Answer:
His results will be skewed because there was more water than stock solution. Which would cause the percentage solution to be less than 50% therefore the density would be less than the actual value.
Explanation:
The solution will have percentage less than that of 50%. Therefore the density would be less than the actual value.
Suppose there should be 50 mL of the solution, and he added 60 mL. So 10 mL of the solution is added more.
Suppose the mass of the solute is m.
Originally, the density is = [tex]$\frac{m}{50}$[/tex] [tex]\left(\frac{\text{mass}}{\text{volume}}\right)[/tex]
Now after adding extra 10 mL , the density becomes [tex]$\frac{m}{60}$[/tex].
Therefore, [tex]$\frac{m}{50}>\frac{m}{60}$[/tex]
So the density decreases when we add more solution.
Sodium is a highly reactive metal and
chlorine is a toxic gas, but when they
come together the resulting material,
sodium chloride, is essential for life.
Which of the following is true when
sodium and chlorine are brought into
contact with one another?
Answer:
NaCl
Explanation:
[tex]na + cl > nacl[/tex]
This is also a salt
which effect of long-term environmental change is the driving force behind evolution?
Answer:
climate change
Explanation:
climate change is driving force of evolution because when the climate is changed the animal and human need to adapt to it's natural change.
Which best expresses the uncertainty of the measurement 32.23 cm?
A.) ±0.05 cm
B.) 0.1 cm
C.) 1%
D.) ±0.01 cm?
Answer:
D.) ±0.01 cm?
Explanation:
Since 32.23 cm has two decimal places, the uncertainty is taken as one-half the last decimal pace.
The last decimal place is 0.03. Half of this is 0.03 cm/2 = 0.015 cm.
Since we cannot go below two decimal places, we ignore the 5 in 0.015 cm.
So, we have our uncertainty as 0.01 cm.
So, the best expression of the uncertainty in the measurement 32.23 cm is ± 0.01 cm.
So, the answer is D. which is ± 0.01 cm.
How many milliliters of a 0.40%(w/v) solution of nalorphine must be injected to obtain a dose of 1.5 mg?
Answer:
0.375mL of solution of nalorphine must be injected
Explanation:
A solution of 0.40% (w/v) contains 0.40g of solute (In this case, nalorphine), in 100mL of solution. To obtain 1.5mg of nalorphine = 1.5x10⁻³g of nalorphine are needed:
1.5x10⁻³g * (100mL / 0.40g) =
0.375mL of solution of nalorphine must be injectedhow many moles of neon gas have a volume of 0.84 L and a pressure of 4.6 atm at 222k
Answer:
n = 0.21 moles
Explanation:
Given that,
Volume, V = 0.84 L
Pressure, P = 4.6 atm
T = 222 K
We need to find the number of moles of Neon gas. We know that,
PV = nRT
Where
n is the number of moles
R i the gas constant, R = 0.08206 L-atm/mol-K
Put all the values,
[tex]n=\dfrac{PV}{RT}\\\\n=\dfrac{4.6\times 0.84}{0.08206 \times 222}\\\\n=0.21\ \text{moles}[/tex]
So, there are 0.21 moles of Neon gas.
Place the following in order of increasing molar entropy at 298 K.
a. C3H8 < SO < CO2
b. CO2 < C3H8 < SO
c. C3H8 < CO2 < SO
d. SO < CO2 < C3H8
e. CO2 < SO < C3H8
Answer:
SO < CO2 < C3H8
Explanation:
Entropy refers to the degree of disorderliness of a system. The standard molar entropy of a substance refers to the entropy of 1 mole of the substance vunder standard conditions.
The molar entropy depends on the number of microstates in the system which in turn depends on the number of atoms in the molecule.
C3H8 has 11 atoms and hence the highest number of microstates followed by CO2 having three atoms and least of all SO having only two atoms.
In an experiment, you added a base, NaOH, one mL at a time to 50 mL acetate buffer and recorded the pH. For the first 6 mL NaOH the pH increased from 4.5 to 4.9. At the 7th mL the pH was 6.6 and by the 8th mL the pH was 10.7. Knowing what you do about titrating acetate buffer with acid, is this experimental result what you expected or is it not expected
Answer:
yes the experimental result is the expected result .
Explanation:
When Titrating acetate buffer with acid the PH will decrease gradually from a more neutral PH to a more acidic level and this is because buffer solutions are prepared with weak acids and its conjugate base.
The results gotten from the continuous addition of base NaOH to the acetate buffer is the expected result because the base is been absorbed by the buffer solution and it is converted to a conjugate base of the buffer solution which will gradually increase the PH level of the solution as more conjugate base is formed due to the addition of more NaOH.
describe how lyophobic sols are synthesize by dispersion method
Explanation:
For preparing lyophobic sol, the substance in bulk is broken down into particles of colloidal dimensions (Dispersion) or aggregating smaller particles into particles of colloidal dimensions (condensation).
At 50.0 oC, a reinforced tank contains 675.5 grams of gaseous argon and 465.0 g of gaseous molecular chlorine with a total pressure of 4.00 atm. Calculate the following:
a. How many moles of Ar are in the tank?
b. How many moles of Cl, are in the tank?
c. Total moles of gas in the tank.
d. The mole fraction of Ar.
e. The mole fraction of Cl2.
f. The Partial Pressure of Ar.
g. The Partial Pressure of Cl2.
Answer:
For (a): The moles of Ar is 16.94 moles
For (b): The moles of [tex]Cl_2[/tex] is 16.94 moles
For (c): The total number of moles in a tank is 23.47 moles
For (d): The mole fraction of Ar is 0.722
For (e): The mole fraction of [tex]Cl_2[/tex] is 0.278
For (f): The partial pressure of Ar is 2.888 atm
For (g): The partial pressure of [tex]Cl_2[/tex] is 1.112 atm
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
For (a):Given mass of Ar = 675.5 g
Molar mass of Ar = 39.95 g/mol
Plugging values in equation 1:
[tex]\text{Moles of Ar}=\frac{675.5g}{39.95g/mol}=16.91 mol[/tex]
For (b):Given mass of [tex]Cl_2[/tex] = 465.0 g
Molar mass of [tex]Cl_2[/tex] = 70.9 g/mol
Plugging values in equation 1:
[tex]\text{Moles of }Cl_2=\frac{465.0g}{70.9g/mol}=6.56 mol[/tex]
For (c):Total moles of gas in the tank = [16.91 + 6.56] mol = 23.47 mol
Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:
[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex] .....(2)
where n is the number of moles
For (d):Moles of Ar = 16.94 moles
Total moles of gas in the tank = 23.47 mol
Putting values in equation 2, we get:
[tex]\chi_{Ar}=\frac{16.94}{23.47}\\\\\chi_{Ar}=0.722[/tex]
For (e):Total mole fraction of the system is always 1
Mole fraction of [tex]Cl_2[/tex] = [1 - 0.722] = 0.278
Raoult's law is the law used to calculate the partial pressure of the individual gases present in the mixture.
The equation for Raoult's law follows:
[tex]p_A=\chi_A\times p_T[/tex] .....(3)
where [tex]p_A[/tex] is the partial pressure of component A in the mixture and [tex]p_T[/tex] is the total partial pressure of the mixture
For (f):We are given:
[tex]\chi_{Ar}=0.722\\p_T=4.00atm[/tex]
Putting values in equation 3, we get:
[tex]p_{Ar}=0.722\times 4.00atm\\\\p_{Ar}=2.888atm[/tex]
For (g):We are given:
[tex]\chi_{Cl_2}=0.278\\p_T=4.00atm[/tex]
Putting values in equation 3, we get:
[tex]p_{Cl_2}=0.278\times 4.00atm\\\\p_{Cl_2}=1.112atm[/tex]
What size volumetric flask would you use to create a 1.00M solution using 166.00 g of KI?
Answer:
A 1 liter volumetric flask should be used.
Explanation:
First we convert 166.00 g of KI into moles, using its molar mass:
Molar mass of KI = Molar mass of K + Molar mass of I = 166 g/mol
166.00 g ÷ 166 g/mol = 1 mol KIThen we calculate the required volume, using the definition of molarity:
Molarity = moles / litersLiters = moles / molarity
1 mol / 1.00 M = 1 LAmmonium sulfate (NH4)2SO4 is made by reacting 25.0 L of 3.0 mol/L H2SO4 with 3.1× 103 L of NH3 at a pressure of 0.68 atm and a temperature of 298 K according to the following reaction .
NH3(g) + H2SO4(aq) → (NH4)2SO4 (aq)
How many grams of ammonium sulfate are produced?
Answer: The mass of [tex](NH_4)_2SO_4[/tex] produced is 9910.5 g
Explanation:
For [tex]H_2SO_4[/tex]:Molarity is calculated by using the equation:
[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex] ......(1)
Molarity of [tex]H_2SO_4[/tex] = 3.0 M
Volume of solution = 25.0 L
Putting values in equation 1, we get:
[tex]\text{Moles of }H_2SO_4=(3.0mol/L\times 25.0L)=75mol[/tex]
For [tex]NH_3[/tex]:The ideal gas equation is given as:
[tex]PV=nRT[/tex] .......(2)
where,
P = pressure of the gas = 0.68 atm
V = volume of gas = [tex]3.1\times 10^3L[/tex]
n = number of moles of gas = ? moles
R = Gas constant = 0.0821 L.atm/mol.K
T = temperature of the gas = 298 K
Putting values in equation 2, we get:
[tex]0.68atm\times 3.1\times 10^3L=n\times 0.0821L.atm/mol.K\times 298K\\\\n=\frac{0.68\times 3.1\times 10^3}{0.0821\times 298}=86.16mol[/tex]
For the given chemical equation:
[tex]NH_3(g)+H_2SO_4(aq)\rightarrow (NH_4)_2SO_4(aq)[/tex]
By stoichiometry of the reaction:
If 1 mole of [tex]H_2SO_4[/tex] reacts with 1 mole of [tex]NH_3[/tex]
So, 75 moles of [tex]H_2SO_4[/tex] will react with = [tex]\frac{1}{1}\times 75=75mol[/tex] of [tex]NH_3[/tex]
As the given amount of [tex]NH_3[/tex] is more than the required amount. Thus, it is present in excess and is considered as an excess reagent
Thus, [tex]H_2SO_4[/tex] is considered a limiting reagent because it limits the formation of the product.
By the stoichiometry of the reaction:
If 1 mole of [tex]H_2SO_4[/tex] produces 1 mole of [tex](NH_4)_2SO_4[/tex]
So, 75 moles of [tex]H_2SO_4[/tex] will produce = [tex]\frac{1}{1}\times 75=75mol[/tex] of [tex](NH_4)_2SO_4[/tex]
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We know, molar mass of [tex](NH_4)_2SO_4[/tex] = 132.14 g/mol
Putting values in above equation, we get:
[tex]\text{Mass of }(NH_4)_2SO_4=(75mol\times 132.14g/mol)=9910.5g[/tex]
Hence, the mass of [tex](NH_4)_2SO_4[/tex] produced is 9910.5 g
Identify the isoelectronic elements.
i. Cl-, F-, Br-, I-, At-
ii. Ne, Ar, Kr, Xe, He
iii. N3-, S2-, Br-, Cs+, Sr2+
iv. N3-, O2-, F-, Na+, Mg2+
v. Li+, Na+, K+, Rb+,Cs+
Answer:
iv. N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺
Explanation:
Isoelectronic elements are those that have the same number of electrons. So, if at least 2 elements differ in their number of electrons, the series is not of isoelectronic elements.
To know the number of electrons we will consider the atomic number and add electrons if it is an anion and subtract electrons it is a cation.
Identify the isoelectronic elements.
i. Cl⁻, F⁻, Br⁻, I⁻, At⁻. NO. Cl⁻ has 18 electrons (17+1) and F⁻ has 10 electrons (9+1). ii. Ne, Ar, Kr, Xe, He. NO. Ne has 10 electrons and Ar has 18. iii. N³⁻, S²⁻, Br⁻, Cs⁺, Sr²⁺. NO. N³⁻ has 10 electrons (7+3) and S²⁻ has 18 (16+2).iv. N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺. YES. They all have 10 electrons v. Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺. NO. Li⁺ has 2 electrons (3-1) and Na⁺ has 10 (11-1).Predict the products from theses reaction, and balance the equations. Include phase symbols.
Reaction : K(s)+Cl2(g)⟶
Reaction :Cu(s)+O2(g)⟶
Answer:
2 K(s) + Cl₂(g) ⟶ 2 KCl(s)
2 Cu(s) + O₂(g) ⟶ 2 CuO(s)
Explanation:
Both reactions are synthesis reactions (two substances combine to form another).
Reaction: K(s) + Cl₂(g) ⟶
The product is the binary salt KCl. The balanced chemical equation is:
2 K(s) + Cl₂(g) ⟶ 2 KCl(s)
Reaction: Cu(s) + O₂(g) ⟶
The most likely product is the metal oxide CuO. The balanced chemical equation is:
2 Cu(s) + O₂(g) ⟶ 2 CuO(s)
A buffer solution contains 0.298 M ammonium chloride and 0.478 M ammonia. If 0.0560 moles of hydroiodic acid are added to 225 mL of this buffer, what is the pH of the resulting solution?
Answer:
pH = 8.87
Explanation:
Hydroiodic acid, HI, is a strong acid that reacts with ammonia, NH3, to produce ammonium ion, NH⁴⁺. That means the moles of HI added = moles of NH3 consumed and moles of NH4⁺ produced.
Initial moles NH₄⁺:
0.225L * (0.298mol/L) = 0.06705 moles
Initial moles NH3:
0.225L * (0.478mol/L) = 0.10755 moles
After the reaction the moles are:
0.10755moles NH3 - 0.0560moles = 0.05155 moles NH3
0.06705moles NH4+ + 0.0560moles = 0.12305 moles NH4+
Using H-H equation for weak bases:
pOH = pKb + log ([NH4+] / [NH3])
pKb for ammonia is 4.75, [NH4+] could be the moles of NH4+ = 0.12305mol,
[NH3] = 0.05155moles
Replacing:
pOH = 4.75 + log (0.12305mol / 0.05155moles)
pOH = 5.13
pH = 14-pOH
pH = 8.87What is the speed of a wave with a frequency of 2 Hz and a wavelength of 87m (subject is science) pls answer fast
Answer:
43.5
Explanation:
Hope that helps
Can the properties of different substances in a mixture be used to separate them?
Answer: here you go
Explanation:
Physical properties of the substances in a mixture are different, so this allows the substances to be separated. Think about the example of a mixture of salt water.
Answer:
Mixtures can be classified on the basis of particle size into three different types: solutions, suspensions, and colloids. The components of a mixture retain their own physical properties. These properties can be used to separate the components by filtering, boiling, or other physical processes.
Explanation:
HELP ME PLZ AND THANKS I WILL MARK YOU AS BRAINLIEST!!!
Answer:
See explanation.
Explanation:
Hello there!
In this case, since this problem is about gas laws, more specifically about the Gay-Lussac's one since the volume is said to be constant, we can use the following equation for its solution for the final pressure, P2:
[tex]\frac{P_2}{T_2} = \frac{P_1}{T_1}[/tex]
[tex]P_2= \frac{P_1T_2}{T_1}\\\\P_2 =\frac{12.0atm*450K}{300K}\\\\P_2= 18.0atm[/tex]
Thus, we fill in the table as follows:
Initial Final
Pressure 12.0 atm 18.0 atm
Volume 4.0 L 4.0 L
Temperature 300K 450K
Regards!
There are four different starting molecules that one might use to synthesize the illustrated alkyl halide as the major product using an electrophilic addition reaction. Please draw all four of them.
Answer:
Explanation:
An electrophilic addition reaction occurs when an electrophile attacks a substrate, with the end result being the inclusion of one or many comparatively straightforward molecules along with multiple bonds.
In the given question, the hydrogen bromide provides the electrophile while the bromide is the nucleophile. The mechanism proceeds with the attack of the electrophile on the carbon, followed by deprotonation. This process is continued with a formation of carbocation and the bromide(nucleophile) finally bonds to the carbocation to form a stable product.
The first diagram showcases the possible various starting molecules for the synthesis while the second diagram illustrates their mechanism.
During a chemical reaction, an iron atom became the ion Fe2+. What happened to the iron atom?
Explanation:
Iron atom is been oxidised as it losses 2 electron to form 2 + ion.
What are the lengths of the diagonals of the kite?
The answer ( 13 and 8 )
x²=5²+12²
x²=25+144
x²=169
x=13
x²=5²+6²
x²=25+36
x²=61
x=7.8
x=8
When an electron moves up to higher energy levels, the atom Choose... a photon of light whereas the atom Choose... a photon of light when an electron drops to a lower energy level. The photons emitted from an atom appear as
Answer:
Explanation:
When an electron moves from a lower energy level to a higher energy level, energy is absorbed by the atom. When an electron moves from a higher to a lower energy level, energy is released and photon is emitted.
this emitted photon is depicted as a small wave-packet being expelled by the atom in a well-defined direction.
explain why seeing a gas doesn not always indicate that there was a chemical change.
you have 4.600x 10^1 ml of a kcl solution which has been made up in 6.0000x10^-1 g/ml solution.you are asked to determine the %v/v/v of the kcl solution.
Answer: The %v/v of the given KCl solution is 7.6%.
Explanation:
Given: Volume of solute = [tex]4.6 \times 10^{1} ml[/tex]
Volume of solution = [tex]6.0 \times 10^{-1} g/ml[/tex]
Formula used to calculate %v/v is as follows.
[tex]\frac{volume of solute}{volume of solution} \times 100[/tex]
Substitute the values into above formula as follows.
[tex]\frac{volume of solute}{volume of solution} \times 100\\\frac{4.6 \times 10^{1}}{6.0 \times 10^{-1}} \times 100\\= 7.6[/tex]
Thus. we can conclude that the %v/v of the given KCl solution is 7.6%.
A tank at is filled with of sulfur tetrafluoride gas and of sulfur hexafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction of each gas. Round each of your answers to significant digits.
The question is incomplete, the complete question is:
A 7.00 L tank at [tex]21.4^oC[/tex] is filled with 5.43 g of sulfur hexafluoride gas and 14.2 g of sulfur tetrafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas. Round each of your answers to significant digits.
Answer: The mole fraction of sulfur hexafluoride is 0.221 and that of sulfur tetrafluoride is 0.779
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
For sulfur hexafluoride:Given mass of sulfur hexafluoride = 5.43 g
Molar mass of sulfur hexafluoride = 146.06 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of sulfur hexafluoride}=\frac{5.43g}{146.06g/mol}=0.0372mol[/tex]
For sulfur tetrafluoride:Given mass of sulfur tetrafluoride = 14.2 g
Molar mass of sulfur tetrafluoride = 108.07 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of sulfur tetrafluoride }=\frac{14.2g}{108.07g/mol}=0.1314mol[/tex]
Total moles of gas in the tank = [0.0372+ 0.1314] mol = 0.1686 mol
Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:
[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex] .....(2)
where n is the number of moles
Putting values in equation 2, we get:
[tex]\chi_{SF_6}=\frac{0.0372}{0.1686}=0.221[/tex]
[tex]\chi_{SF_4}=\frac{0.1314}{0.1686}=0.779[/tex]
Hence, the mole fraction of sulfur hexafluoride is 0.221 and that of sulfur tetrafluoride is 0.779
Which of the following is considered a standard unit of length in the United States?
O square inch
O acre
O cubic yard
O yard
Answer:
Yard . I hope this helped:))
Identify the Lewis acid and Lewis base from among the reactants in each of the following equations.
Part A
Ag+(aq)+2NH3(aq)⇌Ag(NH3)2+(aq)
a. Ag+ is the Lewis acid and NH3 is the Lewis base.
b. Ag+ is the Lewis base and NH3 is the Lewis acid.
Part B
AlBr3+NH3⇌H3NAlBr3
a. AlBr3 is the Lewis base and NH3 is the Lewis acid.
b. AlBr3 is the Lewis acid and NH3 is the Lewis base.
Part C
Cl−(aq)+AlCl3(aq)⇌AlCl4−(aq)
a. AlCl3 is the Lewis acid and Cl− is the Lewis base.
b. AlCl3 is the Lewis base and Cl− is the Lewis acid.
Answer:
Part A
Ag+ is the Lewis acid and NH3 is the Lewis base.
Part B
AlBr3 is the Lewis acid and NH3 is the Lewis base.
Part C
AlCl3 is the Lewis acid and Cl− is the Lewis base.
Explanation:
A Lewis acid is any specie that accepts a lone pair of electrons. Ag^+, AlBr3 and AlCl3 all accepted lone pairs of electrons according to the three chemical reaction equations shown. Hence, they are Lewis acids.
A Lewis base donates a lone pair of electrons. They include neutral molecules having lone pair of electrons such as NH3 or negative ions such as Cl- .
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.
g The boiling of water is a Question 4 options: chemical and physical damage chemical change because a gas (steam) is given off chemical change because heat is needed for the process to occur physical change because the water merely disappears physical change because the gaseous water is chemically the same as the liquid
Answer:
physical change because the gaseous water is chemically the same as the liquid
Explanation:
Matter can be defined as anything that has mass and occupies space. Any physical object that is found on earth is typically composed of matter. Matter are known to be made up of atoms and as a result has the property of existing in states.
Generally, matter exists in three (3) distinct or classical phases and these are; solid, liquid and gas.
A physical change can be defined as a type of change that only affects the physical form of a chemical substance (matter) without having any effect on its chemical properties. Thus, a physical change would only affect the physical appearance and properties of a chemical substance (matter) but not its chemical properties.
This ultimately implies that, a physical change result in a change of matter from one form or phase (liquid, solid or gas) to another without a corresponding change in chemical composition.
Hence, the boiling of water is considered to be a physical change because the gaseous water is chemically the same as the liquid i.e there isn't any changes in chemical composition of water when boiling.
what would be the mass of 44.8 L of CO2 gas at STP?
show work if possible
Answer:
[tex]m=88.02g[/tex]
Explanation:
Hello there!
In this case, for this ideal gas law problem, it turns out necessary for us to remember that one mole of any gas is contained in 22.4 L at STP and therefore, we can use the following ratio to calculate the moles in 44.8 L of CO2:
[tex]\frac{1mol}{22.4L} =\frac{x}{44.8L}\\\\x= \frac{1mol*44.8L}{22.4L}=2mol[/tex]
Finally, since the molar mass of CO2 is 44.01 g/mol, we calculate the mass as follows:
[tex]m=2mol*\frac{44.01g}{1mol}\\\\m=88.02g[/tex]
Regards!
What would happen to the Earth's hydrosphere if there were no atmosphere?
