Answer:
2 mol H₂O
Explanation:
With the reaction,
2H₂(g) + O₂(g) → 2 H₂O(g)1.55 moles of O₂ would react completely with ( 2*1.55 ) 3.1 moles of H₂. There are not as many moles of H₂, thus H₂ is the limiting reactant.
Now we calculate the moles of H₂O produced, starting from the moles of limiting reactant:
2.00 mol H₂ * [tex]\frac{2molH_2O}{2mol H_2}[/tex] = 2 mol H₂OThe number of moles of H₂O that could be produced in the reaction is 2.00 moles
From the question,
We are to determine the number of moles of H₂O that could be produced.
The given balanced chemical equation is
2 H₂(g) + O₂(g) → 2 H₂O(g)
This means,
2 moles of H₂(g) reacts with 1 mole of O₂(g) to produce 2 moles of H₂O
From the given information
Number of moles of H₂ present = 2.00 moles
and
Number of moles of O₂ present = 1.55 moles
Since,
2 moles of H₂(g) reacts with 1 mole of O₂(g) to produce 2 moles of H₂O
Then,
The 2.00 moles of H₂(g) will react with 1.00 mole of O₂(g) to produce 2.00 moles of H₂O
∴ 2.00 moles of H₂O will be produced during the reaction
(NOTE: Only 1.00 mole of the O₂(g) will react, meaning O₂(g) is the excess reactant)
Hence, the number of moles of H₂O that could be produced in the reaction is 2.00 moles
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A system absorbs 191 kJkJ of heat and the surroundings do 120 kJkJ of work on the system. Part A What is the change in internal energy of the system?
Answer: the change in internal energy of the system is 311 kJ
Explanation:
Given that;
A system absorbs a heat dQ = 191 kJ
And the the surroundings do work on the system dW = 120 kJ
change in internal energy of the system = ?
We take a look at the first law of thermodynamic
Initially the system absorbs energy ( dQ =1 91 kJ) as heat.
Then the surroundings do work (dW = 120kJ) on the system,
Hence the system gains 120 kJ more.
The total energy change is = heat + work
dU = dQ + dW
dU = 191 + 120
dU = 307 kJ
change in internal energy = 311 kJ
Therefore the change in internal energy of the system is 311 kJ
11. Which branch of chemistry is concerned with the behavior of substances?
Biochemistry
Physical
Inorganic
O Organic
Answer:
Physical
Explanation:
A gas has a pressure of 645 Torr at 128 ºC. What is the temperature (ºC) of the pressure increases to 824 Torr at constant volume and moles?
Answer:
512 K
Explanation:
Initial pressure P1 = 645 Torr
Initial Temperature T1 = 128 ºC + 273 = 401 K (Upon converting to kelvin temperature)
Final Pressure P2 = 824 Torr
Final Temperature T2 = ?
The relationship between the variables is given as;
P1 / T1 = P2 / T2
Making T2 subject if formula we have;
T2 = P2T1 / P1
Inserting the values we have;
T2 = 824 * 401 / 645 = 512 K
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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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
What are three ways that we can identify when a chemical reaction is occurring?
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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
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.
A 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
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Explanation:
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
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]
A) A solution of c6h2o4 in c2h5oh has a boiling point of 80.38 Celsius. what is the molality of the solution ?
B) what is the approximation of concentration of dissolved solution in blood, if the osmotic pressure is 7.65 atm at 37Celsius.
a. molality-m=1.623
b. molarity-M=0.3
Further explanationa. Boiling point
[tex]\tt \Delta Tb=Kb\times m[/tex]
Kb for ethanol=1.22 °C/m
Boiling point for ethanol = 78.4 °C
ΔTb=80.38-78.4=1.98
[tex]\tt 1.98=1.22\times m\\\\m=\dfrac{1.98}{1.22}=1.623[/tex]
b. osmotic pressure
[tex]\tt \pi =MRT[/tex]
π = 7.65 atm
R = 0.08205 L/atm K
T = 37 + 273 = 310 K
[tex]\tt M=\dfrac{\pi}{RT}=\dfrac{7.65}{0.08205\times 310}=0.3~M[/tex]
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.
1. What is the flame color of sodium?
yellow-orange
purple
green
blue-green
Orange is the color of Sodium
If 75J of heat are applied to 8.4 L of a gas at 732 mmHg and 298K, what is the final temperature, in K, of the gas? Cp for an ideal gas is 20.79 J/mol*K.
Answer:
309 K
Explanation:
Step 1: Convert the pressure to atm
We will use the conversion factor 1 atm = 760 mmHg.
732 mmHg × 1 atm/760 mmHg = 0.963 atm
Step 2: Calculate the moles (n) of the ideal gas
We will use the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 0.963 atm × 8.4 L/0.0821 atm.L/mol.K × 298 K
n = 0.33 mol
Step 3: Calculate the temperature change
We will use the following expression.
Q = n × Cp × ΔT
ΔT = Q/n × Cp
ΔT = 75 J/0.33 mol × 20.79 J/mol.K
ΔT = 11 K
Step 4: Calculate the final temperature
T = 298 K + 11 K = 309 K
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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Use what you’ve learned about Lewis structures and formal charges to predict which of the following sulfur-containing molecule(s) would be least likely to exist.
SO2
H2S2
SCl2
HS
HSOH
Answer:
its DDDDDDDDDDDDDDDDDD
The structure that is least likely to exist is HS.
Sulfur is an element in group 16. Sulfur is a divalent element that has two lone pairs of electrons. This means that sulfur forms compounds in which it is bonded to two atoms or groups.
All the molecules listed can exist because they consists of structures in which sulfur is bonded to two atoms or groups. The only structure that can not exist is HS because it does not satisfy the valency of sulfur.
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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:
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
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.
at what pressure does water boil at room temperature 25 ec
Answer: approximately 100 degrees Celsius
Which statement about atoms is not correct
Answer:
Atoms are not able to exist independently.
Explanation:
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why should motorist be careful when putting air in the motor car tyres
Answer:
They might explode. I am serious.
A patient had for dinner a 3 oz steak, 3 oz salmon, 1 medium apple, and 1 cup of nonfat milk. Using the
chart below, how many total calories did they consume?
Protein
awbohydrate
15
Fal)
0
0
1
126
0
14
22
D
3
4
Food
Apple, 1 medium
Banana, medium
Beel ground, 3oz
Broccoli 304
Carrots, cup
Chicken no skin, 30
Egg, farge
Mik nanta 1 cup
Potato, baked
Salmon 3 02
11
9
2
20
0
3
16
6
10
12
19
10
10
3
23
0
5
16
22
19
Steak 3 0
10
Answer:
1
Explanation:
because there barely eating and there eating protien and healthy foods
how many atoms does sodium hypochlorite have?
Answer:
NaClO; the contained elements ae Na, Cl, O.
How many π electrons are there in adenine? State which atoms contribute one electron to the π MOs and which atoms contribute two electrons?
Answer:
12 π electrons.
Explanation:
The structure of adenine can be seen below.
From the diagram, the total number of electrons in adenine is 70 electrons where 20 of them are core electrons.
Also from this same structure, we will notice that we have a total number of 12 π electrons.
i.e.
Each one of these five atoms described takes part in the π electron. These are N-1, N-3, N-7 which comprise of 2σ bonds, and also each of three N contributes one π electron.
Similarly, N-9 and N-10 contain 3σ bonds; Hence, each of them donates two electrons.
Thus;
5C = 5π electrons
3N = 3π electrons
2N = 4π electrons
= 12π electrons
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%
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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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.
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