Answer:
True
Explanation:
Evaporation is the process by which a substance changes its state from liquid to gas. evaporation occurs at all temperatures but it's rate increases as temperature increases.
Pure water vapour can be produced by evaporation.
As the liquids are removed, the solids present in solution becomes more concentrated.
Consider equimolar samples of different ideal gases at the same volume and temperature. Gas A has a higher molar mass than gas B.
1. Compare the pressure
a. A > B
b. A= B
c. A < B
2. Compare the speed.
a. A > B
b. A= B
c. A < B
Answer:
I don't understand the question
Explanation:
1. c
2. c
Answer:
1. b. A = B
2. c. A < B
Explanation:
Consider equimolar samples of different ideal gases at the same volume and temperature. (nA = nB)
Gas A has a higher molar mass than gas B. (MA > MB)
1. We can calculate the pressure using the ideal gas equation.
P = n × R × T / V
Since n, T and V are equal, the pressure of both gases are equal as well.
2. We can calculate the root-mean-square of the velocity using the following expression.
v(rms) = √(3RT/M)
There is an inverse relationship between speed and molar mass. Since A has a higher molar mass, it has a lower speed.
Diethyl ether (C2H5 )2O vaporizes at room temperature. If the vapor exerts a pressure of 233 mm Hg in a flask at 25 °C, what is the density of the vapor?
Answer: The density of the given vapor is 0.939 g/L.
Explanation:
Given: Pressure = 233 mm Hg (1 mm Hg = 0.00131579 atm) = 0.31 atm
Temperature = [tex]25^{o}C[/tex] = (25 + 273) K = 298 K
According to the ideal gas equation,
[tex]PV = \frac{m}{M}RT[/tex]
where,
P = pressure
V = volume
m = mass
M = molar mass
R = gas constant = 0.0821 L atm/mol K
T = temperature
This formula can be re-written as follows.
[tex]PM = \frac{m}{V}RT[/tex] (where, [tex]Density = \frac{mass (m)}{Volume (V)}[/tex] )
Hence, formula used to calculate density of diethy ether (molar mass = 74.12 g/mol) vapor is as follows.
[tex]d = \frac{PM}{RT}[/tex]
Substitute values into the above formula as follows.
[tex]d = \frac{PM}{RT}\\= \frac{0.31 atm \times 74.12 g/mol}{0.0821 L atm/mol K \times 298 K}\\= \frac{22.9772}{24.4658}\\= 0.939 g/L[/tex]
Thus, we can conclude that the density of the given vapor is 0.939 g/L.
What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:
2 NaN3 is) 2Na (s) + 3N2 (8)
Answer:
108.6 g
Explanation:
2NaN₃(s) → 2Na(s) + 3N₂(g)First we use the PV=nRT formula to calculate the number of nitrogen moles:
P = 1.00 atmV = 56.0 Ln = ?R = 0.082 atm·L·mol⁻¹·K⁻¹T = 0 °C ⇒ 0 + 273.2 = 273.2 KInputting the data:
1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 Kn = 2.5 molThen we convert 2.5 moles of N₂ into moles of NaN₃, using the stoichiometric coefficients of the balanced reaction:
2.5 mol N₂ * [tex]\frac{2molNaN_3}{3molN_2}[/tex] = 1.67 mol NaN₃Finally we convert 1.67 moles of NaN₃ into grams, using its molar mass:
1.67 mol * 65 g/mol = 108.6 gWhat volume (mL) of the sweetened tea described in Example 1 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example
The question is incomplete, the complete question is:
What volume (mL) of the sweetened tea described in Example 3.14 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example. The example is attached below.
Answer: 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink
Explanation:
We first calculate the number of moles of soft drink in a volume of 10 mL
The formula used to calculate molarity:
[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}[/tex] .....(1)
Taking the concentration of soft drink from the example be = 0.375 M
Volume of solution = 10 mL
Putting values in equation 1, we get:
[tex]0.375=\frac{\text{Moles of sugar in soft drink}\times 1000}{10}\\\\\text{Moles of sugar in soft drink}=\frac{0.375\times 10}{1000}=0.00375mol[/tex]
Calculating volume of sweetened tea:
Moles of sugar = 0.00375 mol
Molarity of sweetened tea = 0.05 M
Putting values in equation 1, we get:
[tex]0.05=\frac{0.00375\times 1000}{\text{Volume of sweetened tea}}\\\\\text{Volume of sweetened tea}=\frac{0.00375\times 1000}{0.05}=75mL[/tex]
Hence, 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink
Under certain conditions, the substance mercury(II) oxide can be broken down to form mercury and oxygen. If 32.2 grams of mercury(II) oxide react to form 29.8 grams of mercury, how many grams of oxygen must simultaneously be formed
Explanation:
This is a decomposition reaction. Firstly, you will want to write the chemical equation out and balance it.
[tex]2Hg_2O->4Hg+O_2[/tex] (The -> is supposed to be an arrow, sorry!)
We see that there's only 1mol of Oxygen made in the products, we can do some simple math to solve for the amount of grams of Oxygen produced according to the amount of the reactant (Hg2O).
[tex]32.2gHg_2O*\frac{1molHg_2O}{417.18gHg_2O}*\frac{1molO_2}{2molHg_2O}*\frac{32gO_2}{1molO_2}[/tex]
I want to break this down, just in case:
The 417.18gHg2O is the molecular mass of the molecule (so I doubled Hg and added 16 to it to get this number).
As we can see in the chemical equation, 1mol Hg2O produces 2mol O because Oxygen is a diatomic molecule (so there will always be two of it when it's by itself).
And finally, in 1mol O2 there are 32g of O2.
** When you do math like this, always make sure that all of your units cancel out except for the units you're looking for. For example, here we're looking for the grams of Oxygen, so after everything else cancels out, we should only have grams O2.
So, 1.23gO2 should be your answer.
Given the amount of camphor (200mg) we are using in this experiment, please determine how many mg of sodium borohydride to use in this reaction. We would like you to use 5.2 molar equivalents of this reagent. This means 5.2 times the mmol of camphor we are using. As an example: for 110.0 mg of camphor,142 mg of NaBH4 would be used (see if you can confirm this result). For complete credit, your work needs to be clearly drawn out!
Answer:
Explanation:
From the given information:
Camphor may be reduced as readily in the presence of sodium borohydride(NaHB4). The resulting compound which is stereoselective requires 1 mole of sodium borohydride (NaHB4) to reduce 1 mole of camphor in this reaction. The reaction is shown below.
Through the reduction process of camphor, the reducing agent can reach the carbonyl face with a one-carbon linkage. The product stereoisomer is known as borneol.
If the molecular weight of camphor = 152.24 g/mol
and it mass = 200 mg
The its no of moles = 200 mg/ 152.24 g/mol
= 1.3137 mmol
Now the amount of the required mmol for NaBH4 to be consumed in the reaction = 5.2 × 1.3137 mmol
= 6.831 mmol
since the molar mass of NaBH4 = 37.83 g/mol
Then, using the same formula:
No of moles = mass/molar mass
mass = No of moles × molar mass
mass = 6.831 mmol × 37.83 g/mol
mass of NaBH4 used = 258.42 mg
atomaticity of chlorine 1) 2, 2)1, 3) 32 , 4) 4.
Answer:
ATOMICITY OF CHLORINE IS 2Explanation:
Atomicity is defined as the total number of atoms present in a molecule.
Which 2 resonance forms destablize the carbocation intermediate if bezonitrile undergoes chlronation at the ortho or para positions
The question is incomplete, the complete question is shown in the image attached
Answer:
A and B
Explanation:
The electrophilic substitution of arenes yields a cation intermediate. The positive charge of the cation is delocalized over the entire ring.
The -CN group directs incoming electrophiles to the ortho/para position. The resonance structures for the chlorination of benzonitrile are shown in the question.
Recall that -CN is an electron withdrawing group. The resonance forms that destablize the carbocation intermediate are those in which the -CN group is directly attached to the carbon atom bearing the positive charge as in structures A and B.
Suppose you perform a titration of an unknown weak acid solution. You start with 4.00 mL of the weak acid and find that it takes 14.2 mL of 0.0500 M NaOH to reach the equivalence point. What is the concentration of the unknown weak acid solution
Answer:
0.1775 M
Explanation:
The reaction that takes place is:
HA + NaOH → NaA + H₂OWhere HA is the unknown weak acid.
At the equivalence point all HA moles are converted by NaOH. First we calculate how many NaOH moles reacted, using the given concentration and volume:
0.0500 M NaOH * 14.2 mL = 0.71 mmol NaOHThat means that in 4.00 mL of the weak acid solution, there were 0.71 weak acid mmoles. With that in mind we can now calculate the concentration:
0.71 mmol HA / 4.00 mL = 0.1775 Mg Arrange the following compounds in order of acidity (highest to lowest): H2O, H3O , HCl A. CH3COOH > HCl > H2O B. H2O > CH3COOH > HCl C. HCl > H2O > CH3COOH D. HCl > CH3COOH > H2O
Answer:
Arrange the following compounds in order of acidity (highest to lowest): H2O, CH3COOH , HCl
A. CH3COOH > HCl > H2O
B. H2O > CH3COOH > HCl
C. HCl > H2O > CH3COOH
D. HCl > CH3COOH > H2O
Explanation:
The given substances are acetic acid, hydrochloric acid, and water.
Since HCl is a strong acid and it undergoes complete ionization.
CH3COOH acetic acid is a weak acid and it undergoes partial dissociation in water.
Pure water is a neutral substance.
Hence, the order of acidity is shown below:
HCl > CH3COOH > H2O.
Among the given options, option D is the correct answer.
If a student drops 2.3g pieces of magnesium into a flask of hydrochloric acid, this reaction occurs: Mg + 2HCl= MgCl2 + H2
How many liters of hydrogen can be produced at a pressure of 2 atm and a temperature of 298 K
Answer:
1.2 L
Explanation:
Step 1: Write the balanced equation
Mg + 2 HCl ⇒ MgCl₂ + H₂
Step 2: Calculate the moles corresponding to 2.3 g of Mg
The molar mass of Mg is 24.31 g/mol.
2.3 g × 1 mol/24.31 g = 0.095 mol
Step 3: Calculate the moles of H₂ produced
0.095 mol Mg × 1 mol H₂/1 mol Mg = 0.095 mol H₂
Step 4: Calculate the volume occupied by the hydrogen
We will use the ideal gas equation.
P × V = n × R × T
V = n × R × T/P
V = 0.095 mol × (0.0821 atm.L/mol.K) × 298 K/2 atm = 1.2 L
g Calculate the percent yield when you start with 0.50 grams of salicylic acidand end with 0.33 grams of aspirin. The molecular weight of salicylic acid is 138.12 g/mol, and the molecular weight of aspirin is 180.157 g/mol.
Explanation:
hope the picture above helps you to understand:)
In a single displacement reaction Zinc can displace ALL but…
Iron
Nickel
Calcium
Lead
Answer:
Calcium
Explanation:
Zinc cannot displace Ca because calcium is above it in the reactivity series
3)O que são políticas públicas?
Answer:
azertyuiopazertyuiiop
Help pls!!!
For groups 13 through 18, the number of valence electrons is equal to the group number
a. minus 10
b. minus the period number
c. plus 1
d. plus the period number
Answer:
a. minus 10
Explanation:
An element in group 13 = Boron ,valence electrons = 3 , therefore, valence electrons in group 13 = group no. -10
An element in group 18 = Neon, valence electrons = 8 , therefore, valence electrons in group 18 = group no. - 10
For groups 13 through 18, the number of valence electrons in an atom is equal to the group number minus 10. Therefore, option (A) is correct.
What is a valence electron?Valence electrons in an atom can be described as the electrons occupying the outer most electron shell of an atom while the electrons in the inner shell are core electrons. Lewis structures can be helpful to calculate the number of valence electrons.
Valence electrons can be filled in several electron shells as they are caused interaction between atoms and responsible for the formation of chemical bonds. Only valence electrons can contribute to the formation of a chemical bond and decide the reactivity of the element.
The general electronic configuration of group 13 is ns²np¹ has three valence electrons. It can be determined as group number - 10 = 13 - 10 = 3.
The general electronic configuration of group 18 is ns²np⁶ has eight valence electrons. It can be determined as group number - 10 = 18 - 10 = 8.
Learn more about valence electrons, here:
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How many molecules of C 2H 5Br will be present if you had 4.52 g of this compound?
How many molecules are present in 1296 g of dinitrogen pentoxide (N2O5)
Answer:
The molar mass of N2O5 is 108 g/mol. 1296g of N2O5 has 1296 / 108 = 12 moles. 1 mole contains 6.022 x 10^23 molecules
If the volume of the gas is increased to 9.6 L , what will the pressure be?
15. You are interested in separating 4-methylbenzoic acid from 1,4-dimethoxybenzene using a procedure similar to the extraction procedure we used in lab. You plan to use sodium bicarbonate instead of sodium hydroxide. a) Show the reaction between salicylic acid and sodium bicarbonate. Label the acid, base, conjugate acid, conjugate base. b) Give the pKa values of the acid and conjugate acid. c) Which base will work better, sodium hydroxide or sodium bicarbonate
Solution :
a). The separation of 4-methylbenzoic acid from 1,4-dimethoxybenzene will work but it will result in lower recovery.
In the reaction of acid-base to form a sodium 4 - methoxy benzoate, that is soluble in the water, 4-methoxy benzoic acid reacts with the sodium bicarbonate to give sodium 4-methoxybenzoate as well as carbonic acid.
b). The pKa for the 4-methoxybenzoic acid is [tex]4.46[/tex], and that of carbonic acid is [tex]6.37[/tex]
c). The Keq for the reaction is [tex]10(6.37 - 4.46) = 101.91[/tex]
Therefore, the equilibrium lies to the right and also the reaction favors the products and the separation works.
But the recovery will be low when compared to the extraction with Sodium hydroxide as the strong base will drive the equilibrium further to the right position, thus neutralizing all the acids virtually. And the weak base will partially neutralize the acid.
The element antimony has two stable isotopes, antimony-121 with a mass of 120.90 amu and antimony-123 with a mass of 122.90 amu. From the atomic weight of Sb = 121.76 one can conclude that: ________
antimony-123 has the highest percent natural abundance
most antimony atoms have a mass of 121.76 amu
antimony-121 has the highest percent natural abundance
both isotopes have the same percent natural abundance
Answer:
antimony-121 has the highest percent natural abundance
Explanation:
percent natural abundance;
121.76 = 120.90 x + 122.90 (1 - x)
121.76 = 120.90 x + 122.90 - 122.90x
121.76 = -2x + 122.90
121.76 - 122.90 = -2x
x= 121.76 - 122.90/ -2
x= 0.57
Where x and 1 - x refers to the relative abundance of each of the isotopes
Percent natural abundance of antimony-121 = 57 %
Percent natural abundance of antimony-123 = (1 - 0.57) = 43%
Let us remember that isotopy refers to a phenomenon in which atoms of the same element have the same atomic number but different mass numbers. This results from differences in the number of neutrons in atoms of the same element.
We can clearly see that antimony-121 has the highest percent natural abundance.
1. A 225-L barrel of white wine has an initial free SO2 concentration of 22 ppm and a pH of 3.70. How much SO2 (in grams) should be added to the barrel to result in the required SO2 level
Answer:
The appropriate answer is "9.225 g".
Explanation:
Given:
Required level,
= 63 ppm
Initial concentration,
= 22 ppm
Now,
The amount of free SO₂ will be:
= [tex]Required \ level -Initial \ concentration[/tex]
= [tex]63-22[/tex]
= [tex]41 \ ppm[/tex]
The amount of free SO₂ to be added will be:
= [tex]41\times 225[/tex]
= [tex]9225 \ mg[/tex]
∵ 1000 mg = 1 g
So,
= [tex]9225\times \frac{1}{1000}[/tex]
= [tex]9.225[/tex]
Thus,
"9.225 g" should be added.
A sample of calcium fluoride was decomposed into the constituent elements. Write a balanced chemical equation for the decomposition reaction. If the sample produced 294 mg of calcium, how many g of fluorine were formed
Answer:
A sample of calcium fluoride was decomposed into the constituent elements. Write a balanced chemical equation for the decomposition reaction. If the sample produced 294 mg of calcium, how many g of fluorine was formed
Explanation:
The balanced chemical equation for the decomposition of calcium fluoride is shown below:
[tex]CaF_2(s)->Ca(s)+F_2(g)[/tex]
The sample produced 294 g of calcium then, how many grams of fluorine is formed?
From the balanced chemical equation,
1 mol of CaF2 forms 1mol of calcium and 1 mol of fluorine.
That is:
40g of calcium and 38.0 g of fluorine are formed.
then,
If 294 g of calcium is formed then how many grams of fluorine is formed?
[tex]294g Ca * 38g F2 / 40g Ca\\=279.3 g F_2[/tex]
Hence, 279.3 g of fluorine will be formed.
Consider the synthesis of water as shown in Model 3. A container is filled with 10,0 g of H, and
5.0 g of Oz
Which reactant (hydrogen or oxygen) is the limiting reactant in this case?
Answer:
Oxygen, O₂ is the limiting reactant
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2H₂ + O₂ —> 2H₂O
Next, we shall determine the masses of H₂ and O₂ that reacted from the balanced equation. This can be obtained as follow:
Molar mass of H₂ = 2 × 1 = 2 g/mol
Mass of H₂ from the balanced equation = 2 × 2 = 4 g
Molar mass of O₂ = 16 × 2 = 32 g/mol
Mass of O₂O from the balanced equation = 1 × 32 = 32 g
SUMMARY:
From the balanced equation above,
4 g of H₂ reacted with 32 g of O₂.
Finally, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
4 g of H₂ reacted with 32 g of O₂.
Therefore, 10 g of H₂ will react with
= (10 × 32)/4 = 80 g of O₂.
From the calculations made above, we can see that a higher mass (i.e 80 g) of O₂ than what was given (i.e 5 g) is required to react completely with 10 g of H₂. Therefore, O₂ is the limiting reactant.
Oxygen has been the limiting reactant in the reaction.
A limiting reactant can be defined as the reactant in the reaction in which the product concentration has been dependent.
The balanced equation for the formation of water has been:
[tex]\rm 2\;H_2\;+\;O_2\;\rightarrow\;2\;H_2O[/tex]
For the formation of reaction to form 2 moles of water, 2 moles of hydrogen reacts with 1 mole of oxygen.
The moles can be calculated as:
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
The moles of Hydrogen in 10 g [tex]\rm H_2[/tex]:
Moles = [tex]\rm \dfrac{10}{2}[/tex]
Moles of hydrogen = 5 mol.
Moles of Oxygen in 5 grams Oxygen:
Moles = [tex]\rm \dfrac{5}{32}[/tex]
Moles of oxygen = 0.156 mol.
For the reaction with 2 moles of Hydrogen 1 mole of Oxygen has been required.
For reacting with 5 mol of Hydrogen, moles of oxygen required are:
Moles of oxygen = [tex]\rm \dfrac{1}{2}\;\times\;5[/tex]
Moles of oxygen required = 2.5 moles.
The available oxygen = 0.156 moles.
Since the moles of oxygen available is lesser than required, the formation of the product has been dependent on the concentration of the oxygen.
Thus, oxygen has been the limiting reactant in the reaction.
For more information about the limiting reactant, refer to the link:
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A sample of 10.6 g of KNO3 was dissolved in 251.0 g of water at 25 oC in a calorimeter. The final temperature of the solution was 21.5 oC. What is the molar heat of solution of KNO3
Answer:
36.55kJ/mol
Explanation:
The heat of solution is the change in heat when the KNO3 dissolves in water:
KNO3(aq) → K+(aq) + NO3-(aq)
As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.
To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:
Moles KNO3 -Molar mass: 101.1032g/mol-
10.6g * (1mol/101.1032g) = 0.1048 moles KNO3
Change in heat:
q = m*S*ΔT
Where q is heat in J,
m is the mass of the solution: 10.6g + 251.0g = 261.6g
S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-
And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C
q = 261.6g*4.184J/g°C*3.5°C
q = 3830.87J
Molar heat of solution:
3830.87J/0.1048 moles KNO3 =
36554J/mol =
36.55kJ/mol
We can use bond-line formulas to represent alkenes in much the same way that we use them to represent alkanes. Consider the following alkene: h5ch5e4 How many carbon atoms are sp2−hybridized in this alkene?
Answer:
2
Explanation:
The number of carbon atoms that are sp²-hybridized in this alkene is 2
Because all the single bonded carbon atoms in the alkene are sp²-hybridized
There are three(3) single formed via sp² orbitals and one ( 1 ) PI bond formed via Pure-P-orbital
attached below is the some part of the solution
Calculate the mass of butane needed to produce 80.0g of carbon dioxide
Answer:
Multiply the number of moles of butane by its molar mass, 58.12g/mol, to produce the mass of butane. Mass of butane = 18.8g.
Suppose you analyze a 30.3 g sample of bleach and determine that there are 2.61 g of sodium hypochlorite present. What is the percent of sodium hypochlorite in the bleach sample
Answer:
8.61 %
Explanation:
The percent of sodium hypochlorite in the bleach sample can be calculated using the following formula:
% of sodium hypochlorite = Mass of sodium hypochlorite / mass of sample * 100%We input the data given by the problem:
% of sodium hypochlorite = 2.61 g / 30.3 g * 100 % = 8.61 %A gas bottle contains 0.650 mol of gas at 730. mmHg pressure. If the final pressure is 1.15 atm, how many moles of gas were added to the bottle
Answer: There are 0.779 moles of gas were added to the bottle.
Explanation:
Given: [tex]n_{1}[/tex] = 0.650 mol, [tex]P_{1}[/tex] = 730 mm Hg (1 mm Hg = 0.00131579 atm) = 0.96 atm
[tex]n_{2}[/tex] = ?, [tex]P_{2}[/tex] = 1.15 atm
Formula used is as follows.
[tex]\frac{P_{1}}{n_{1}} = \frac{P_{2}}{n_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{P_{1}}{n_{1}} = \frac{P_{2}}{n_{2}}\\\frac{0.96 atm}{0.650 mol} = \frac{1.15 atm}{n_{2}}\\n_{2} = 0.779 mol[/tex]
Thus, we can conclude that there are 0.779 moles of gas were added to the bottle.
compound of aspartame is a dipeptide that is often used as a sugar substitute which functional groups are present
Answer:
Carboxyl, primary amine, amide, ester, and phenyl.
Explanation:
The functional groups present in the compound of aspartame are carboxyl, primary amine, amide, ester, and phenyl. Aspartame is an artificial non-saccharide sweetener which is 200 times sweeter than sucrose. This aspartame is commonly used as a sugar substitute in many foods and beverages. It has the trade names such as NutraSweet, Equal, and Canderel.
Determine whether each of the properties described applies to volumetric or graduated glassware. 1. Used for applications in which great accuracy is needed_____.a. volumetricb. graduated 2. Capable of measuring a range of volumes of liquid_____.a. volumetric
b. graduated 3. Designed to measure one specific volume of liquid_____.a. volumetric
b. graduated
Answer:
1) volumetric
2) graduated
3) volumetric
Explanation:
A volumetric glassware is a glassware that is marked at a particular point. A typical example of a volumetric glassware is the volumetric flask. A volumetric glassware is capable of measuring only a specific volume of a liquid.
On the other hand, graduated glassware can measure a range of volumes of liquid. However, a volumetric glassware is still required where a high degree of accuracy is important.
