Answer:
Fraction of association = 0.01
pH = 11
Explanation:
A weak base, B, is in equilibrium with water as follows:
B(aq) + H2O(l) ⇄ BH⁺(aq) + OH⁻(aq)
Where Kb is defined as:
Kb = 1.00x10⁻⁵ = [BH⁺] [OH⁻] / [B]
Some B will react producing BH⁺ and OH⁻:
[BH⁺] = X
[OH⁻] = X
[B] = 0.100M - X
As Kb <<< [B] we can say:
[B] ≈ 0.100M
Replacing:
1.00x10⁻⁵ = [X] [X] / [0.100]
1.00x10⁻⁶ = X²
X = 1x10⁻³M = [BH⁺] = [OH⁻]
The fraction of association is [BH⁺] / [B] = 1x10⁻³M / 0.100M = 0.01
As [OH⁻] = 1x10⁻³M, pOH = -log[OH⁻] = 3
pH = 14- pOH
pH = 11A tank contains 7.0 moles of a mixture of nitrogen gas and oxygen gas. The total
pressure exerted by the gas mixture is 530 kPa. The partial pressure exerted by the
oxygen is 140 kPa. How many moles of NITROGEN gas are present in the tank?
Answer:
5.2 mol
Explanation:
Step 1: Given data
Total number of moles (n): 7.0 molTotal pressure (P): 530 kPaPartial pressure of oxygen (pO₂): 140 kPaStep 2: Calculate the mole fraction of oxygen
We will use the following expression.
pO₂ = P × X(O₂)
X(O₂) = pO₂ / P
X(O₂) = 140 kPa / 530 kPa
X(O₂) = 0.264
Step 3: Calculate the moles of oxygen
We will use the definition of mole fraction of oxygen.
X(O₂) = n(O₂) / n
n(O₂) = X(O₂) × n
n(O₂) = 0.264 × 7.0 mol
n(O₂) = 1.8 mol
Step 4: Calculate the moles of nitrogen
The total number of moles is equal to the sum of moles of the individual gases.
n(O₂) + n(N₂) = n
n(N₂) = n - n(O₂)
n(N₂) = 7.0 mol - 1.8 mol
n(N₂) = 5.2 mol
1. What is the flame color of sodium?
yellow-orange
purple
green
blue-green
Orange is the color of Sodium
A student pipets 5.00 mL of a 5.103 M aqueous NaOH solution into a 250.00 mL volumetric flask and dilutes up to the mark with distilled water. What is the final molarity of the dilute solution?
Answer:
0.102 M.
Explanation:
From the question given above, the following data were obtained:
Volume of stock solution (V1) = 5 mL
Molarity of stock solution (M1) = 5.103 M
Volume of diluted solution (V2) = 250 mL
Molarity of diluted solution (V2) =?
The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:
M1V1 = M2V2
5.103 × 5 = M2 × 250
25.515 = M2 × 250
Divide both side by 250
M2 = 25.515 / 250
M2 = 0.102 M
Thus, the molarity of the diluted solution is 0.102 M.
The dilution of the NaOH results in the dilution of the molarity of the solution. The molarity of the final NaOH solution is 0.102 M.
What is molarity?Molarity is the concentration unit that defines the amount of solute present in a liter of solution.
The dilution of the solution results in a change in the molarity. It can be accessed as:
[tex]\rm M_1V_1=M_2V_2[/tex]
The initial molarity of the solution is, [tex]\rm M_1=5.103\;M[/tex]
The initial volume of the NaOH is, [tex]\rm V_1=5\;mL[/tex]
The final volume of the NaOH is, [tex]\rm V_2=250\;mL[/tex]
Substituting the values for the calculation of final molarity, [tex]\rm M_2[/tex]:
[tex]\rm 5.103\;\M\;\times\;5.0\;mL=M_2\;\times\;250\;mL\\\\M_2=\dfrac{5.103\;\M\;\times\;5.0\;mL}{250\;mL} \\\\M_2=0.102\;M[/tex]
The molarity of the NaOH solution after dilution is 0.102 M.
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What is the difference between Mega and Milli symbols?
Answer:
Mega = 10^6
Milli = 10^-3
Explanation:
Mega is whatever times 10 to 6th power.
Milli is whatever times 10 to the -3rd power.
1. How many atoms are in 0.25 moles of carbon?
2. How many atoms are in 12.3 grams of sodium?
3. How many grams are there in 0.52 moles of boron?
4. How many grams are there in 2.0 moles of HCl?
5. How many moles are in 3.4 grams of HBr?
6. How many grams are there in 4.5x10^10 atoms of NaCl?
7. How many atoms are there in 45.1 grams of MgO?
Answer 1 mole = 6.02 × 1023 atoms
0.25 moles × (6.02 × 1023) = 1.5 × 1023 atoms Carbon
for the first one
Calculate the mass of zinc that reacts with 16.23 g of hydrochloric acid to form 15.2 g of zinc chloride and 18.97 g of hydrogen gas
Answer:
17.94 g
Explanation:
The barometric pressure measured outside an airplane at 3 km ( ft) was 524 mmHg. Calculate the pressure in kPa.
Answer:
69.8609 KPa.
Explanation:
The following data were obtained from the question:
Pressure (in mmHg) = 524 mmHg.
Pressure (in KPa) =?
Thus we can obtain the pressure in KPa as follow:
760 mmHg = 101.325 Kpa
Therefore,
524 mmHg = 524 mmHg × 101.325 KPa / 760 mmHg
524 mmHg = 69.8609 KPa
Thus, 524 mmHg is equivalent to 69.8609 KPa.
Which statement about atoms is not correct
Answer:
Atoms are not able to exist independently.
Explanation:
Brainliest pls
3. Theoretically how many grams of magnesium is required to produce to 5.0 g of
Magnesium oxide?
Answer:
3grams
Explanation:
The reaction for the production of Magnesium dioxide will be
Mg + O2 → MgO
we have 5g of MgO (molar mass 40g)
no of moles of MgO = 5/40 = 0.125
Using unitary method we have
1 mole of Mg require 1 mole of MgO
0.125 Mole of MgO = 0.125mole of Mg
n = given mass /molar mass
0.125 = mass / molar mass
mass = 0.125* 24 = 3grams
in which beakers are the particles moving the most slowly
Answer:
A and D
Explanation:
11. Which branch of chemistry is concerned with the behavior of substances?
Biochemistry
Physical
Inorganic
O Organic
Answer:
Physical
Explanation:
How many atoms are in 1.4 mol of phosphorus trifluoride (PF3)?
Number of atoms in 1.4 mol of Phosphorus trifluoride (PF₃) : 8.428 x 10²³
Further explanationThe mole is the number of particles(molecules, atoms, ions) contained in a substance
1 mol = 6.02.10²³ particles
Can be formulated
N=n x No
N = number of particles
n = mol
No = Avogadro's = 6.02.10²³
1.4 mol of Phosphorus trifluoride (PF₃), number of atoms :
[tex]\tt N=1.4\times 6.02\times 10^{23}\\\\N=8.428\times 10^{23}[/tex]
if you a boy answer this question because............. idk
Answer:
no
Explanation:
A reaction has a theoretical yield of 124.3 g SF6, but only 113.7 g SF6 are obtained in the lab, what is the percent yield of SF6 for this reaction?
Answer:
Percent yield = 91%
Explanation:
Given data:
Theoretical yield of SF₆ = 124.3 G
Actual yield of SF₆ = 113.7 g
Percent yield of SF₆ = ?
Solution:
Formula:
Percent yield = (actual yield / theoretical yield)× 100
By putting values,
Percent yield = (113.7 g/ 124.3 g) × 100
Percent yield = 0.91 × 100
Percent yield = 91%
Which of the following are not decomposition reactions?
A. KCIO3 → KCI + O2
B. CaO + H2O + Ca(OH)2
C. 2Mg + O2 + 2Mgo
O D. NaOH + Na2O + H2O
The following are not decomposition reactions :
B. CaO + H2O ⇒ Ca(OH)2
C. 2Mg + O2 ⇒ 2MgO
Further explanationThe decomposition reaction is a chemical reaction shows the decomposition of a compound into its constituent elements or compounds
General formula :
AB⇒A+B
A. KCIO3 → KCI + O2
decomposition reaction
B. CaO + H2O ⇒ Ca(OH)2
synthesis/combination reaction
C. 2Mg + O2 ⇒ 2MgO
synthesis/combination reaction
D. 2NaOH ⇒ Na2O + H2O
decomposition reaction
how many atoms does sodium hypochlorite have?
Answer:
NaClO; the contained elements ae Na, Cl, O.
at what pressure does water boil at room temperature 25 ec
Answer: approximately 100 degrees Celsius
A solid substance was tested in the laboratory. The test results are listed below: 1.) dissolves in water 2.) is an electrolyte 3.) melts at a high temperature. Based on these results, the solid substance could be
Answer:
NaCl may be i guess so....
Balance the following equation.
C3H8(g) + O2(g) → CO2(g) + H2O(l)
Explanation:
C3H8+5O2----->3CO2+4H2O
Answer:
C3H8+5O2----->3CO2+4H2O
Explanation:
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Answer:
I play!!!!!!!!
Explanation:
DO NOT TROLL AND ANSWER WITHOUT THE CORRECT ANSWER! I EXPECT THE CORRECT ANSWER NOT A TROLL SO IF YOU DONT HAVE THE ANSWER DO NOT WRITE ANYTHING. IF YOU TROLL I WILL REPORT YOU
A weather balloon with a volume of 3.40774 L
is released from Earth’s surface at sea level.
What volume will the balloon occupy at an
altitude of 20.0 km, where the air pressure is
10 kPa?
Answer in units of L.
Answer: The volume occupied at an altitude of 20.0 km is 34.5289 L
Explanation:
Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
[tex]P\propto \frac{1}{V}[/tex] (At constant temperature and number of moles)
where,
[tex]P_1[/tex] = initial pressure of gas = 101.325 kPa ( sea level)
[tex]P_2[/tex] = final pressure of gas = 10 kPa
[tex]V_1[/tex] = initial volume of gas = 3.40774 L
[tex]V_2[/tex] = final volume of gas = ?
Now put all the given values in the above equation, we get the final pressure of gas.
[tex]101.325\times 3.40774=10\times V_2[/tex]
[tex]V_2=34.5289L[/tex]
Therefore, the volume occupied at an altitude of 20.0 km is 34.5289 L
why should motorist be careful when putting air in the motor car tyres
Answer:
They might explode. I am serious.
A 50.0 mL solution of Ba(OH)2 is combined with a 150 mL solution of 0.20 M HCl. If the resulting solution has a hydroxide ion concentration of 0.12 M, what was the concentration of Ba(OH)2 in the original solution?
Answer:
0.54M of Ba(OH)2
Explanation:
When Ba(OH)2 reacts with HCl, BaCl2 and H2O are produced as follows:
Ba(OH)2 + 2HCl → BaCl2 + 2H2O
The remanent hydroxide ion is because not all Ba(OH)2 reacts. Thus, we need to find moles of Ba(OH)2 that doesn't react and moles of Ba(OH)2 that reacts. The ratio between total moles and volume of the Ba(OH)2 solution = 0.050L is the molarity of the original solution
Moles of Ba(OH)2 that doesn't react:
50mL + 150mL = 0.200L * (0.12 mol OH- / L) = 0.024 moles OH-
2 moles of OH- are in 1 mole of Ba(OH)2:
0.024 moles OH- * (1mol Ba(OH)2 / 2 mol OH-) = 0.012 moles Ba(OH)2
Moles of Ba(OH)2 that react:
0.150L * (0.20mol / L) = 0.030 moles HCl
2 moles of HCl react per mole of Ba(OH)2:
0.030 moles HCl * (1mol Ba(OH)2 / 2 mol HCl) = 0.015 moles Ba(OH)2
Total moles:
0.012mol + 0.015mol = 0.027mol Ba(OH)2 in 50mL
0.027mol Ba(OH)2 / 0.0500L =
0.54M of Ba(OH)2
In the electron dot diagram for sulfur monoxide, each atom of sulfur will have how many unshared pairs of electrons an each atom of oxygen will have how many unshared pairs of electrons.
Answer:
Both sulphur and oxygen have two unshared pairs of electrons
Explanation:
The compound sulphur monoxide has the formula S=O. It is quite analogous to the diatomic molecules of the group 16 elements.
We must remember that each group element has six electrons in their outermost shell. Two of these are lone pairs leaving only two electrons available for bond formation.
These two electrons are used to form the sigma and pi bonds in S=O, leaving two lone pairs on each of sulphur and oxygen atoms.
Which of these lead (II) salts will dissolve to the greatest extent in water?
a. PbSO4, Ksp = 1.7x10^-8
b. PbI2, Ksp = 6.5x10^-9
c. PbCrO4, Ksp = 1.8x10^-14
d. PbS, Ksp = 2.5x10^-27
e. Pb3(AsO4)2, Ksp = 4.0x10^-36
Answer:
a. PbSO4, Ksp = 1.7x10^-8.
Explanation:
Hello!
In this case, since the solubility product indicates how likely a solid is able to ionize and consequently dissolve in water, we can infer that the larger the solubility product Ksp, the more ions are able dissolve in water; therefore the proper answer goes with the largest Ksp, which is a. PbSO4, Ksp = 1.7x10^-8 since the power goes closer to 1 than the other options.
Best regards!
Suppose you are titrating vinegar, which is an acetic acid solution, of unknown strength with sodium hydroxide according to the equation HC2H3O2 + NaOH → H2O + NaC2H3O2 If you require 32.17 mL of 0.116 M NaOH solution to titrate 10.0 mL of HC2H3O2 solution, what is the concentration of acetic acid in the vinegar? (M)
Answer:
0.373 M
Explanation:
The reaction is
HC₂H₃O₂ + NaOH → H₂O + NaC₂H₃O₂So 1 mol of acetic acid reacts with 1 mol of sodium hydroxide.
First we calculate the moles of NaOH:
0.116 M * 32.17 mL = 3.732 mmol NaOHBecause the stoichiometric ratio is 1:1 :
3.732 mmol NaOH = 3.732 mmol Acetic AcidNow we calculate the concentration of acetic acid, dividing the moles by the volume:
3.732 mmol Acetic Acid / 10.0 mL = 0.373 M32.17 mL of 0.116 M NaOH are required to titrate 10.0 mL of 0.373 M acetic acid.
What is titration?Titration is a common laboratory method of quantitative chemical analysis to determine the concentration of an identified analyte.
Let's consider the neutralization reaction between acetic acid and sodium hydroxide.
HC₂H₃O₂ + NaOH → H₂O + NaC₂H₃O₂
We can calculate the concentration of acetic acid in vinegar using the following expression.
Ca × Va = Cb × Vb
Ca = Cb × Vb / Va
Ca = 0.116 M × 32.17 mL / 10.0 mL = 0.373 M
where,
Ca is the concentration of the acid.Va is the volume of the acid.Cb is the concentration of the base.Vb is the volume of the base.32.17 mL of 0.116 M NaOH are required to titrate 10.0 mL of 0.373 M acetic acid.
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Draw the product or products that are obtained from the reaction of HCl with 1-butene and with 2-butene. Show the relative stereochemistry if the product has two or more stereocenters.
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What are three ways that we can identify when a chemical reaction is occurring?
1. 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g) a. Determine the volume (mL) of 15.0 M sulfuric acid needed to react with 45.0 g of aluminum to produce aluminum sulfate. b. Determine the % yield if 112 g of aluminum sulfate is produced under the above conditions.
Answer:
a. 167 mL b. 39.27 %
Explanation:
a. From the chemical equation. 2 mole of Al reacts with 3 mole H₂SO₄ to produce 1 mol Al₂(SO₄)₃.
Now, we calculate the number of moles of Al in 45.0 g Al.
We know number of moles, n = m/M where m = mass of Al = 45.0 g and M = molar mass of Al = 26.98 g/mol.
So n = 45.0 g/26.98 g/mol = 1.668 mol
Since 2 mole of Al reacts with 3 mole H₂SO₄, then 1.668 mole of Al reacts with x mole H₂SO₄. So, x = 3 × 1.668/2 mol = 2.5 mol
So, we have 2.5 mol H₂SO₄.
Now number of moles of H₂SO₄, n = CV where C = concentration of H₂SO₄ = 15.0 M = 15.0 mol/L and V = volume of H₂SO₄.
V = n/C
= 2.5 mol/15.0 mol/L
= 0.167 L
= 167 mL of 15.0 M H₂SO₄ reacts with 45.0 g Al to produce aluminum sulfate.
b. From the chemical reaction, 2 mol Al produces 1 mol Al₂(SO₄)₃
Therefore 1.668 mol Al will produce x mol Al₂(SO₄)₃. So, x = 1 mol × 1.668 mol/2 mol = 0.834 mol
So, we need to find the mass of 0.834 mol Al₂(SO₄)₃. Now molar mass Al₂(SO₄)₃ = 2 × 26.98 g/mol + 3 × 32 g/mol + 4 × 3 × 16 g/mol = 53.96 g/mol + 96 g/mol + 192 g/mol = 341.96 g/mol.
Also number of moles of Al₂(SO₄)₃, n = mass of Al₂(SO₄)₃,m/molar mass Al₂(SO₄)₃, M
n =m/M
So, m = nM = 0.834 mol × 341.96 g/mol = 285.2 g
% yield = Actual yield/theoretical yield × 100 %
Actual yield = 112 g, /theoretical yield = 285.2 g
So, % yield = 112 g/285.2 g × 100 %
= 0.3927 × 100 %
= 39.27 %
The volume (mL) of 15.0 M sulfuric acid needed to react with 45.0 g of aluminum is 166mL and % yield of the reaction is 39.46%.
How do we calculate moles?Moles of any substance will be calculated by using the below formula as:
n = W/M, where
W = given mass
M = molar mass
Given chemical reaction is :
2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g)
Moles of 45g of Al will be calculated as:
n = 45g / 27g/mol = 1.66 mole
From the stoichiometry of the reaction, it is clear that:
1.66 moles of Al = react with 3/2×1.66=2.49 moles of H₂SO₄
By using the formula of molarity we can calculate the volume of H₂SO₄ as:
M = n/V
V = (2.49) / (15) = 0.166L = 166mL
Again from the stoichiometry it is clear that:
1.66 moles of Al = produces 1/2×1.66= 0.83 moles of Al₂(SO₄)₃
Mass of 0.83 moles of Al₂(SO₄)₃ = (0.83mol)(341.96g/mol) = 283.82 g
Given actual yield of Al₂(SO₄)₃ = 112g
% yield will be calculated as:
Percent yield = (Actual yield/Theoretical yield) × 100
% yield = (112/283.82) × 100 = 39.46%
Hence required values are discussed above.
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Suppose you light a Bunsen burner and notice that the flame is very yellow and too short. To get a blue flame, you should ________further. To increase the size of the flame, you should _____________the _____________further.
Answer:
In the given case, to get a blue flame, one should open the air inlet further. To increase the size of the flame one should open the gas regulator further. When one opens the air inlet, more amount of oxygen goes within, and thus, one can get a more intense form of blue flame.
When one opens the gas regulator more concentration of gas goes and the larger the size of flame one gets.
To regulate a bunsen burner follow this procedure
To get a blue flame, you should rotate the wheel of the burner anticlockwise further to reduce the level of oxygen flow.
To increase the size of the flame, you should rotate the wheel clockwise further
The bunsen burner is an apparatus in the laboratory used to carry out experiment when heating of substance is required, hence it is used to add heat to substances, it has a wheel used to regulate the flow of air which is required for combustion, it has a hose which is used to tap gas to the burner
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