Answer:
See explanation
Explanation:
The average speed of the molecules of a gas depends on the temperature of the gas and its molar mass and not on the volume of the gas.
The average velocity of a gas is given by; vrms=√3RTM
R= gas constant
T= Absolute temperature
M= molar mass of the gas
Where the temperature of the gas is held constant, the average velocity of gas molecules depends on the molar mass of the gas. Hence, if a sample of gas is slowly compressed to one-half of its original volume with no change in temperature, the average speed of the molecules in the sample of gas remains the same.
t the centers of some stars, fluorine-19 can undergo a fusion reaction with a proton (a hydrogen-1 nucleus) to produce two different nuclei. One of the products is an alpha particle (a helium-4 nucleus). What is the other
Answer:
Explanation:
¹⁹F₉ + ¹H₁ = ⁴He₂ + ¹⁶X₈
The atomic weight of fluorine is 19 and atomic weight of hydrogen is 1 . Total becomes 20 . Helium , the product has atomic weight of 4 . Therefore atomic weight of product X must be 16 . Similarly we can calculate atomic number of product X . It is found to be 8 .
Hence atomic weight and number of product will be 16 and 8 respectively.
It is similar to oxygen atom
Hence the other product will be oxygen.
What is the volume of 1.5 moles of gas at STP ?
0 9.02 L
0 20.0 L
0 33.6 L
0 22.4L
The volume of 1.5 moles of gas at STP is 33.6 L.
Volume of the gas at STPThe volume of the gas at STP is calculated as follows;
I mole of gas at STP = 22.4 L
1.5 moles of the gas at STP = ?
= 1.5 moles x 22.4 L/mole
= 33.6 L
Thus, the volume of 1.5 moles of gas at STP is 33.6 L.
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The numerical value of 0.001 is written with the prefix:
Answer:
Milli, m.
Explanation:
Hey there!
In this case, according to the given information, it turns out possible for us to answer to this question by bearing to mind the attached file whereas the most common prefixes and their factors are shown both in standard and scientific notation.
In such a way, we will be able to infer that the prefix related to the numerical value of 0.001 is milli, m, for example 1 mm which is 0.001 m.
Regards!
Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 2.9 g of sulfuric acid is mixed with 3.53 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
1.07 g of water.
Explanation:
A reaction between an acid and a base makes water and a salt as product.
Our reaction is:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Reactants are the acid and the base. Which is the limiting?
2.9 g . 1mol /98 g = 0.0296 moles of acid
3.53 g . 1mol / 40 g = 0.088 moles of base
2 moles of base react to 1 mol of acid
0.088 moles may react to (0.088 . 1)/2 = 0.044 moles of acid
And we only have 0.0296, sulfuric acid is the limiting
Ratio is 1:2. 1 mol of acid can produce 2 moles of water.
Our 0.0296 moles may produce (0.0296 . 2) /1 = 0.0592 moles of water.
We convert moles to mass:
0.0592 mol . 18g /mol = 1.07 g
Which statement bets describes the liquid state of matter
Answer:
It has no fixed shape and therefore takes the shape of the part of the container in which it is placed
What is the relationship (formula) and proportionality
between frequency, and wavelength?
Answer:
Frequency and wavelength are inversely proportional. c=f⋅λ (The speed of light is directly proportional to f and λ)
The standard enthalpies of combustion of fumaric acid and maleic acid (to form carbon dioxide and water) are - 1336.0 kJ moJ-1 and - 1359.2 kJ moJ-1, respectively. Calculate the enthalpy of the following isomerization process:
maleic acid ----> fumaric acid
Answer:
Explanation:
maleic acid ⇒ fumaric acid
ΔHreaction = ΔHproduct - ΔHreactant
ΔHproduct = -1336.0 kJ mol⁻¹
ΔHreactant = - 1359.2 kJ mol⁻¹.
ΔHreaction = -1336.0 kJ mol⁻¹ - ( - 1359.2 kJ mol⁻¹.)
= 1359.2 kJ mol⁻¹ -1336.0 kJ mol⁻¹
= 23.2 kJ mol⁻¹ .
Enthalpy of isomerization from maleic to fumaric acid is 23.2 kJ per mol.
Chromium-51 is a radioisotope that is used to assess the lifetime of red blood cells The half-life of chromium-51 is 27.7 days. If you begin with 39.7 mg of this isotope, what mass remains after 48.2 days have passed?
Answer:
11.9g remains after 48.2 days
Explanation:
All isotope decay follows the equation:
ln [A] = -kt + ln [A]₀
Where [A] is actual amount of the isotope after time t, k is decay constant and [A]₀ the initial amount of the isotope
We can find k from half-life as follows:
k = ln 2 / Half-Life
k = ln2 / 27.7 days
k = 0.025 days⁻¹
t = 48.2 days
[A] = ?
[A]₀ = 39.7mg
ln [A] = -0.025 days⁻¹*48.2 days + ln [39.7mg]
ln[A] = 2.476
[A] = 11.9g remains after 48.2 days
What is the molality of a glucose solution prepared by dissolving 16.7 g of glucose, C6H12O6, in 133.6 g of water
Answer:
0.696 m
Explanation:
We'll begin by calculating the number of mole in 16.7 g of C₆H₁₂O₆. This can be obtained as follow:
Mass of C₆H₁₂O₆ = 16.7 g
Molar mass of C₆H₁₂O₆ = (6×12) + (12×1) + (6×16)
= 72 + 12 + 96
= 180 g/mol
Mole of C₆H₁₂O₆ =?
Mole = mass / molar mass
Mole of C₆H₁₂O₆ = 16.7 / 180
Mole of C₆H₁₂O₆ = 0.093 mole
Next, we shall convert 133.6 g of water to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
133.6 g = 133.6 g × 1 Kg / 1000 g
133.6 g = 0.1336 Kg
Thus, 133.6 g is equivalent to 0.1336 Kg.
Finally, we shall determine the molality of the solution. This can be obtained as illustrated below:
Mole of C₆H₁₂O₆ = 0.093 mole
Mass of water = 0.1336 Kg
Molality =?
Molality = mole / mass of water (in Kg)
Molality = 0.093 / 0.1336
Molality = 0.696 m
Therefore, the molality of the solution is 0.696 m
Please help thank you
Answer:
[tex]K=1.7x10^{-3}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly setting up the equilibrium expression for the given reaction, in agreement to the law of mass action:
[tex]K=\frac{[NO]^2}{[N_2][O_2]}[/tex]
Next, we plug in the given concentrations on the data table to obtain:
[tex]K=\frac{(0.034)^2}{(0.69)(0.98)}\\\\K=1.7x10^{-3}[/tex]
Regards!
For each of the scenarios, determine if the ionic strength of the solution would increase, decrease, or not change.
a. If a solution of HNO3 were added to a solution of KOH , the ionic strength of the KOH solution would:_________
1. Increase
2. Decrease
3. Not change
b. If a dilute solution of KOH were added to a solution of CaCl2 (Ca(OH)2 (s) is formed), the ionic strength would:
1. Increase
2. Decrease
3. Not change
Answer:
Increase
Decrease
Explanation:While in solution, ionic substances produce ions. The ions in solution determine the conductivity of the solution.
The ionic strength of a solution shows the concentration of ions in a given solution. The more the number of ions in the solution, the greater the ionic strength of the solution and vice versa.
When HNO3 is added to a solution of KOH, the number of ions in the solution increases and so does the ionic strength of the solution.
When KOH is added to a solution of CaCl2 then Ca(OH)2 is formed. The formation of a solid precipitate decreases the concentration of ions in solution as well as the ionic strength of the solution.
Write the net ionic equation for the reaction that occurs when equal volumes of 0.546 M aqueous acetylsalicylic acid (aspirin) and sodium acetate are mixed. It is not necessary to include states such as (aq) or (s).
Answer:
[tex]C_9H_8O_4+C_2H_3O_2^-\rightarrow C_2H_4O_2+C_9H_7O_4^-[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to figure out the required net ionic equation by firstly writing out the complete molecular equation between aspirin and sodium acetate:
[tex]C_9H_8O_4+NaC_2H_3O_2\rightarrow C_2H_4O_2+NaC_9H_7O_4[/tex]
Whereas acetic acid and sodium acetylsalicylate are formed. Now, we write the complete ionic equation whereby sodium acetate and sodium acetylsalicylate are ionized because they are salts yet neither aspirin nor acetic acid are ionized as they are weak acids:
[tex]C_9H_8O_4+Na^++C_2H_3O_2^-\rightarrow C_2H_4O_2+Na^++C_9H_7O_4^-[/tex]
Finally, for the net ionic equation we cancel out the sodium spectator ions to obtain:
[tex]C_9H_8O_4+C_2H_3O_2^-\rightarrow C_2H_4O_2+C_9H_7O_4^-[/tex]
Regards!
How many moles of gas occupy a volume of 101.3L?
Answer:
V= n VmV: gas volume , n : The number of moles of gas , Vm : molar volume*The molar volume of any gas at standard conditions of temperature and pressure is 22.4 L/mol
V= 101.3 L , n=? , Vm = 22.4 L/mol
V=n Vm101.3 = n × 22.4
n=101.3 / 22.4
n = 4.52 mol
I hope I helped you ^_^
Answer:
[tex]\boxed {\boxed {\sf 4.522 \ mol}}[/tex]
Explanation:
We are asked to find how many moles of gas occupy a volume of 101.3 liters.
1 mole of any gas at STP (standard temperature and pressure) has a volume of 22.4 liters. We can use this information to make a proportion.
[tex]\frac {1 \ mol}{22.4 \ L}[/tex]
We are converting 101.3 liters to moles, so we multiply the proportion by that value.
[tex]101.3 \ L *\frac {1 \ mol}{22.4 \ L}[/tex]
The units of liters (L) cancel.
[tex]101.3 *\frac {1 \ mol}{22.4}[/tex]
[tex]\frac {101.3}{22.4} \ mol[/tex]
Divide.
[tex]4.52232143 \ mol[/tex]
The original value of liters (101.3 L) has 4 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 3 in the ten-thousandths place to the right tells us to leave the 2 in the thousandths place.
[tex]4.522 \ mol[/tex]
101.3 liters of gas is equal to approximately 4.522 moles of gas.
A dehydration reaction starting with 3.0 g cyclohexanol produces 1.9 g cyclohexene. Calculate the theoretical yield for this reaction. Report your answer with two significant figures.
Answer:
77%
Explanation:
First we convert 3.0 g of cyclohexanol (C₆H₁₂O) to moles, using its molar mass:
Molar mass of C₆H₁₂O = 100.158 g/mol3.0 g ÷ 100.158 g/mol = 0.030 molThen we convert 1.9 g of cyclohexene (C₆H₁₀) to moles, using its molar mass:
Molar mass of C₆H₁₀ = 82.143 g/mol1.9 g ÷ 82.143 g/mol = 0.023 molFinally we calculate the theoretical yield:
0.023 mol / 0.030 mol * 100% = 77%Sulfur trioxide dissolves in water, producing H2SO4. How much sulfuric acid can be produced from 10.1 mL of water (d= 1.00 g/mL) and 23.9 g of SO3? How much of the reagent in excess is left over?
Answer:
29.2 g of H₂SO₄ are produced
0.263 moles of water remain after the reaction goes complete.
Explanation:
We make the reaction in the first step:
Reactants are water and SO₃
H₂O + SO₃ → H₂SO₄
Let's determine moles of reactants:
23.9 g . 1 mol / 80.06g = 0.298 moles
We apply density, to determine mass of water:
D = m/ V so m = D . V
1 g/mL . 10.1 mL = 10.1 g
moles of water are: 10.1 g . 1 mol/ 18g = 0.561 moles
As ratio is 1:1, for 0.298 moles of SO₃ we need the same amount of water, and we have 0.561 moles. Then, water is the excess reagent and sulfur trioxide is the limiting.
0.561 - 0.298 = 0.263 moles of water that remain after the reaction goes complete.
As ratio is 1:1, again, 0.298 moles of SO₃ can produce 0.298 moles of acid.
We determine the mass: 0.298 mol . 98.06 g /mol = 29.2 g
here is the question
Answer:
1. Nitrate ions, NaNO3 - Sodium nitrate.
2. Sulphide ions, K2S - Potassium sulphide.
3. Sulphate ions, CaSO4 - Calcium sulphate.
4. Hydrogensulphite ions, NaHSO3 - Sodium hydrogensulphite.
5. Carbonate ions, CaCO3 - Calcium carbonate.
6. Hydrogencarbonate ions, KHCO3 - Potassium hydrogencarbonate.
7. Phosphite ions, PH3 - Hydrogen phosphite.
8. Nitride ions, NH3 - Hydrogen nitride ( ammonia ).
9. Ethanoate ions, CH3COONa - Sodium ethanoate.
10. Methanoate ions, HCOONa - Sodium methanoate.
11. Fluoride ions, HF - Hydrogen fluoride.
12. Chloride ions, KCl - Potassium chloride.
13. Bromide ions, HBr - Hydrogen bromide.
14. Iodide ions, NaI - Sodium iodide.
15. Phosphate ions, K3PO3 - potassium phosphate.
A sample of a compound is analyzed and found to contain 0.420 g nitrogen, 0.480g oxygen, 0.540 g carbon and 0.135 g hydrogen. What is the empirical formula of the compound? a. C2H5NO b. CH3NO c. C3H9N2O2 d. C4HN3O4 e. C4H13N3O3
Answer:
c. C3H9N2O2
Explanation:
The empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a molecule. To solve this question we need to convert the mass of each atom to moles. With the moles we can find the ratio as follows:
Moles N -Molar mass: 14.01g/mol-
0.420g N * (1mol/14.01g) = 0.0300 moles N
Moles O -Molar mass: 16g/mol-
0.480g O * (1mol/16g) = 0.0300 moles O
Moles C -Molar mass: 12.01g/mol-
0.540g C * (1mol/12.01g) = 0.0450 moles C
Moles H -Molar mass: 1.0g/mol-
0.135g H * (1mol/1g) = 0.135moles H
Dividing in the moles of N (Lower number of moles) the ratio of atoms is:
N = 0.0300 moles N / 0.0300 moles N = 1
O = 0.0300 moles O / 0.0300 moles N = 1
C = 0.0450 moles C / 0.0300 moles N = 1.5
H = 0.135 moles H / 0.0300 moles N = 4.5
As the empirical formula requires whole numbers, multiplying each ratio twice:
N = 2, O = 2, C = 3 and H = 9
And the empirical formula is:
c. C3H9N2O2
Identify each of the following properties as more typical of organic or inorganic compound
a. contains Li and F
b. is a gas at room temperature
c. contains covalent bonds
d. produces ion in water
this is the difference between organic and inorganic if this doesn't help you can research more on it
What type of energy does a skier stopped at the top of a hill have because of
his or her position?
A. Kinetic energy
B. Gravitational potential energy
C. Heat energy
D. Chemical energy
Answer:
B
Explanation:
Complete the table by assigning variable or fixed to the shape and volume of solids, liquids, and gases.
You are currently in a labeling module. Turn off browse mode or quick nav, Tab to items, Space or Enter to pick up, Tab to move, Space or Enter to drop.
Shape Volume
solids
liquids
gases
Answer Bank
Fixed or variable
The properties of solids, liquids and gases regarding their shapes and volumes are:
Shape Volume
Solids Fixed Fixed
Liquids Variable Fixed
Gases Variable Variable
Solids have strong attraction forces between their molecules. Thus, the molecules are closely packed with little movement. As a consequence, both shape and volume are fixed.
In liquids, attraction and repulsion forces are similar. They have a little more movement than the solid state. Then, they do have a fixed volume but they adopt the shape of the container.
Gases have very weak attraction forces between their molecules. They move very freely and expand trying to occupy as much volume as possible. So, they have a variable volume and shape (adopt the shape of the container).
You can learn more about states of matter here:
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In recrystallization from boiling water of benzoic acid contaminated with acetanilide, you begin with an impure sample of 5 grams. If the % composition of the acetanilide impurity in the sample is 6.3 %, what is the minimum amount in mL of solvent (water) required for the recrystallization
This question is incomplete, the complete question is;
In recrystallization from boiling water of benzoic acid contaminated with acetanilide, you begin with an impure sample of 5 grams. If the % composition of the acetanilide impurity in the sample is 6.3 %, what is the minimum amount in mL of solvent (water) required for the recrystallization
Compound Solubility in water at 25°C Solubility in water at 100°C
Benzoic Acid 0.34 g/100mL 5.6 g/100mL
Acetanilide 0.53 g/100mL 5.5 g/100mL
Answer:
The minimum amount in mL of solvent (water) required for the recrystallization is 91 mL
Explanation:
Given the data in the question;
mass of sample = 5 g
percentage composition of the acetanilide impurity = 6.3%
mass of the acetanilide in impure sample will be;
⇒ 6.3% × 5 g = 0.315 g
Mass of benzoic acid in impure sample;
⇒ 5 g - 0.315 g = 4.685 g
now, solubility in water at 100°C for benzoic acid = 5.6 g/100mL
hence 4.685 g of benzoic acid is soluble in x mL
x = [ 100 mL × 4.685 g ] / 5.6 g
x = 83.66 ≈ 84 mL
Also, solubility in water at 100°C for acetanilide = 5.5 g/100mL
hence 0.315 g of benzoic acid is soluble in x mL
x = [ 100 mL × 0.315 g ] / 5.5 g
x = 5.727 ≈ 6 mL
So, the minimum amount in mL of solvent (water) required for the recrystallization will be;
⇒ 85 mL + 6 mL = 91 mL
The minimum amount in mL of solvent (water) required for the recrystallization is 91 mL
What is the mass of a piece of iron if its density is 1.98 g/mL and its volume is 2.45 mL?
0.80 g
4.858
1.248
5.998
2.71 g
Answer:
4.858 g
Explanation:
Start with the formula
density = [tex]\frac{mass}{volume}[/tex]
density = 1.98 g/mL
volume = 2.45 mL
mass = ??
rearrange the formula to solve for mass
(density) x (volume) = mass
Add in the substitutes and solve for mass
1.98 g/mL x 2.45 mL = 4.858 g
How much energy is released when 31.0 g of water freezes? The heat of fusion for water is 6.02 kJ/mol.
Express your answer in kilojoules to three significant figures.
Answer:
The molar heat of fusion for a given substance basically tells you how much heat is required to melt one mole of that substance at its melting point from two angles.
Explanation:
:)
Find the oxidation number of:
A. sulfur in S032-
B. nickel in NiO2
c. iron in Fe(OH)2
Answer:
A. 4+.
B. 4+
C. 2+.
Explanation:
Hey there!
In this case, according to the given substances, it turns out possible for us to find the oxidation number of each element by applying the concept of charge balance in all of them as shown below:
A. sulfur in S032- : overall charge is 2- and the oxidation number of oxygen is 2-, thus:
[tex]x-6=-2\\\\x=6-2\\\\x=4+[/tex]
B. nickel in NiO2 : overall charge is 0 and the oxidation number of oxygen is 2-, thus:
[tex]x-4=0\\\\x=4+[/tex]
C. iron in Fe(OH)2: overall charge is 0 and the oxidation state of the OH ion is 1-, thus:
[tex]x-2=0\\\\x=2+[/tex]
Regards!
A 12.37 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 13.197 g. Identify the empirical formula of the new oxide
Answer:
MoO2
Explanation:
The empirical formula is defined as the simplest whole number ratio of atoms present in a molecule.
To solve this question we need to find the moles of Mo2O3. Twice these moles = Moles of Mo. With the moles of Mo we can find its mass.
The difference in masses between mass of new oxide and mass of Mo = Mass of oxygen. With the mass of oxygen we can find its moles and the empirical formula as follows:
Moles Mo2O3 -Molar mass: 239.9g/mol-
12.37g * (1mol / 239.9g) = 0.05156 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.1031 moles of Mo
Mass Mo -95.95g/mol-:
0.1031 moles of Mo * (95.95g/mol) = 9.895g of Mo
Mass oxygen in the oxide:
13.197 - 9.895g = 3.302g Oxygen
Moles oxygen -Molar mass: 16g/mol-:
3.302g Oxygen * (1mol / 16g) = 0.206 moles O
Now, the ratio of moles O / moles Mo is:
0.206 moles O / 0.1031 moles Mo = 2
That means there are 2 moles of O per mole of Mo and the empirical formula of the new oxide is:
MoO2The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. Decide, in each case, whether or not an insoluble precipitate is formed.
a. K2S and NH4Cl
b. CaCl2 and NH4CO3
c. Li2S and MnBr2
d. Ba(NO3)2 and Ag2SO4
e. RbCO3 and NaCl
Answer:
a) [tex]K_{2} S[/tex] and [tex]NH_{4} Cl[/tex] :
There are no insoluble precipitate forms.
b) [tex]Ca Cl_{2}[/tex] and [tex](NH_{4} )_{2} Co_{3}[/tex] :
There are the insoluble precipitates of [tex]CaCo_{3}[/tex] forms.
c) [tex]Li_{2}S[/tex] and [tex]MnBr_{2}[/tex] :
There are the insoluble precipitates of [tex]MnS[/tex] forms.
d) [tex]Ba(No_{3} )_{2}[/tex] and [tex]Ag_{2} So_{4}[/tex] :
As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.
e) [tex]Rb_{2}Co_{3}[/tex] and [tex]NaCl[/tex]:
There are no insoluble precipitates forms.
Explanation:
a)
Solubility rule suggests:- [tex]K_{2} S[/tex] ⇒ soluble, [tex]NH_{4} Cl[/tex] ⇒ soluble.
KCl ⇒ soluble, [tex](NH_{4})_{2} S[/tex] ⇒ soluble.
There are no insoluble precipitate forms.
b)
Solubility rule suggests:- [tex]Ca Cl_{2}[/tex] ⇒ soluble, [tex](NH_{4} )_{2} Co_{3}[/tex] ⇒ soluble.
[tex]CaCo_{3}[/tex] ⇒ insoluble, [tex]NH_{4} Cl[/tex] ⇒ soluble.
There are the insoluble precipitates of [tex]CaCo_{3}[/tex] forms.
c)
Solubility rule suggests:- [tex]Li_{2}S[/tex] ⇒ soluble, [tex]MnBr_{2}[/tex] ⇒ soluble.
[tex]LiBr[/tex] ⇒ soluble, [tex]MnS[/tex] ⇒ insoluble.
There are the insoluble precipitates of [tex]MnS[/tex] forms.
d)
Solubility rule suggests:- [tex]Ba(No_{3} )_{2}[/tex] ⇒ soluble, [tex]Ag_{2} So_{4}[/tex] ⇒insoluble.
As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.
e)
Solubility rule suggests:- [tex]Rb_{2}Co_{3}[/tex] ⇒ soluble, [tex]NaCl[/tex] ⇒ soluble.
[tex]RbCl[/tex] ⇒ soluble, [tex]Na_{2} Co_{3}[/tex] ⇒ soluble.
There are no insoluble precipitates forms.
In order to dry wet clothes, we spread them on a clothline. This is because (i) spreading increases surface area (ii) clothes become brighter when spread
Answer:
this is because spreading it makes more sunlight hit the cloth which results in it drying faster
The second law of thermodynamics requires that spontaneous processes generate entropy, either in the system or in the surroundings. What is the thermodynamic driving force for dissolving a solid in a liquid if it is an endothermic process (which reduces the entropy of the surroundings)
Answer:
See explanation
Explanation:
Truly, the second law of thermodynamics requires that spontaneous processes generate entropy, either in the system or in the surroundings.
When a solid is dissolved in a liquid, the solid dissociates into ions. These ions increases the number of particles and hence the entropy of the system thereby making the process spontaneous.
Hence, the dissolution of a substance via an endothermic process is spontaneous because of increase in the number of particles which in turn increases the entropy of the system.
Please help answering 46)
Answer:
CO2 is a trigonal planar.
A gas at 273K temperature has a pressure of 590 MM Hg. What will be the pressure if you change the temperature to 273K? 
Explanation:
here's the answer to your question
