g of water absorbed 35 J of heat?
Explanation:
its 399929939932929292
In an sp2 hybridized atom, we saw that there was one unhybridized 2p orbital. How many unhybridized 2p orbitals remain on an atom that has sp3 hybrid orbitals?
A. One.B. Two.C. Three.D. None.
Answer:
None
Explanation:
We must recall that the number of hybrid orbitals formed must be equal to the number of atomic orbitals that were combined. Hence, if we have an sp3 hybridized orbital, there are four hybrid orbitals formed.
Since there are four hybrid orbitals formed from the three p and one s orbital that combined, there is no unhybridized orbital left.
The number of unhybridized 2p orbitals that remains on an atom that has sp3 hybrid orbital is : ( D ) None
The hybridization of an s orbital with three p orbitals will yield four sp3 hybrid orbitals, which means that there will be no unhybridized 2p orbitals left.
This is because during the hybridization, the number of hybrid orbitals yielded must be equal to the combined number of atomic orbitals that yielded the Hybrid orbitals.
Hence we can conclude that the number of unhybridized 2p orbitals that remain on an atom that has sp3 hybrid orbital is zero ( None )
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What two types of elements make up an Ionic bond ?
Answer:
Metals and Non-metals
Explanation:
Which step of mitosis involves the nucleus spitting and nuclear membranes forming around each new nucleus??
Answer:
telophase
Explanation:
Find ΔHrxn for the following reaction:
2PbS(s)+3O2(g)→2PbO(s)+2SO2(g)
Answer:
ΔH°rxn = -827.5 kJ
Explanation:
Let's consider the following balanced equation.
2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)
We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]
ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]
ΔH°rxn = -827.5 kJ
The standard enthalpy of the reaction is -827.5 kJ/mol
The standard enthalpy of reaction [tex]\mathbf{\Delta H^0_{rxn}}[/tex] is the enthalpy change that happens in a system whenever one mole of the matter is converted through a chemical process under normal conditions.
The given reaction can be expressed as:
2PbS(s) + 3O₂(g) → 2PbO(s) + 2SO₂(g)
The standard enthalpy can be represented by the equation:
[tex]\mathbf{\Delta H^0_{rxn} = \sum \Delta _f ^0(products) - \sum \Delta _f^0(reactants)}[/tex]
At standard conditions, the standard enthalpies of formation of the given species are:
ΔH°f(PbO(s)) = 219 kJ/molΔH°f(SO₂(g)) = -296.83 kJ/molΔH°f(PbS(s)) = 100.4 kJ/molΔH°f(O₂(g)) = 0 kJ/mol
∴
[tex]\mathbf{ \Delta H^0_{rxn} = \Big[2 mol \times \Delta H^0_f(PbO(s)) + 2 mol \times \Delta H^0_ f(SO_2(g) )\Big] - \Big[2 mol \times \Delta H^0_f (PbS(s))} + \mathbf{ 3 mol \times \Delta H^0_f(O_2(g) )\Big] }}[/tex]
[tex]\mathbf{\Delta H^0rxn = [2 mol \times (-217.32 kJ/mol) + 2 mol \times (-296.83)] - [2 mol \times (-100.4)} \\ \mathbf{+ 3 mol \times 0 kJ/mol]}}[/tex]
[tex]\mathbf{\Delta H^0rxn = -827.5 \ kJ/mol}}[/tex]
Therefore, we can conclude that the standard enthalpy of the reaction is -827.5 kJ/mol
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Which statement describes the effect of sorting and recombining genes in sexual reproduction?
Group of answer choices
offspring receive only recessive genes
offspring receive only dominant genes
offspring receive the same set of genes
offspring receive a unique combination of genes
A major component of gasoline is octane C8H18. When liquid octane is burned in air it reacts with oxygen O2 gas to produce carbon dioxide gas and water vapor. Calculate the moles of octane needed to produce 1.4mol of water. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits. 1.8mol
Answer:
0.2 moles of octane
Explanation:
The equation of the reaction is;
C8H18(g) + 25/2 O2(g) ------> 8CO2(g) + 9H2O(l)
From the equation;
1 mole of octane yields 9 moles of water
x moles of octane yields 1.4 moles of water
x= 1.4 × 1/9
x= 0.16 moles of octane
The amount of UVA radiation hitting a surface at sea level in a lightly clouded day is about 70W/m2. About half of that can be absorbed by the skin. A typical carbon- carbon bond requires 348 kJ/mol to break. A person lies on the beach for about 1 hour without sunscreen (i.e. fully exposed to UVA radiation). Estimate the number of C-C bonds broken in this person’s back (about 0.18 m2) over that period. Assume that the average wavelength of UVA is 335 nm.
Answer:
Explanation:
energy of solar radiation = 70 W / m²
energy absorbed in 1 hour by an area of .18 m²
= 70 x .5 x .18 x 60 x 60 J
= 22.68 x 10³ J
bond energy of i mole bond = 348 x 10³ J
bond energy of 6.02 x 10²³ bonds = 348 x 10³ J
bond energy of one bond = 57.8 x 10⁻²⁰ J
No of bonds broken by energy 22.68 x 10³
= 22.68 x 10³ / 57.8 x 10⁻²⁰
= .3923 x 10²³
= 39.23 x 10²⁰ .
Someone plz hello me ASAP it would be appreciated on question 9 btw
Answer:
C.
Explanation:
The water is being stored for the time being, and potential energy is, essentially, defined like that. Once the energy is utilized, it will turn into kinetic energy.
The escape velocity from Earth’s surface is 1.12*10^4 meters per second. At this speed, how many kilometers would a rocket travel in 3 minutes
Answer:
2016 Km.
Explanation:
The following data were obtained from the question:
Speed (S) = 1.12×10⁴ m/s
Time (t) = 3 mins
Distance (d) =.?
Next, we shall convert 3 mins to seconds. This can be obtained as follow:
1 min = 60 s
Therefore,
3 mins = 3 min × 60 s / 1 min
3 mins = 180 s
Next, we shall determine the distance travelled by the rocket. This is illustrated below:
Speed (S) = 1.12×10⁴ m/s
Time (t) = 180 s
Distance (d) =.?
Speed (S) = distance (d) /time (t)
S = d/t
1.12×10⁴ = d/ 180
Cross multiply
d = 1.12×10⁴ × 180
d = 2016000 m
Finally, we shall convert 2016000 m to Km. This can be obtained as shown below:
1000 m = 1 Km
Therefore,
2016000 m = 2016000 m × 1 Km / 1000 m
2016000 m = 2016 Km
Therefore, the rocket will travel 2016 Km in 3 mins
helpppp me asap please help me asap please
If theoretical mass of CH3OH produced in the reaction below was 85.3grams, and the actual mass produced during the experiment was 80.1 gA. Calculate the percent yield of this reaction. B. Give one reason why the numbers are different. COfg) + 2 H2(g) -> CH;OH(I)
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Research Hypatia's achievements in the world
of science.
What is she most known for?
Write down three interesting facts about her
life.
Answer:
See explanation
Explanation:
Hypatia is popular for her work in mathematics. She also did some work in the area of astronomy. Her well know work in mathematics is her ideas about conic sections.
She was born the Theon of Alexandria and she was a professional mathematician in her life time.
She was the greatest mathematician of her time and she was telling leader of the Neoplatonist school of philosophy in Alexandria. By so doing, she conquered the culture of sexism in her time.
She was trained by her father in mathematics and eventually replaced him. She was the last major mathematician in the Alexandrian tradition.
