How many grams of KNO3 can dissolve in 100g of water at 20°C?
Answer:
30 grams of KNO3 can be dissolved.
Explanation:
Hello there!
In this case, since the solubility is defined as the maximum amount of solute that can be dissolved in a certain amount of solvent, usually 100 grams of water as function of the temperature, we will need to recall the graph for the solubility of KNO3 as shown on the attached file.
Thus, by identifying the curve for KNO3, we realize that at a temperature of 20 °C, the solubility is about 30 grams; which means 30 grams can be dissolved in 100 grams of water at 20 °C.
Regards!
In the following reaction, Zn is Zn(s) H2SO4(aq) --> ZnSO4(aq) H2(g) A.Reduced B.Oxidized C.This is not a redox reaction D.An oxidizing agent
Answer: The correct option is B) oxidized
Explanation:
Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.
The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.
A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.
For the given chemical reaction:
[tex]Zn+H_2SO_4+S\rightarrow ZnSO_4+H_2[/tex]
On the reactant side:
Oxidation number of H = +1
Oxidation number of Zn = 0
Oxidation number of S = +6
Oxidation number of O = -2
On the product side:
Oxidation number of H = 0
Oxidation number of Zn = +2
Oxidation number of S = +6
Oxidation number of O = -2
As the oxidation number of Zn is increasing from 0 to +2. Thus, it is getting oxidized. Similarly, the oxidation number of H is decreasing from +1 to 0. Thus, it is getting reduced.
Hence, the correct option is B) oxidized
Suppose you need to prepare 21.0 mL of formate buffer with a ratio of 4 of [sodium formate]/[formic acid] by mixing 0.10 M formic acid and 0.10 M sodium formate. How many milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid)
Answer: A volume of 20.49 milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid).
Explanation:
Given: Total volume of the buffer = 21.0 mL
[tex]\frac{[HCOONa]}{[HCOOH]} = 4[/tex] ... (1)
It is assumed that the volume of HCOONa is x. Hence, volume of HCOOH is (21.0 - x) mL.
Hence,
[HCOONa] = Molarity [tex]\times[/tex] Volume
= 0.10 [tex]\times[/tex] x
= 0.1x mmol
Similarly, [HCOOH] = Molarity [tex]\times[/tex] Volume
= 0.10 [tex]\times[/tex] (21.0 - x) mmol
Using equation (1),
[tex]\frac{[HCOONa]}{[HCOOH]} = 4\\\frac{0.1x}{(21.0 - x)} = 4\\0.1x = 84.0 - 4x\\4.1x = 84.0\\x = 20.49 mL[/tex]
As x is the volume of sodium formate. Hence, 20.49 mL of sodium formate is required to make the buffer.
Thus, we can conclude that a volume of 20.49 milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid).
If the electromagnet in the PhET simulation is disconnected from the battery, the compass needle will
A. Not move
B. Flip directions
C. Point north
D. Point south
You have been observing an insect that defends itself from enemies by secreting a caustic liquid. Analysis of the liquid shows it to have a total concentration of formate plus formic acid (Ka =1.8 x 10^-4) of 1.45 M; the concentration of formate ion is 0.015 M. What is the pH of the secretion?
Answer:
pH = 1.76
Explanation:
The mixture of formate and formic acid produce a buffer. The pH of the buffer is obtained using H-H equation as follows:
pH = pKa + log [A-] / [HA]
Where pH is the pH of the buffer:
pKa = -log Ka = 3.74
[A-] is molar concentration of conjugate base (Formate) = 0.015M
[HA] is molar concentration of weak acid (Formic acid) = 1.45M - 0.015M = 1.435M
Replacing:
pH = 3.74 + log [0.015M] / [1.435M]
pH = 1.76A stain is made up of molecules that do not have charged regions. What soap/solvent combination should I use?Are there multiple?
Answer:
The answer is below
Explanation:
Yes, there are multiple combinations.
Firstly, a solvent or soap combination could be used as a polar solvent with soap. In this way, the nonpolar region of the soap molecules would mix directly with the stain while the solvent would surround the soap-stain micelle.
Secondly, another method or way is to have a combination of a nonpolar solvent with no soap. In this method, however, it is the solvent that would mix directly with the stain.
Compare the solubility of calcium sulfite in each of the following aqueous solutions:
a. 0.10 M Ca(CH3COO)2
b. 0.10 M K2SO3
c. 0.10 M NaNO3
d. 0.10 M KCH3COO
1. More soluble than in pure water.
2. Similar solubility as in pure water.
3. Less soluble than in pure water.
Answer:
0.10 M Ca(CH3COO)2- Less soluble than in pure water.
0.10 M K2SO3- Less soluble than in pure water.
0.10 M NaNO3 - More soluble than in pure water.
0.10 M KCH3COO- Similar solubility as in pure water.
Explanation:
We have to cast our minds back to the idea of common ions effect. If any ion is already present in solution, the presence of that ion in solution prevents any solute containing a common ion with the solution from dissolving in that solution. In order words, the presence of a common ion makes a solute less soluble in a solvent than it is in pure water.
For instance, 0.10 M Ca(CH3COO)2 and K2SO3 both contain Ca^2+ and SO3^2- ions respectively which are also contained in the solute calcium sulfite.
The presence of these common ions in solution makes calcium sulfite less soluble in these solutions than it is in pure water because the equilibrium position for the dissolution of the solute lies towards the left hand side.
However, calcium sulfite is more soluble in 0.10 M NaNO3 than in pure water due to displacement reaction between the ions in solution.
The solubility of calcium sulfite and 0.10 M KCH3COO in pure water is quite comparable.
Consider the following electron configurations to answer the questions that follow: (i) 1s2 2s2 2p6 3s1 (ii) 1s2 2s2 2p6 3s2 (iii) 1s2 2s2 2p6 3s2 3p1 (iv) 1s2 2s2 2p6 3s2 3p4 (v) 1s2 2s2 2p6 3s2 3p5 The electron configuration belonging to the atom with the highest second ionization energy is ________.
Option (i) would have the highest 2nd Ionization Energy.
Option (i) is Sodium.
Can be Written as 2, 8 , 1
For its 1st Ionization energy... It'd be extremely easy to remove that Electron cos its on the outermost shell.
Now After Removing that Electron...
Sodium's Electronic Configuration Reduces to that of Neon Which is 2, 8.
Neon has a very stable Octet.
It would take an ENORMOUS amount of energy to break its Octet stability... that is... Remove 1 electron from its Octet.
So
Option (i) [Sodium] has the highest 2nd Ionization Energy
A 0.15 M solution of BaCl2 contains: Group of answer choices 0.30 M Ba2 ions and 0.30 M Cl- ions. 0.15 M Ba2 ions and 0.15 M Cl- ions. 0.30 M Ba2 ions and 0.15 M Cl- ions. 0.15 M Ba2 ions and 0.30 M Cl- ions. none of the above
Answer:
0.15 M Ba⁺² ions and 0.30 M Cl⁻ ions
Explanation:
The dissociaton of barium chloride is as follows:
BaCl₂ → Ba²⁺ + 2Cl⁻
By observing the stoichiometric coefficients, we can tell that the number of moles of Ba²⁺ is the same as the number of moles of BaCl₂, while the number of moles of Cl⁻ is the double of that.
A chemistry student needs 15.0 g of methanol for an experiment. She has available 320. g of 44.4% w/w solution of methanol in water. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button. Round your answer to 3 significant digits.
Answer:
33.8 g Solution
Explanation:
A chemistry student needs 15.0 g of methanol for an experiment. The concentration of ethanol in the solution is 44.4% w/w, that is, there are 44.4 g of methanol every 100 g of solution. The mass of solution that would contain 15.0 g of methanol is:
15.0 g Methanol × 100 g Solution/44.4 g Methanol = 33.8 g Solution
Since 33.8 g are required and 320. g are available, there is enough solution for the requirements.
Consider the following events that take place when rip currents occur.
A. Waves travel to the beach.
B. Waves are trapped by the sandbars.
C. Waves reach the shore and go back to the ocean.
D. Waves speed up and flow between the sandbars.
E. Waves are broken by the sandbars.
Which list shows the order of events in the production of rip currents?
g The most stable nucleon in terms of binding energy per nucleon is 56Fe. If the atomic mass of Fe-56 is 55.9349 amu, calculate the binding energy in J/nucleon and in Mev/nucleon for Fe-56. The atomic mass of a proton is 1.00782 amu and the atomic mass of a neutron is 1.00866 amu. Would the binding energy per nucleon for U-235 be larger or smaller than that of Fe-56
Answer:
The binding energy per nucleon of U-235 is lesser than that Fe-56
Explanation:
The binding energy refers to the energy required to hold the nucleons together in the nucleus of an atom.
It also corresponds to the energy that must be supplied in order to disintegrate the nucleus of an atom.
The binding energy per nucleon of elements depends on the number of nucleons present in the nucleus of the atom of that element. It is defined as the binding energy of the nucleus divided by the number of nucleons.
U-235 contains more nucleons than Fe-56, the binding energy per nucleon of U-235 is less than that of Fe-56. This is further confirmed by the fact that the greater the number of protons in the nucleus, the greater the coulumbic repulsion in the nucleus and the lesser the nuclear force of attraction between nucleons.
Atoms of element X have 12 times the mass of atoms of oxygen, and atoms of oxygen have 16 times the mass of atoms of hydrogen. What is the mass of atoms of element X on a scale on which atoms of hydrogen have a mass of 1?
A. 16/12
B. 12/16
С. 12 x 16
D. 16-12
E. 1216
Answer:
C. 12 × 16
Explanation:
Let mH be the mass of atoms of hydrogen and mO be the mass of atoms of oxygen. The atoms of oxygen have 16 times the mass of atoms of hydrogen. That is,
mO = 16 mH [1]
Let mX be the mass of the atoms of X. Atoms of element X have 12 times the mass of atoms of oxygen. That is,
mX = 12 mO [2]
If we substitute [1] in [2], we get
mX = 12 × 16 mH
If mH is 1, we get,
mX = 12 × 16 × 1
mX = 12 × 16
Compound A has molecular formula C4H10. Compound A gives two monochlorides, B and C, on photochemical chlorination. Treatment of either of these monochlorides with potassium tert-butoxide gives the same alkene (C4H8) as the product, but B leads to just one isomer of the alkene, D, whereas C gives D and another isomer of the alkene, E. Treatment of monochlorides B and C with aqueous ethanol gives products F and G, respectively, both of which are of molecular formula C4H10O. What are the chemical names of compounds A-G
Answer:
Explanation:
Radical chlorination of butane in the presence of light gives rise to the formation of two isomeric monochlorides B (1-chlorobutane) and C (2-chlorobutane). Both upon reaction with a bulky base (potassium tert-butoxide) give rise to D(but-1-ene) and E(but-2-ene) respectively, this is because the bulky base abstracts the less hindered proton. Compound B and C were later treated with aqueous C2H5OH which results in the hydrolysis of alkyl halides to produce Compound F(butan-1-ol) and compound G (butan-2-ol) respectively.
The diagrammatic expression of the whole reaction is shown in the attached image below.
Select the keyword or phrase that will best complete each sentence about the three general rules of for drawing Lewis structures.a. main group b. hafnium c. valence d. hydrogen e. core f. transition metal1. Draw only__the electrons. 2. Give every___element (except hydrogen) an octet of electrons. 3. Give each____two electrons.
Answer:
1. Draw only the valence electrons.
2. Give every element main group element (except hydrogen) an octet of electrons.
3. Give each hydrogen two electrons.
Explanation:
Lewis structures are used to describe the arrangement or configurations of the valence electrons of molecules and polyatomic ions involved in electronic bonding. A Lewis structure consists of the symbols of the elements in the molecule surrounded by dots with each dot representing each of the elements valence electrons. Also, the electrons shared between two elements are shown by dots between the two elements and these electrons are known as shared electron pairs. The valence electrons on atom that is not involved in bonding is known as lone pairs.
The three general rules for drawing Lewis structures are:
1. Draw only the valence electrons. Only the valence electrons of the atoms of elements are shown since they are the only electrons involved in chemical bonding.
2. Give every element main group element (except hydrogen) an octet of electrons. The complete eight valence electrons of the noble gases is associated with their stability. Thus, the main group elements show a tendency to form enough bonds to obtain eight valence electronsmin order to achieve stability. This is known as the octet rule. However, since the maximum number of valence electrons for elements in the first period of the period table is two, the noble gas helium has completely-filled valence shell containing two electrons known as a duplet. Hydrogen belongs to the first period and is therefore an exception tomthe octet rule.
3. Give each hydrogen two electrons. Hydrogen attains a duplet structure in accordance with the structure of helium
In an oxidation-reduction reaction, the substance oxidized always ________. a. takes on oxygen atoms b. gives up hydrogen atoms c. gains electrons d. loses electrons e. becomes a charged species
Answer:
Explanation:
Let us define oxidation and reduction.
Oxidation and reduction deal with the transferring of electrons between reactants. The reactant that loses electrons is oxidized, while the reactant that gains electrons is reduced.
In an oxidation-reduction reaction, the substance oxidized always loses electrons and the correct option is option D.
An oxidation-reduction (redox) reaction is a type of chemical reaction that involves the transfer of electrons between species. It consists of two half-reactions: an oxidation half-reaction, where a species loses electrons and increases its oxidation state, and a reduction half-reaction, where a species gains electrons and decreases its oxidation state.
In an oxidation half-reaction, the species that is oxidized is called the reducing agent or the electron donor, as it donates electrons. In a reduction half-reaction, the species that is reduced is called the oxidizing agent or the electron acceptor, as it accepts electrons.
Thus, the ideal selection is option D.
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Helppp!
Is astatine (At), atomic number 85, a metal or nonmetal? Why?
Answer:
At is a metalloid
Explanation: Meaning that it displays some of the characteristics of a metal but can also display characteristics of nonmetals. Silicon is one such element. Elemental Silicon has a dark lustrous metallic appearance but it is quite brittle. Some authorities include Astatine as a halogen as it has seven electrons in its valence shell placing it just below iodine in period 17 VIIA. Additionally, because Astatine is only ever found naturally as the result of radioactive decay from more dense elements it is the rarest naturally occurring element in Earth's crust.
I hope this helps at least a little.
Regards,
JR Holt
Chemist retired
How many grams of solid sodium cyanide should be added to 1.00 L of a 0.119 M hydrocyanic acid solution to prepare a buffer with a pH of 8.809
Answer:
1.62 g
Explanation:
Given that:
Concentration of HCN = 0.119 M
Assuming the ka 4.00 × 10⁻¹⁰
The pKa of HCN (hydrocyanic acid) = -log (Ka)
= - log ( 4.00 × 10⁻¹⁰)
= 9.398
pH of buffer = 8.809
Using Henderson Hasselbach equation:
[tex]pH = pKa + log \dfrac{[conjugate\ base ]}{acid}[/tex]
[tex]pH = pKa + log \dfrac{[CN^-]}{[HCN]}[/tex]
[tex]8.809 = 9.398 +log \dfrac{[CN^-]}{[HCN]}[/tex]
[tex]log \dfrac{[CN^-]}{[HCN]}= 8.809 - 9.398[/tex]
[tex]log \dfrac{[CN^-]}{[HCN]}= -0.589[/tex]
[tex]\dfrac{[CN^-]}{[HCN]}= 0.2576[/tex]
[CN^-] = 0.2576[HCN]
[CN^-] = 0.2756 (0.119) L
[CN^-] = 0.033 M
∴
The amount of NaCN (sodium cyanide) is calculated as follows:
[tex]= 1.00 L \times \dfrac{0.033 \ mol \ NacN }{1 \ L } \times \dfrac{49.01 \ g}{1 \ mol \ of \ NacN}[/tex]
= 1.62 g
study the reaction given below in which excess magnesium ribbon (Mg)reacts with 50cm of a diluted sulphuric acid solution at room temperature
Questions
what Changes can be made to the following substance to increase the rate of reaction?
5.1.1 Magnesium
5.1.2 Sulphuric acid
Answer:
Magnesium reacts with dilute hydrochloric acid in a conical flask which is ... One student can add the magnesium ribbon to the acid and stopper the flask, ... 50 cm3 of 1M hydrochloric acid is a six-fold excess of acid.
What is represented by the chemical formula
PbCl2(s)?
A) a substance
B) a solution
C) a homogeneous mixture
D) a heterogeneous mixture
A hemil reation
Answer:
d
Explanation:
cuaase that it sirhal
The chemical formula PbCl₂(s) represented by:
A) a substance
The chemical formula PbCl₂(s) represents a substance. A substance is a single, pure chemical entity with a definite composition. It can be an element, a compound, or an alloy.
A solution is a homogeneous mixture of two or more substances. A homogeneous mixture is a mixture in which the components are evenly distributed throughout the mixture.
A heterogeneous mixture is a mixture in which the components are not evenly distributed throughout the mixture.
A reaction is a process in which one or more substances are transformed into one or more new substances.
Therefore, the chemical formula PbCl₂(s) represents a substance, and the answer is (A).
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Match all the terms that apply to each element. Some terms may apply to more than one element.
a. Main group element
b. halogen
c. noble gas
d. inner transition element
e. alkaline earth
f. Non-metal
g. Alkali metal
h. Metalloid
i. Metal
1. Titanium
2. Gold
3. Argon
4. Thallium
Answer:
Titanium - Metal
Gold - Metal
Argon - noble gas
Thallium - Metal
Explanation:
A metal is a species that easily looses electrons to yield a positive ion. Metals are found in the left hand side of the periodic table.
Thallium is a groups 13 metal, gold is a group 11 metal while titanium is a 4 metal.
However, argon is a noble gas hence it belongs to groups 18 of the periodic table.
Titanium -metal
gold - metal,
Argon - noble gas
Thallium -metal, transition metal
Explanation:
In modern periodic table :
Metals are listed from the middle to left-hand side of the periodic tableNonmetals are listed on the upper right-hand side of the periodic tableMetalloids divide the periodic table in a zig-zag line with nonmetals on the right and metals on the left.The alkali metals are listed under group 1The alkaline earth metals are listed under group 2The noble gases are listed under group 18The transition metals are listed from group 3 to group 12.The halogens are listed under group 17The inner transition elements (f-block) are listed at the bottom separately.So, from this, we can match the given elements as:
Titanium -metal
gold - metal,
Argon - noble gas
Thallium -metal. transition metal
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g Suppose you are titrating vinegar, which is an acetic acid solution of unknown concentration, with a sodium hydroxide solution according to the equation H C 2 H 3 O 2 + N a O H ⟶ H 2 O + N a C 2 H 3 O 2 If you require 30.01 mL of 0.1798 M N a O H solution to titrate 10.0 mL of H C 2 H 3 O 2 solution, what is the molar concentration of acetic acid in the vinegar? Type answer:
Answer: The molar concentration of acetic acid in the vinegar is 0.539 M.
Explanation:
The formula used is:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] and [tex]V_1[/tex] are the concentration and volume of base.
[tex]M_2[/tex] and [tex]V_2[/tex] are the concentration and volume of an acid.
Given:
Molar concentration of NaOH = 0.1798 M
Volume of NaOH = 30.01 mL
Volume of acetic acid = 10.0 mL
Now putting all the given values in the above formula, we get:
[tex]M_1V_1=M_2V_2\\\\0.1798M\times 30.01mL=M_2\times 10.0mL\\\\M_2=0.539M[/tex]
Thu, the molar concentration of acetic acid in the vinegar is 0.539 M.
Desclony When Fe(NO3)2 dissolves in water, what particles are present in the solution?
Answer:
hjfhgjgjgjghgjgjhgjhhjh
Which statement best describes how the tilt of Earth's rotational axis relates to the amount of solar energy received?
A. When a hemisphere is tilting away from the sun, the whole planet receives less direct sunlight,
B. When a hemisphere is tilting toward the sun, that hemisphere receives less direct sunlight.
C. When a hemisphere is tilting toward the sun, the whole planet receives less direct sunlight.
D. When a hemisphere is tilting away from the sun, that hemisphere receives less direct sunlight.
Explanation:
A. Around December 21, the Northern Hemisphere tilts the farthest away from the Sun. This is called the northern winter solstice, and it is when we have the least amount of daylight of any time of the year.
B.Solstices and shifting solar declinations are a result of Earth's 23.5° axial tilt as it orbits the sun. Throughout the year, this means that either the Northern or Southern Hemisphere is tilted toward the sun and receives the maximum intensity of the sun's rays.
C.The tilt of the Earth's axis also defines the length of daylight. Daylight hours are shortest in each hemisphere's winter. Between summer and winter solstice, the number of daylight hours decreases, and the rate of decrease is larger the higher the latitude. The fewer sunlight hours the colder the nights
D.The second solstice occurs on December 21 or 22 when the north pole is tilting 23.5 degrees away from our Sun and the south pole is inclined toward it. This is the shortest day of the year in the northern hemisphere — the northern hemisphere winter solstice.
How do enzymes function in biological reactions?
Answer:
it binds molecules like a chemical bond-breaking
Explanation:
Enzymes perform the critical task of lowering a reaction's activation energy—that is, the amount of energy that must be put in for the reaction to begin. Enzymes work by binding to reactant molecules and holding them in such a way that the chemical bond-breaking and bond-forming processes take place more readily.
An ideal gas is made up of gas particles that...
(A) ...have volume.
(B) ... attract each other.
(C)...can be liquified.
(D) ... are in random motion.
Answer:
Its D Are in random motion
i hope it will help you if not soo sorry :)
Which of the following expressions is the correct equilibrium constant expression for the equilibrium between dinitrogen tetroxide and nitrogen dioxide? N subscript 2 O subscript 4 space left parenthesis g right parenthesis space rightwards arrow over leftwards arrow 2 space NO subscript 2 space left parenthesis straight g right parenthesis
Answer:
Correct answer:
The equilibrium expression is defined as the ratio of the concentration of products divided by the concentration of the reactants. Each reactant or product is raised to the power corresponding to its coefficient in the balanced chemical equation.
Explanation:
A weather balloon containing helium with a volume of 410.0 L rises in the atmosphere and is cooled from 17∘
∘
C to -27∘
∘
C. The pressure on the gas is reduced from 110.0 kPa to 25.0 kPa. What is the volume of the gas at the lower temperature and pressure? What law did you use to solve?
Answer:
V₂ = 1500 Liters ( 2 sig. figs.)
Explanation:
Given the following gas law variables:
P₁ = 110.0KPa P₂ = 25KPa
V₁ = 410 Liters V₂ = ?
T₁ = 17°C ( = 290K) T₂ = -27°C ( = 248K)
P₁V₁/₁T₁ = P₂V₂/T₂ => V₂ = V₁(P₁/P₂)(T₂/T₂)
V₂ = 410L(110.0KPa/25KPa)(248k/290K) = 1542 L (calc. ans.)
V₂ = 1500 Liters ( 2 sig. figs.)
Match each description below:
a. Will react with water.
b. Will react with steam or acid, but not water.
c. Will react with acid only
d. Will not tract with water, steam, or acid.
1. Cr
2. Cu
3. K
d. Sn
Answer:
a) will react with water: K
b) will react with steam or acid, but not water: Cr & Sn
d) will not react with water, steam, or acid: Cu
Explanation:
K reacts violently with water.
Cr will react with steam to form an oxide + H gas and will react with most acids
Sn will react with steam to form SnO₂ + H gas and, though it does not react as rapidly as other metals in acid, it dissolves easily in concentrated acids
Cu is an extremely nonreactive metal, which is what makes it so suitable for wiring
carbon dioxide gas evolve during the fermentation of sugar which was collected at 22.5°C and 0.945 ATM after perfect strangers in the volume was found to be 25.0 ML how many grams of carbon dioxide were collected
Answer:
0.043 grams
Explanation:
We can find the mass of carbon dioxide as follows:
[tex] m = n*M [/tex]
Where:
n: is the number of moles
M: is the molar mass = 44.01 g/mol
First, we need to calculate the number of moles. We can use the Ideal gas equation:
[tex] PV = nRT [/tex]
Where:
P: is the pressure = 0.945 atm
V: is the volume = 25.0 mL
R: is the gas constant = 0.082 L*atm/(K*mol)
T: is the tempearture = 22.5 °C
[tex]n = \frac{PV}{RT} = \frac{0.945 atm*25 mL*\frac{1 L}{1000 mL}}{0.082 L*atm/K*mol*(22.5 + 273) K} = 9.75 \cdot 10^{-4} moles[/tex]
Hence, the mass is:
[tex]m = 9.75 \cdot 10^{-4} moles*44.01 g/mol = 0.043 g[/tex]
Therefore, 0.043 grams were collected.
I hope it helps you!