Answer:
Biosphere
Explanation:
because of the rain it gets from huge water sources.
I hope that was useful.
Describe the buffer capacity of the acetic acid buffer solution in relation to the addition of both concentrated and dilute acids and bases.
Answer:
The answer is in the explanation
Explanation:
Acetic acid, CH₃COOH, is a weak acid that will produce a buffer when its conjugate base, CH₃COO⁻, acetate ion, is added to the solution.
That is because a buffer is the mixture of a weak acid and its conjugate base or vice versa.
When an acid (HX) is added to the solution, the acetate ion will react producing acetic acid, thus:
CH₃COO⁻ + HX → CH₃COOH + X⁻
For this reason, the pH doesn't change abruptly because H⁺ ions are not produced.
Now, if a base (BOH) is added to the buffer, CH₃COOH will react producing acetate ion and water, thus:
CH₃COOH + BOH → CH₃COO⁻ + H₂O + B⁺.
In the same way, there are not produced free OH⁻ and the pH doesn't change significantly.
How many moles of CO2 can be produced by the complete reaction of 1.0 g of lithium carbonate with excess hydrochloric acid (balanced chemical reaction is given below)? Li2CO3(s) + 2HCl(aq) --> 2LiCl(aq) + H2O(l) + CO2(g) Question 1 options: 1.7 g 1.1 g 0.60 040 g
Answer:Mass of CO2 = 0.60g
Explanation:
Given the chemical rection
Li2CO3(s) + 2HCl(aq) --> 2LiCl(aq) + H2O(l) + CO2(g
No of moles = mass / molar mass
molar mass Li2CO3 = Molecular mass calculation: 6.941 x 2 + 12.0107 + 15.9994 x 3 =
= 73.8909 g/mol
therefore Number of moles Li2CO3 = 1.0g / 73.89 g/mol
= 0.0135 moles Li2CO3
From our given Balanced equation, shows that
Li2CO3(s) + 2HCl(aq) --> 2LiCl(aq) + H2O(l) + CO2(g
1 mole Li2CO3 produces 1 mole CO2
therefore 0.0135 mol Li2CO3 will produce 0.0135 moles of CO2
Also
No of moles = mass / molar mass
Mass = No of moles x molar mass
molar mass of CO2=12.0107 + 15.9994 x 2=44.0095 g/mol
Mass of CO2= 0.0135 X 44.0095 g/mol =0.594≈0.60g
How many grams of POCl3 are produced when 225.0 grams of P4O10 and 675.0 grams of PCl5 react? This is the balance equation P4O10 + 6PCl5 → 10POCl3
Answer:
900g of POCl₃
Explanation:
Hello,
To solve this question, we'll require the equation of reaction.
P₄O₁₀ + 6PCl₅ → 10POCl₃
Molar mass of P₄O₁₀ = 283.886 g/mol
Molar mass of PCl₅ = 208.24 g/mol
Molar mass of POCl₃ = 153.33 g/mol
But Number of moles = mass / molar mass
Mass = molar mass × number of moles
Mass of POCl₃ = 153.33 × 10 = 1533.3g
Mass of PCl₅ = 208.24 × 6 = 1249.44g
Mass of P₄O₁₀ = 283.886 × 1 = 283.886g
From the equation of reaction,
283.886g of P₄O₁₀ + 1249.44g of PCl₅ produces 1533.33g of POCl₃
I.e 1533.33g of reactants produces 1533.33g of product (law of conservation of mass)
Therefore, (225g of P₄O₁₀ + 675g of PCl₅) = 900g will give x g of POCl₃.
1533.33g of reactants = 1533.33g of products
900g of reactants = x g of products
x = (900 × 1533.33) / 1533.33
x = 900g of POCl₃
What are 3 characteristics of chemical reactions
Answer:
Evolution of gas.
Formation of a precipitate.
Change in color.
Explanation:
a solid x when heated gives up a brown gas. If x is soluble in excess sodium hydroxide solution but insoluble in excess ammonium hydroxide solition. What is X?
Answer:
Lead (ii) nitrate
Explanation:
It is soluble in sodium hydroxide but insoluble in aqueous ammonia
When heated it produces nitrogen (IV) oxide that isbrown in colour
During the winter months, many locations experience snow and ice storms. It is a common practice to treat roadways and sidewalks with salt, such as NaCl . If a 11.3 kg bag of NaCl is used to treat the sidewalk, how many moles of NaCl does this bag contain
Answer:The moles of NaCL in the 11.3kg bag is 193.36moles
Explanation:
Given that a bag of NaCl = 11.3kg
1kg = 1000g
therefore 11.3 kg = 11,300g
Remember that
No of moles = mass of subatance/ molar mass of substance
The molar mass of NaCl = Na + Cl= 22.989769 + 35.453 =58.442769≈ 58.44g/mol
No of moles = mass of subatance/ molar mass of substance
= 11300g/ 58.44g/mol = 193.36 moles
The moles of NaCL in the 11.3kg bag is 193.36 moles.
When comparing the two chair conformations for a monosubstituted cyclohexane ring, which type of substituent shows the greatest preference for occupying an equatorial position rather than an axial position
Answer:
See the explanation
Explanation:
In this case, we have to keep in mind that in the monosubstituted product we only have to replace 1 hydrogen with another group. In this case, we are going to use the methyl group [tex]CH_3[/tex].
In the axial position, we have a more steric hindrance because we have two hydrogens near to the [tex]CH_3[/tex] group. If we have more steric hindrance the molecule would be more unstable. In the equatorial positions, we don't any interactions because the [tex]CH_3[/tex] group is pointing out. If we don't have any steric hindrance the molecule will be more stable, that's why the molecule will the equatorial position.
See figure 1
I hope it helps!
The diagram below shows that the periodic table is divided into different blocks.
A periodic table is shown. The main table consists of seven rows; two additional rows are shown below. In each block, the first column is labeled and the remaining columns are empty. The s-block is shaded in yellow and comprises the first two columns, plus one cell at the far side of the table. The first column has seven rows with entries 1 s, 2 s, 3 s, 4 s, 5 s, 6 s, and 7 s. A lone cell labeled 1 s appears at the top far right corner, aligned with the 1 s cell in the first column. The d-block is shaded in blue and contains 10 columns and 3 or 4 rows. The first column is directly to the right of the s-block. The first entry in the first d-block column aligns with the 4 s block, and is labeled 3d; further entries in that column are 4 d, 5 d, and 6 d. The first three columns in the block are four entries long; the remaining columns are three entries long, losing the bottom entry. The p-block is shaded in orange, and has 6 columns and 5 rows. The top row aligns with the 2 s block; entrie
Elements that have complete valence electron shells are mostly found in the
s block.
d block.
p block.
Answer:
p block.
Explanation:
jus took the test
Answer:
c p block
Explanation:
If one contraction cycle in muscle requires 55 kJ55 kJ , and the energy from the combustion of glucose is converted with an efficiency of 35%35% to contraction, how many contraction cycles could theoretically be fueled by the complete combustion of one mole of glucose? Round your answer to the nearest whole number.
Answer:
18 moles
Explanation:
Here the combustion of one mole of glucose ----> carbon dioxide + water, releases 2870 kilojoules / moles.
_______________________________________________________
With one contraction cycle requiring 55 kilojoules,
2870 / 55 ≈ 52.18
And with the efficiency being 35 percent,
52.1818..... * 0.35 = ( About ) 18 moles
Hope that helps!
hno3 express your answer using four significant figures
Answer:
[tex]M=63.02g/mol[/tex]
Explanation:
Hello,
In this case, for the calculation of the molar mass of nitric acid, we should employ the following formula, knowing that there is one hydrogen atom, one nitrogen atom and three oxygen atoms:
[tex]M=m_H+m_N+3*m_O[/tex]
Now, we use the atomic mass of each atom to compute the total molar mass:
[tex]M=1.008g/mol+14.01g/mol+3*16.00g/mol\\\\M=63.02g/mol[/tex]
Best regards.
an unknown metal of mass 512 g at a temperature of 15C is dropped into 325 g of water held in a 100 g aluminum container at the temperature of 98 c. is the system reaches thermal equilibrium at 78C determine th despicfic heat of the metal
Answer:
The specific heat capacity of the metal is 0.843J/g°C
Explanation:
Hello,
To determine the specific heat capacity of the metal, we have to work on the principle of heat loss by the metal is equals to heat gained by the water.
Heat gained by the metal = heat loss by water + calorimeter
Data,
Mass of metal (M1) = 512g
Mass of water (M2) = 325g
Initial temperature of the metal (T1) = 15°C
Initial temperature of water (T2) = 98°C
Final temperature of the mixture (T3) = 78°C
Specific heat capacity of metal (C1) = ?
Specific heat capacity of water (C2) = 4.184J/g°C
Heat loss = heat gain
M2C2(T2 - T3) = M1C1(T3 - T1)
325 × 4.184 × (98 - 78) = 512 × C1 × (78 - 15)
1359.8 × 20 = 512C1 × 63
27196 = 32256C1
C1 = 27196 / 32256
C1 = 0.843J/g°C
The specific heat capacity of the metal is 0.843J/g°C
The reaction of 15 moles carbon with 30 moles O2 will
result in a theoretical yield of moles CO2.
Answer:
The amount of Co2 generated is 15 moles
Explanation:
It bears the same ratio with Carbon meaning oxygen was used in excess
On a hot summer day, the density of air at atmospheric pressure at 35.5°C is 1.1970 kg/m3. (a) What is the number of moles contained in 1.00 m3 of an ideal gas at this temperature and
Complete question:
On a hot summer day, the density of air at atmospheric pressure at 35.5°C is 1.1970 kg/m3. (a) What is the number of moles contained in 1.00 m3 of an ideal gas at this temperature and pressure.
Answer:
The number of moles contained by an ideal gas at this temperature and pressure is 41.32 moles.
Explanation:
Given;
density of dry air, ρ = 1.1970 kg/m³
temperature of the air, T = 35.5°C = 273 + 35.5 = 308.5 K
air volume, V = 1 m³
Apply ideal gas law for dry to calculate the air pressure;
[tex]P = \rho R_dT[/tex]
where;
P is the air pressure
ρ is the air density
Rd is gas constant for dry air = 287 J/kg/K
P = 1.197 x 287 x 308.5 = 105,981.78 Pa
(a) Now, determine the number of moles contained by an ideal gas at this temperature and pressure, by applying ideal gas law;
PV = nRT
where;
P is the pressure of the gas (Pa)
V is the volume of the gas (m³)
n is number of gas moles
R is gas constant = 8.314 m³.Pa / mol.K
T is temperature (K)
n = (PV) / (RT)
n = (105,981.78 x 1) / (8.314 x 308.5)
n = 41.32 moles
Therefore, the number of moles contained by an ideal gas at this temperature and pressure is 41.32 moles.
The number of moles of an ideal gas at this temperature and pressure is 41.5 moles.
Given that;
Density of dry air = 1.1970 kg/m3
Pressure of dry air = ?
Temperature of dry air = 35.5°C + 273 = 308.5 K
Hence;
P = Density × gas constant of dry air × Temperature
P = 1.1970 kg/m3 × 287.1 J/Kg/K × 308.5 K
P = 106019 Pa or 1.05 atm
Using the ideal gas equation;
PV = nRT
n = PV/RT
n = 1.05 atm × 1000 L/0.082 atmL/K.mol × 308.5 K
n = 41.5 moles
Learn more: https://brainly.com/question/11897796
Enter your answer in the provided box. On a cool, rainy day, the barometric pressure is 739 mmHg. Calculate the barometric pressure in centimeters of water (cmH2O) (d of Hg = 13.5 g/mL; d of H2O = 1.00 g/mL).
Answer:
997.65cmH2O
Explanation:
Barometric pressure = 739 mmHg
density of Hg = 13.5 g/ml
density of water (H2O) = 1.00 g/ml
Calculate Barometric pressure in centimetres of water ( cmH20)
equate the barometric pressure of Hg and water
739 * 13.5 * 9.8 = x * 1 * 9.81
x ( barometric pressure of water in mmH2O ) = 739 *13.5 / 1 = 9976.5mmH2O
in cmH2O = 997.65cmH2O
Harvey kept a balloon with a volume of 348 milliliters at 25.0˚C inside a freezer for a night. When he took it out, its new volume was 322 milliliters, but its pressure was the same. If the final temperature of the balloon is the same as the freezer’s, what is the temperature of the freezer?
Answer:
[tex]T2=276K[/tex]
Explanation:
Given:
Initial volume of the balloon V1 = 348 mL
Initial temperature of the balloon T1 = 255C
Final volume of the balloon V2 = 322 mL
Final temperature of the balloon T2 =
To calculate T1 in kelvin
T1= 25+273=298K
Based on Charles law, which states that the volume of a given mass of a ideal gas is directly proportional to the temperature provided that the pressure is constant. It can be applied using the below formula
[tex](V1/T1)=(V2/T2)[/tex]
T2=( V2*T1)/V1
T2=(322*298)/348
[tex]T2=276K[/tex]
Hence, the temperature of the freezer is 276 K
Answer: 276 kelvins
Explanation:
What are the concentrations of Cu2+, NH3, and Cu(NH3)42+ at equilibrium when 18.8 g of Cu(NO3)2 is added to 1.0 L of a 0.800 M solution of aqueous ammonia? Assume that there is no volume change upon the addition of the solid, and that the reaction goes to completion and forms Cu(NH3)42+.
Answer:
Explanation:
Cu(NO₃)₂ + 4NH₃ = Cu(NH₃)₄²⁺ + 2 NO₃⁻
187.5 gm 4M 1 M
187.5 gm reacts with 4 M ammonia
18.8 g reacts with .4 M ammonia
ammonia remaining left after reaction
= .8 M - .4 M = .4 M .
187.5 gm reacts with 4 M ammonia to form 1 M Cu(NH₃)₄²⁺
18.8 g reacts with .4 M ammonia to form 0.1 M Cu(NH₃)₄²⁺
At equilibrium , the concentration of Cu²⁺ will be zero .
concentration of ammonia will be .4 M
concentration of Cu(NH₃)₄²⁺ formed will be 0.1 M
A certain element consists of two stable isotopes. The first has a mass of 62.9 amu and a percent natural abundance of 69.1 %. The second has a mass of 64.9 amu and a percent natural abundance of 30.9 %. What is the atomic weight of the element?
Answer:
63.518
Explanation:
The following data were obtained from the question:
Mass of Isotope A = 62.9 amu
Abundance of isotope A (A%) = 69.1%
Mass of isotope B = 64.9 amu
Abundance of isotope B (B%) = 30.9%
Atomic weight of the element =..?
The atomic weight of the element can be obtained as follow:
Atomic weight = [(Mass of A x A%)/100] + [(Mass of B x B%) /100]
Atomic weight = [(62.9 x 69.1)/100] + [(64.9 x 30.9)/100]
Atomic weight = 43.4639 + 20.0541
Atomic weight = 63.518
Therefore, the atomic weight of the element is 63.518.
A solution is prepared by mixing 5.00 mL of 0.100 M HCl and 2.00 mL of 0.200 M NaCl. What is the molarity of chloride ion in this solution?
Answer:
0.129 M
Explanation:
0.100 M HCl = 0.100 mol/L solution HCl
5.00 mL = 0.00500 L solution HCl
0.100 mol/L HCl * 0.00500 L = 0.000500 mol HCl
HCl ------> H+ + Cl-
1 mol 1 mol
0.000500 mol 0.000500 mol
0.200 M NaCl = 0.200 mol/L solution NaCl
2.00 mL = 0.00200 L solution NaCl
0.200 mol/L NaCl*0.00200 L = 0.000400 mol NaCl
NaCl ------> Na+ + Cl-
1 mol 1 mol
0.000400 mol 0.000400 mol
Chloride ion altogether (0.000500 mol + 0.000400 mol) =0.000900 mol
Solution altogether (0.00500 L+0.00200 L) = 0.00700L
Molarity (Cl-)= solute/solution = 0.000900 mol/0.00700L = 0.129 mol/L=
= 0.129 M
Determine the quantity (g) of pure MgSO4 in 2.4 g of MgSO4•7H2O. Show your work.
Answer: The quantity of pure [tex]MgSO_4[/tex] in 2.4 g of [tex]MgSO_4.7H_2O[/tex] is 1.17 g
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar Mass}}=\frac{2.4g}{246g/mol}=0.0098moles[/tex]
As 1 mole of [tex]MgSO_4.7H_2O[/tex] contains = 1 mole of [tex]MgSO_4[/tex]
Thus 0.0098 moles of [tex]MgSO_4.7H_2O[/tex] contains = [tex]\frac{1}{1}\times 0.0098=0.0098mole[/tex] of [tex]MgSO_4[/tex]
Mass of [tex]MgSO_4=0.0098mol\times 120g/mol=1.17g[/tex]
Thus the quantity of pure [tex]MgSO_4[/tex] in 2.4 g of [tex]MgSO_4.7H_2O[/tex] is 1.17 g
describe how would you use chromatography to show whether blue ink contains a single purple dye or a mixture of dyes
Explanation:
if the solution placed on the chromatography is pure there will be formation of one spot from the baseline and will go farthest to the front line unlike the impure one
With ink chromatography, a small amount of ink is added to the paper, one end is submerged in water, and the different colors of the ink are revealed as the water moves up the paper. All of this is made possible by the water base and variety of salabilities or densities that make up ink.
What is chromatography ?Separating mixture's constituent parts by chromatography is a method. The mixture is dissolved in a material known as the mobile phase to start the process, which then transports it through a material known as the stationary phase.
A little dot of the ink to be separated is placed at one end of a strip of filter paper to perform ink chromatography. The paper strip's opposite end is submerged in a solvent. The solvent moves up the paper strip, dissolving the chemical combination as it goes and pulling it up the paper.
Throughout the experiment, the dyes are pulled along by the mobile phase (water) as it gently advances up the stationary phase (paper).
Thus, With ink chromatography, a small amount of ink is added to the paper, one end is submerged in water.
To learn more about chromatography, follow the link;
https://brainly.com/question/11960023
#SPJ5
how many grams are there in 9.4x10^25 molecules of H2
Answer:
You start with 9.4 x 1025 molecules of H2.
You know that an Avogadro's number of molecules of H2 has a mass of 2.0 g.
To solve, 9.4 x 1025 molecules H2 x (2.0 g H2 / 6.023 x 1023 molecules H2) = 312. g H2
Explanation:
the reaction below is at equilibrium. What would happen if more carbon were added ?
Answer:
B
Explanation:
If more carbon were added, the equilibrium position would shift to produce more products.
What is equilibrium?In a chemical reaction chemical equilibrium is defined as the state at which there is no further change in concentration of reactants and products.
If the concentration of one (or more) of the reactants or products is increased the equilibrium will shift to decrease the concentration.
Or if the temperature is decreased the equilibrium will shift to increase the temperature by favouring the exothermic reaction.
Hence, if more carbon were added, the equilibrium position would shift to produce more products.
Learn more about equilibrium here:
https://brainly.com/question/5537989
#SPJ2
The acetic acid/acetate buffer system is a common buffer used in the laboratory. Write the equilibrium equation for the acetic acid/acetate buffer system. The formula of acetic acid is CH3CO2H .
Answer:
CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺
Explanation:
A buffer is defined as the mixture of a weak acid and its conjugate base or vice versa.
For the acetic acid buffer, CH₃CO₂H is the weak acid and its conjugate base is the ion without H⁺, that is CH₃CO₂⁻. The equilibrium equation in water knowing this is:
CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺In the equilibrium, the acid is dissociated in the conjugate base and the hydronium ion.
The acetic acid/acetate buffer system is a common buffer used in the laboratory, the equilibrium equation for the acetic acid/acetate buffer system. The formula of acetic acid is CH3CO2H -
CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺
An acid buffer is a solution that contains roughly the same concentrations of a weak acid and its conjugate base.
an acetate buffer contains roughly equal concentrations of acetic acid and acetate ion.Both are in chemical equilibrium with each other.The equation is:
CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺
where CH₃CO₂H - acetic acid
and, CH₃CO₂⁻ acetate ion
Thus, CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺ is the equilibrium equation for the acetic acid/acetate buffer system.
Learn more:
https://brainly.com/question/3435382
A protein's secondary structure arises due to interactions between amino acids. Two common secondary structures found in many proteins are alpha helices and beta sheets. Classify the following statements as pertaining to alpha helices or beta sheets.
1. Keep their shape due to hydrogen bonds between adjacent polypeptides chains
2. form long thin structures important for transmembrane proteins
3. flat zig zag like structure
4. a rigid spring like structure around a central axis
Answer:
Alpha helixes:
- form long thin structures important for transmembrane proteins
- have a rigid spring like structure around a central axis
Beta sheets:
- Keep their shape due to hydrogen bonds between adjacent polypeptides chains
- have a flat zig zag like structure
Explanation:
Alpha helices and beta pleated sheets are two types of secondary structure found in proteins.
Alpha helix: In this structure, the polypeptide backbone is tightly wound around an imaginary central axis drawn longitudinally through the center with the R groups of the amino acids protruding outward from the helical backbone. This structure looks like a spring and could either be a left-handed or right-handed helix, though the left-handed helix has not been observed in proteins. Each turn of the helix includes about 3.6 amino acid residues.
The alpha helix is stabilized by a hydrogen bond between the hydrogen atom attached to the electronegative nitrogen atom of a peptide linkage and the electronegative carbonyl oxygen atom of the fourth amino acid on the amino-terminal side of that peptide bond.
Alpha-helices due to their structure, are the most common transmembrane proteins- protein structure element that crosses biological membranes.
Beta sheets: In the beta conformation, the backbone of the polypeptide chain is extended into a zigzag structure. The zigzag polypeptide chain can be arranged side by side to form a structure resembling a series of pleats known as beta sheets. Hydrogen bonds formed between adjacent segments of the polypeptide chain functions to stabilize the structure.
The beta comformation in the form of turns is common in globular proteins.
A solution is prepared by dissolving 33.0 milligrams of sodium chloride in 1000. L of water. Assume a final volume of 1000. liters. Calculate the following values listed below.
a. Molarity of NaCl
b. Molarity of sodium ions
c. Molarity of chloride ions
d. Osmolarity of the solution
e. Mass percent of NaCl
f. Parts per million of sodium chloride
g. Parts per billion of sodium chloride
h. Look at your answers to parts e, f, and g.
Which one of these values is the most convenient or easiest to say and use/understand when discussing concentration.
Answer:
a. Molarity of NaCl solution = 5.64 * 10⁻⁷ mol/L
b. molarity of Na⁺ = 5.64 * 10⁻⁷ mol/L
c. molarity of Cl⁻ = 5.64 * 10⁻⁷ mol/L
d. Osmolarity = 1.128 osmol
e. mass percent of NaCl = 3.30 * 10⁻⁶ %
f. parts per million NaCl = 0.033 ppm NaCl
g. parts per billion of NaCl = 33 ppb of NaCl
h. From the values obtained from e, f and g, the most convenient to use and understand is parts per billion as it has less of a fractional part to deal with especially since the solute concentration is very small.
Explanation:
Molarity of a solution = number of moles of solute (moles)/volume of solution (L)
where number of moles of solute = mass of solute (g)/molar mass of solute (g/mol)
a. Molarity of NaCl:
molar mass of NaCl = 58.5 g/mol, mass of NaCl = 33.0/1000) g = 0.033g
number of moles of NaCl = 0.033/58.5 = 0.000564 moles
Molarity of NaCl solution = 0.000564/1000 = 5.64 * 10⁻⁷ mol/L
b. Equation for the dissociation of NaCl in solution: NaCl ----> Na⁺ + Cl⁻
From the above equation I mole of NaCl dissociates to give 1 mole of Na⁺ ions,
Therefore molarity of Na⁺ = 1 * 5.64 * 10⁻⁷ mol/L = 5.64 * 10⁻⁷ mol/L
c. From the above equation I mole of NaCl dissociates to give 1 mole of Cl⁻ ions,
therefore molarity of Cl⁻ = 1 * 5.64 * 10⁻⁷ mol/L = 5.64 * 10⁻⁷ mol/L
d. From the above equation, dissociation of NaCl in water produces 1 mol Na⁺ and 1 mole Cl⁻.
Total number of particles produced = 2
Osmolarity of solution = number of particles * molarity of siolution
Osmolarity = 2 * 5.64 * 10⁻⁷ mol/L = 1.128 osmol
e. mass of percent of NaCl = {mass of NaCl (g)/ mass of solution (g)} * 100
density of water = 1 Kg/L
mass of water = 1 Kg/L * 1000 L = 1000 kg
1Kg = 1000 g
Therefore mass of solution in g = 1000 * 1000 = 1 * 10⁶ g
mass percent of NaCl = (0.033/1 * 10⁶) * 100 = 3.30 * 10⁻⁶ %
f. Parts per million of NaCl:
parts per million = 1 mg of solute/L of solution
One thousandth of a gram is one milligram and 1000 ml is one liter, so that 1 ppm = 1 mg per liter = mg/Liter.
Since the density of water is 1kg/L = 1,000,000 mg/L
1mg/L = 1mg/1,000,000mg or one part in one million.
parts per million NaCl = 33.0/1000 L = 0.033 ppm NaCl
g. Parts per billion = 1 µg/L of solution
1 g = 1000 µg
therefore, 33.0 mg = 33.0 * 1000 µg = 3.30 * 10⁴ µg
parts per billion of NaCl = 3.30 * 10⁴ µg/1000 L = 33 ppb of NaCl
h. From the values obtained from e, f and g, the most convenient to use and understand is parts per billion as it has less of a fractional part to deal with especially since the solute concentration is very small.
"Strike anywhere matches contain the compound tetraphosphorus trisulfide, which burns to form tetraphosphorus decaoxide and sulfur dioxide gas. How many milliliters of sulfur dioxide, measured at 751 torr and 21.0°C, can be produced from burning 0.869 g of tetraphosphorus trisulfide?"
Answer:
Can be produced 288mL of SO₂
Explanation:
Based in the reaction:
P₄S₃ + 8O₂ → P₄O₁₀ + 3SO₂
Where 1 mole of tetraphosphorus trisulfide reacts producing 3 moles of sulfur dioxide gas.
0.869g of tetraphosphorus trisulfide (Molar mass of P₄S₃: 220.09g/mol) are:
0.869g P₄S₃ ₓ (1mol / 220.09g) = 3.948x10⁻³ moles of P₄S₃
As 3 moles of SO₂ are produced per mole of P₄S₃:
3.948x10⁻³ moles of P₄S₃ ₓ (3 moles SO₂ / 1 mole P₄S₃) = 0.0118 moles SO₂
Using PV = nRT
V = nRT / P
Where n are 0.0118 moles, R gas constant (0.082atmL/molK), T absolute temperature (21.0°C + 273.15K = 294.15K), and P pressure (751torr / 760 = 0.988atm).
Replacing:
V = 0.0118molₓ0.082atmL/molKₓ294.15K / 0.988atm
V = 0.288L = 288mL
18. Investigue por qué en el bloque "d" se aplica la fórmula n-1 y en el bloque "f" n-2.
Answer:
La razón de esto es porque los elementos en el bloque d tienen sus electrones d en la capa (n-1) (penúltima capa) y puede usar electrones de la capa (n-1) (penúltima capa) para formar enlaces químicos.
Los elementos en el bloque f tienen sus electrones f en sus capas (n-2) y pueden llegar a usar electrones de la capa (n-2) (capa antes de la penúltima capa) para enlaces químicos.
Explanation:
Para explicar esto, comencemos con los elementos de bloque syp; Los elementos de bloque syp usan electrones que tienen el mismo número cuántico principal para enlaces químicos. Por ejemplo, el aluminio utiliza los electrones 3s más externos y el carbono los electrones 2s y 2p cuando forman enlaces químicos. A pesar de que el carbono (bloque p) utiliza electrones de dos conjuntos de orbitales (2s, 2p) para la unión covalente, su número cuántico principal es el mismo (2) ya que todavía provienen de la misma capa.
Pero, los metales de transición (elementos de bloque d) usan electrones del orbital "s" de la capa más externa y los orbitales "d" de la penúltima capa forman enlaces químicos. Por ejemplo, los elementos de la primera serie de transición como el manganeso, el cobre y el hierro pueden usar los electrones 4s más externos y los electrones 3d de la penúltima capa, con diferentes números cuánticos principales (4 y 3). Eso significa que los metales de transición pueden usar tanto el orbital ‘ns’ más externo como los orbitales d (n-1) de la penúltima caparazón.
Luego, para los metales de transición internos (elementos de bloque f), tienen sus electrones f en sus capas (n-2) y pueden llegar a usar electrones de la capa (n-2) (capa antes de la penúltima capa) para enlaces químicos.
¡¡¡Espero que esto ayude!!!
The atomic mass of gallium is 69.72 . The density of iron is 7.87 . The atomic mass of iron is 55.847 . Calculate the number of gallium atoms in one ton (2000 pounds) of gallium. (Enter your answer to three significant figures.)
Answer:
the atomic mass of any elemet contains avogardo numberof atoms
In case of Gallium,
69.72 gram is atomic mass and it cotnains around 6.023*10^23 atoms of Gallium
but, 2000 punds = 907184.7 grams
907184.7 gram of gallium contains= 6.023*10^23* 907184/69.72
= 79 *10^26 atoms
Explanation:
Why does a chemical change occur when copper is heated?
Answer:
When copper is heated, it decomposes to form copper oxide and carbon dioxide. It is an endothermic reaction, which means that it absorbs heat. When heated, copper is easily bent or molded into shapes.
Explanation:
am i correct if not correct me please
Answer:
D. Hund's rule
Explanation:
Not sure, but I would go with Hund's since it talks about filing electrons in each orbital before you can pair them up. The reason sulfur has lower ionization is because it has one set of paired electrons which makes the orbital unstable whereas phosphorus has 3 unpaired e's which means it is more stable. Thus it is easier to remove electron from sulfur hence lower ionization energy.