The process of watering the crops is called irrigation.
Any two methods of irrigation are:
(i) Sprinkler system:This system is used on the uneven land where less water is available. The perpendicular pipes, having rotating nozzles on top, are joined to the main pipeline at regular intervals. Water is allowed to flow through main pipe under pressure, which escapes from the rotating nozzles. In this way water gets sprinkled on the crop.
(ii) Drip irrigation:This system is used to save water as it allows the water to flow drop by drop at the roots of the plants. It is the best technique for watering fruit plants, gardens and trees. Water is not wasted at all.
what would happen if the number of protons and electrons in the Atom did change
Calculate the boiling point of a 3.5 % solution (by weight) of sodium chloride in water.
Kb of H2O = 0.512 oC/M
Answer: The boiling point of the solution is [tex]101.02^oC[/tex]
Explanation:
We are given:
3.5 % (by weight) NaCl
This means that 3.5 g of NaCl is present in 100 g of solution
Mass of solvent = Mass of solution - Mass of solute
Mass of solvent (water) = (100 - 3.5) g = 96.5 g
Elevation in the boiling point is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent.
The expression for the calculation of elevation in boiling point is:
[tex]\text{Boiling point of solution}-\text{boiling point of pure solvent}=i\times K_b\times m[/tex]
OR
[tex]\text{Boiling point of solution}-\text{Boiling point of pure solvent}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)
where,
Boiling point of pure solvent (water) = [tex]100^oC[/tex]
Boiling point of solution = ?
i = Vant Hoff factor = 2 (for NaCl)
[tex]K_b[/tex] = Boiling point elevation constant = [tex]0.512^oC/m[/tex]
[tex]m_{solute}[/tex] = Given mass of solute (NaCl) = 3.5 g
[tex]M_{solute}[/tex] = Molar mass of solute (NaCl) = 36.5 g/mol
[tex]w_{solvent}[/tex] = Mass of solvent (water) = 96.5 g
Putting values in equation 1, we get:
[tex]\text{Boiling point of solution}-(100)=2\times 0.512\times \frac{3.5\times 1000}{36.5\times 96.5}\\\\\text{Boiling point of solution}=(1.02+100)^oC\\\\\text{Boiling point of solution}=101.02^oC[/tex]
Hence, the boiling point of the solution is [tex]101.02^oC[/tex]
Write a balanced half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution. Be sure to add physical state symbols where appropriate.
Answer:
2 H₂O(l) + 2 e⁻ ⇒ H₂(g) + 2 OH⁻(aq)
Explanation:
Let's consider the unbalanced half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution.
H₂O(l) ⇒ H₂(g)
First, we will perform the mass balance. We will balance oxygen atoms by multiplying H₂O by 2 and adding 2 OH⁻ to the right side.
2 H₂O(l) ⇒ H₂(g) + 2 OH⁻(aq)
Then, we perform the charge balance by adding 2 electrons to the left side.
2 H₂O(l) + 2 e⁻ ⇒ H₂(g) + 2 OH⁻(aq)
A gas sample containing a constant number of gas molecules has a volume of 2.70 L at a constant pressure and a temperature of 25.0o C. What would be the volume (in Liters) of this gas sample at 75.0o C? Round your answer to 3 sig fig
Answer:
[tex]\boxed {\boxed {\sf 8.10 \ L}}[/tex]
Explanation:
This question asks us find the volume of a gas sample given a change in temperature. Since the pressure remains constant, we only are concerned with the variables of temperature and volume.
We will use Charles's Law. This states the volume of a gas is directly proportional to the temperature of a gas. The formula is:
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
The gas starts at a volume of 2.70 liters and a temperature of 25.0 degrees Celsius.
[tex]\frac {2.70 \ L}{25.0 \textdegree C}=\frac{V_2}{T_2}[/tex]
The temperature is increased to 75.0 degrees Celsius, but the volume is unknown.
[tex]\frac {2.70 \ L}{25.0 \textdegree C}=\frac{V_2}{75.0 \textdegree C}[/tex]
We are solving for the volume at 75 degrees Celsius, so we must isolate the variable V₂.
It is being divided by 75.0 °C. The inverse operation of division is multiplication, so we multiply both sides of the equation by 75.0 °C.
[tex]75.0 \textdegree C *\frac {2.70 \ L}{25.0 \textdegree C}=\frac{V_2}{75.0 \textdegree C} * 75.0 \textdegree C[/tex]
[tex]75.0 \textdegree C *\frac {2.70 \ L}{25.0 \textdegree C}= V_2[/tex]
The units of degrees Celsius (° C) cancel.
[tex]75.0 *\frac {2.70 \ L}{25.0}= V_2[/tex]
[tex]75.0 *0.108 \ L = V_2[/tex]
[tex]8.1 \ L = V_2[/tex]
The original measurements have 3 significant figures, so our answer must have the same. Currently, the answer has 2. If we add another 0, the value of the answer does not change, but the number of sig figs does.
[tex]8.10 \ L = V_2[/tex]
The volume of this gas sample at 75.0 degrees Celsius is 8.10 Liters.
give one use of zinc
What is the molarity of a solution containing 150 g of zinc sulfate (ZnSO4) per liter?
Answer:
0.93 M
Step-by-step Explanation:
First, we have to calculate the molar mass (MM) of ZnSO₄ by using the molar mass of each chemical element:
MM(ZnSO₄) = 65.4 g/mol Zn + 32 g/mol S + (16 g/mol x 4) = 161.4 g/mol
Then, we divide the mass of ZnSO₄ into its molar mass to obtain the number of moles:
moles ZnSO₄ = mass/MM = 150 g/(161.4 g/mol)= 0.93 mol
Since the molarity of a solution expresses the number of moles of solute per liter of solution, we calculate the molarity (M) as follows:
M = moles ZnSO₄/1 L = 0.93 mol/1 L = 0.93 M
In aqueous solution the Ni2" ion forms a complex with four ammonia molecules. Write the formation constant expression for the equilibrium between the hydrated metal ion and the aqueous complex. Under that, write the balanced chemical equation for the first step in the formation of the complex K,=________.
Answer:
The correct equation is "[tex]\frac{[Ni(H_2O)_3 (NH_3)]^{2+}}{[Ni(H_2O)_4]^{2+} [NH_3]}[/tex]".
Explanation:
According to the question,
Throughout an aqueous solution, [tex]Ni^{2+}[/tex] exist as [tex][Ni(H_2O)_4]^{2+}[/tex]
So,
⇒ [tex][Ni(H_2O)_4]^{2+} + 4NH_3 \rightleftharpoons [Ni(NH_3)_4]^{2+} + H_2O[/tex]
⇒ [tex]K_f = \frac{[Ni(NH_3)_4]^{2+}}{[Ni(H_2O)_4^{2+}] [NH_3]^4}[/tex]
Here, we have excluded [tex][H_2O][/tex] as concentration of water will be const.
Now,
This formation of [tex][Ni(NH_3)_4]^{2+}[/tex] proceeds via several steps,
Step 1:
⇒ [tex][Ni(H_2O)_4]^{2+}+NH_3 \rightleftharpoons [Ni(H_2O)_3 (NH_3)]^{2+} + H_2O[/tex]
⇒ [tex]K_1 = \frac{[Ni(H_2O)_3 (NH_3)]^{2+}}{[Ni(H_2O)_4]^{2+} [NH_3]}[/tex]
A 2.584 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of dioxygen, producing 5.874 g CO2 and 2.404 g H2O. What mass of oxygen is contained in the original sample?a. 0.7119 g.b. 0.8463 g.c. 0.29168 g.d. 0.1793 g.e. 0.6230 g.
Answer:
a. 0.7119 g
Explanation:
To solve this question we need to know that all carbon of the compound will react producing CO2 and all Hydrogen producing H2O.
Thus, we can find the mass of C and the mass of H and by difference regard to the 2.584g of the compound we can find the mass of oxygen as follows:
Moles CO2 = Moles C -Molar mass: 44.01g/mol-
5.874g CO2 * (1mol/44.01g) = 0.1335 moles CO2 = 0.1335 moles C
Mass C -Molar mass: 12.01g/mol-:
0.1335 moles C * (12.01g /mol) = 1.6030g C
Moles H2O -Molar mass: 18.01g/mol-
2.404gH2O * (1mol / 18.01g) = 0.1335 moles H2O * (2mol H / 1mol H2O) = 0.267 moles H
Mass H -Molar mass: 1g/mol-
0.267 moles H * (1g/mol) = 0.2670g H
Mass Oxygen =
Mass O = 2.584g compound - 1.6030g C - 0.2670g H
Mass O = 0.714g O ≈
a. 0.7119 g
The absolute temperature of a gas is increased four times while maintaining a constant volume. What happens to the
pressure of the gas?
Olt decreases by a factor of four.
O It increases by a factor of four.
It decreases by a factor of eight
It increases by a factor of eight.
Answer:
The pressure increases by a factor of four.
Explanation:
Let's consider a gas at a given temperature and pressure (T₁, P₁). The absolute temperature of a gas is increased four times (T₂ = 4 T₁) while maintaining a constant volume. We can assess the effect on the pressure (P₂) by using Gay Lussac's law.
P₁/T₁ = P₂/T₂
P₂ = P₁ × T₂/T₁
P₂ = P₁ × 4 T₁/T₁
P₂ = 4 P₁
The pressure increases by a factor of four.
0.28 M Ca(NO3)2
Express your answer using two significant figures.
Answer:
Mass=Moles × RFM
Mass= 0.28M× 164
Mass= 45.92 grammes
Sobre ações relacionadas ao aquecimento global, assinale somente as alternativas corretas:
a) ( x) As ações humanas não influenciam no aumento da temperatura do planeta.
b) ( ) As mudanças climáticas são intensificadas pela emissão de gases das atividades humanas.
c) ( ) A queima de combustíveis fósseis e de florestas são as principais ações humanas que liberam gases que intensificam o efeito estufa.
d) ( ) O efeito estufa é um fenômeno natural.
e) ( ) Se as águas dos oceanos ficarem mais quentes, os furacões não terão tanta força.
Answer:
123456788012346778901234567890
At 35 C, a sample of gas has a volume of 256 ml and a pressure of 720.torr. What would the volume
be if the temperature were changed to 22 C and the pressure to 1.25 atmospheres?
Answer: Volume would be 196.15 mL if the temperature were changed to [tex]22^{o}C[/tex] and the pressure to 1.25 atmospheres.
Explanation:
Given: [tex]T_{1} = 35^{o}C = (35 + 273) K = 308 K[/tex], [tex]V_{1}[/tex] = 256 mL,
[tex]P_{1}[/tex] = 720 torr (1 torr = 0.00131579 atm) = 0.947368 atm
[tex]T_{1} = 22^{o}C = (22 + 273) K = 295 K[/tex], [tex]P_{2} = 1.25 atm[/tex]
Formula used to calculate volume is as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{1 atm \times 256 mL}{308 K} = \frac{1.25 atm \times V_{2}}{295 K}\\V_{2} = 196.15 mL[/tex]
Thus, we can conclude that the volume would be 196.15 mL if the temperature were changed to [tex]22^{o}C[/tex] and the pressure to 1.25 atmospheres.
You need to produce a buffer solution that has a pH of 5.50. You already have a solution that contains 10 mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution?
Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
What kind of element is Phosphorus is
Answer:
NON-METAL
Explanation:
Phosphorus is a non-metal that sits just below nitrogen in group 15 of the periodic table. This element exists in several forms, of which white and red are the best known.
How does the type of material an object is made of affect its ability to absorb or release thermal energy?
Answer:
Different materials will change temperature at different rates when exposed to the same amount of thermal energy. This is because each substance has its own specific heat.
Explanation:
Calculate the specific heat of a piece of wood if 2000 g of wood absorbs 75,250 J of heat, and its temperature changes from 30°C to 50°C.
A 37.63
B) 0.53
C) 1.88
D
752.50
Answer:
C
Explanation:
The specific heat capacity=quantity of heat in joule/mass×change in temperature
from this question the quantity of heat is 75250,the mass is 2000 and the change in temperature is 50-30
which is 20
therefore
c=75250/2000×20
c=75250/40000
c=1.88
I hope this helps
from kinatic point of view explain the change from solid to liquied based on the effect of change of tempreture.
Answer:
Temperature affects the kinetic energy in a gas the most, followed by a comparable liquid, and then a comparable solid. The higher the temperature, the higher the average kinetic energy, but the magnitude of this difference depends on the amount of motion intrinsically present within these phases.
Explanation:
Liquids have more kinetic energy than solids. When a substance increases in temperature, heat is being added, and its particles are gaining kinetic energy. Because of their close proximity to one another, liquid and solid particles experience intermolecular forces. These forces keep particles close together.
Ethylene glycol (C2H6O2) is mixed with water to make auto engine coolants. How many grams of C2H6O2 are in 5.00 L of a 6.00 M aqueous solution
Answer:
1860g.
Explanation:
It is known that the molar mass of C2H6O2 is 62.08 g/mol.,
Now to solve for the number of moles of solute, one must multiply both
sides by the volume:
moles of solute = (6.00 M)(5.00 L) = 30.0 mol
Notice since the definition of molarity is mol/L, the
product M × L gives mol, a unit of amount.
Use the molar mass of C3H8O3, one can convert mol to g:
Mass m =30 mol × 62.08 g/mol
m = 1860g.
Hence, there are 1,860 g of C2H6O2 in the specified amount of
engine coolant.
Leaming Task 1:
Distinguish the process as spontaneous or non-spontaneous process. Write S it spontaneous and NSi non-spontaneous
on the bionk.
1. Melling ofice
2 Ruisting of ton
3. Marble going down the spiral.
4. Going up the
& Keeping the food fresh from spolage
Solution :
Spontaneous Process
A spontaneous process is defined as the process that occurs without the help of any external aid or inputs. A spontaneous process is a natural process which occurs naturally in the environment.
Non Spontaneous process
A non spontaneous process is a process which does not occur naturally. Some inputs are provided for the process to occur. Energy from external source is applied into the process to start the process.
The following processes are :
1. Melling of ice ---- Spontaneous
2 Rusting of iron --- Spontaneous
3. Marble going down the spiral. --- spontaneous
4. Going up the hill ---- Non spontaneous
5. Keeping the food fresh from spoilage --- Non spontaneous
You prepare a solution by dissolving 25.3 g sucrose (C12H22O11) 705 mL of water. Calculate the molarity of the solution.
Answer:
0.105 M
Explanation:
First we convert 25.3 grams of sucrose into moles, using sucrose's molar mass:
Molar Mass of C₁₂H₂₂O₁₁ = 342.3 g/mol25.3 g C₁₂H₂₂O₁₁ ÷ 342.3 g/mol = 0.0739 mol C₁₂H₂₂O₁₁Now we calculate the molarity of the solution, using the given volume and the calculated number of moles:
Converting 705 mL ⇒ 705 mL / 1000 = 0.705 LMolarity = 0.0739 mol / 0.705 L = 0.105 MA sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipitate. (a) Write the molecular equation for the reaction. (b) What is the limiting reactant in the reaction? (c) How many grams of precipitate potentially form?
Answer:
0.21 g
Explanation:
The equation of the reaction is;
NaCl(aq) + AgNO3(aq) -----> NaNO3(aq) + AgCl(s)
Number of moles of NaCl= 0.0860 g /58.5 g/mol = 0.00147 moles
Number of moles of AgNO3 = 30/1000 L × 0.050 M = 0.0015 moles
Since the reaction is 1:1, NaCl is the limiting reactant.
1 mole of NaCl yields 1 mole of AgCl
0.00147 moles of NaCl yields 0.00147 moles of AgCl
Mass of precipitate formed = 0.00147 moles of AgCl × 143.32 g/mol
= 0.21 g
How many atoms are in protons
Answer:
the number of protons in a atom is unique to each element
Explanation:
protons are about 99.86% as massive neutrons. The number of protons in a atom is unique to each element .For example carbon atoms have six protons in an atom is referred to as the atomic number of that element
Is ribose a reducing or non reducing sugar?
Ribose is a reducing sugar. A reducing sugar is a carbohydrate that can undergo a redox reaction, in which it donates electrons to another chemical species.
This is usually observed when the sugar opens its ring structure to form an aldehyde or ketone functional group.
Ribose, a five-carbon sugar, can form an open-chain structure with an aldehyde functional group. In this form, it can donate electrons and act as a reducing agent in certain chemical reactions, such as the reduction of other compounds like Benedict's reagent during laboratory tests for reducing sugars.
Learn more about Ribose, here:
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Help me, please
Help me, please
Answer:.......
Explanation:
A hydrocarbon contains only the elements____?
Explanation:
elements are carbons and hydrogen
Answer:
Carbon and Hydrogen.
Explanation:
It’s in the name Hydro (H) Carbon (C)
Which technique is best suited to each application?
a. In the second week of a four week biochemistry experiement, you have 50 fractions collected from a gel filtration column to determine which fractions contain lactate dehydrogenase. You are given only 400 uL of 0.100 mg/mL lactate dehydrogenase to prepare your calibration curve. 96-well microplate
b. Your environmental lab has 2000 samples to be analyzed for trace ammonia by next week. discrete analyzer.
c. Twenty water samples must be analyzed for Cl-, NH3, PO3-, and So during each work shift. flow injection analysis colorimeter.
d. Your professor heard you will be hiking the Appalachian Trail next summer. She asks you to collect 100-mL water samples from the ten streams with the highest concentration of phosphate.
Answer:
a. discrete analyzer
b. 96 well microplate
c. flow injection analysis
d. colorimeter
Explanation:
96 well microplates are instruments designed for sample collection and throughput screening. If an environment lab has collected 2000 samples then 96 well microplate is best suited application. Discrete analyzer is automated chemical analyzer which performs test on samples kept in discrete cells. Flow injection analysis is approach used for chemical analysis. It injects a plug of sample into a flowing carrier stream. Colorimeter is a device which measures absorbance of wavelength of light by a specific solution.
An HCl solution has a concentration of 0.09714 M. Then 10.00 mL of this solution was then diluted to 250.00 mL in a volumetric flask. The diluted solution was then used to titrate 250.0 mL of a saturated AgOH solution using methyl orange indicator to reach the endpoint.
Required:
a. What is the concentration of the diluted HCl solution?
b. If 7.93 mL of the diluted HCl solution was required to reach the endpoint, what is the concentration of OH- in solution?
Answer:
A. Concentration of diluted acid = 0.00389 M
B. Concentration of OH- in AgOH solution = 0.00012 M
Explanation:
A. Using the dilution formula: C1V1 = C2V2
Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
From the data provided, C1 = 0.09714 M, V1 = 10.0 mL, V2 = 250.0 mL and C2 = ?
Making C2 subject of the formula above; C2 = C1V1/V2
C2 = 0.09714 M × 10 / 250 = 0.00389 M
B. Equation of the neutralization reaction is given below:
HCl + AgOH ---> AgCl + H₂O
From the equation, 1 mole of acid neutralizes 1 mole of base
Using the titration formula; CaVa/CbVb = na/nb
Where Ca is the concentration of the acid HCl = 0.00389 M
Va is the volume of acid = 7.93 mL
Cb is the concentration of base, AgOH = ?
Vb is volume of base = 250.0 mL
na/nb = mole ratio of acid and base = 1
Making Cb subject of the formula in the equation above; Cb = CaVa/Vb
Cb = 0.00389 M × 7.93 / 250
Cb = 0.00012 M
The reversible reaction: 2SO2(g) O2(g) darrw-tn.gif 2SO3(g) has come to equilibrium in a vessel of specific volume at a given temperature. Before the reaction began, the concentrations of the reactants were 0.060 mol/L of SO2 and 0.050 mol/L of O2. After equilibrium is reached, the concentration of SO3 is 0.040 mol/L. What is the equilibrium concentration of O2
Answer:
0.030 M
Explanation:
Step 1: Make an ICE chart
2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)
I 0.060 0.050 0
C -2x -x +2x
E 0.060-2x 0.050-x 2x
Step 2: Find the value of x
The concentration of SO₃ at equilibrium is 0.040 mol/L. Then,
2x = 0.040
x = 0.020
Step 3: Calculate the concentration at equilibrium of O₂
[O₂] = 0.050 - x = 0.050 - 0.020 = 0.030 M
The equilibrium concentration of oxygen is 0.030 M.
A reversible reaction is a reaction that can move either in the forward or in the reverse direction. We have the reaction; 2SO2(g) + O2(g) ⇄ 2SO3(g). We can now set up the ICE table as shown below;
2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)
I 0.060 0.050 0
C -2x -x +2x
E 0.060-2x 0.050-x 2x
At equilibrium;
2x = 0.040 M
x = 0.040 M/2 = 0.020 M
For oxygen;
0.050-x
0.050 M - 0.020 M = 0.030 M
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When heated, carboxylic salts in which there is a good leaving group on the carbon beta to the carbonyl group undergo decarboxylation/elimination to give an alkene. Draw the structures of the products expected when this compound is heated.
Answer:
i dont know mate
Explanation:
A mixture of hydrogen and argon gases, at a total pressure of 980 mm Hg, contains 0.291 grams of hydrogen and 5.62 grams of
argon. What is the partial pressure of each gas in the mixture?
PH2
Par
mm Hg
mm Hg
Answer:
Partial pressure of H₂ = 499 mmHg
Partial pressure of Ar = 481 mmHg
Explanation:
We'll begin by calculating the number of mole of each gas. This can be obtained as follow:
For Hydrogen:
Molar mass of H₂ = 2 × 1 = 2 g/mol
Mass of H₂ = 0.291 g
Mole of H₂ =?
Mole = mass /molar mass
Mole of H₂ = 0.291/ 2
Mole of H₂ = 0.1455 mole
For Argon:
Molar mass of Ar = 40 g/mol
Mass of Ar = 5.62 g
Mole of Ar =?
Mole = mass /molar mass
Mole of Ar = 5.62 / 40
Mole of Ar = 0.1405 mole
Next, we shall determine the mole fraction of each gas. This can be obtained as follow:
Mole of H₂ = 0.1455 mole
Mole of Ar = 0.1405 mole
Total mole = 0.1455 + 0.1405
Total mole = 0.286 mole
Mole fraction of H₂ (nₕ₂) = mole of H₂ / total mole
Mole fraction of H₂ (nₕ₂) = 0.1455/0.286
Mole fraction of H₂ (nₕ₂) = 0.509
Mole fraction of Ar (nₐᵣ) = mole of Ar / total mole
Mole fraction of Ar (nₐᵣ) = 0.1405/0.286
Mole fraction of Ar (nₐᵣ) = 0.491
Finally, we shall determine the partial pressure of each gas. This can be obtained as follow:
For Hydrogen:
Mole fraction of H₂ (nₕ₂) = 0.509
Total pressure (Pₜ) = 980 mmHg
Partial pressure of H₂ (Pₕ₂) =?
Pₕ₂ = nₕ₂ × Pₜ
Pₕ₂ = 0.509 × 980
Partial pressure of H₂ (Pₕ₂) = 499 mmHg
For Argon:
Partial pressure of H₂ (Pₕ₂) = 499 mmHg
Total pressure (Pₜ) = 980 mmHg
Partial pressure of Ar (Pₐᵣ) =?
Pₜ = Pₕ₂ + Pₐᵣ
980 = 499 + Pₐᵣ
Collect like terms
980 – 499 = Pₐᵣ
481 = Pₐᵣ
Partial pressure of Ar (Pₐᵣ) = 481 mmHg
SUMMARY:
Partial pressure of H₂ (Pₕ₂) = 499 mmHg
Partial pressure of Ar (Pₐᵣ) = 481 mmHg
