Answer:
At 468.57 atm pressure the gas volume would change to 28 L.
Explanation:
Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant.
This law can be expressed mathematically as:
P · V = k
where P is pressure, V is volume and k is a constant.
This formula can be used to determine the change in pressure or volume during an isothermal transformation (i.e. the temperature is constant) as follows:
P1 · V1 = P2 · V2
In this case,
P1= 410 atmV1= 32 LP2= ?V2= 28 LReplacing:
410 atm*32L= P2*28 L
Solving:
[tex]P2=\frac{410 atm*32 L}{28L}[/tex]
P2=468.57 atm
At 468.57 atm pressure the gas volume would change to 28 L.
When heated above 200°C, sodium bicarbonate decomposes quickly to form sodium
carbonate, water, and carbon dioxide. Find the AH for the reaction, and classify the reaction
as endothermic or exothermic.
2NaHCO3(s) → Na, CO3(s) + H20(1) + CO2(9)
21-948 kJ/mol) → (-1311 kJ/mol) + (-286 kJ/mol) + (-394 kJ/mol)
AH = {AH, (products) - EAH, (reactants)
Flats comments
AH =
AH =
Circle one: endothermic or exothermic
The AH for the reaction is 1043 kJ/mol and the reaction is endothermic.
Explanation:The reaction you provided:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
AH = -948 kJ/mol - (-1311 kJ/mol + -286 kJ/mol + -394 kJ/mol)
AH = -948 kJ/mol + 1311 kJ/mol + 286 kJ/mol + 394 kJ/mol
AH = 1043 kJ/mol
The value of AH is 1043 kJ/mol, which means the reaction is endothermic. This is because the AH value is positive, indicating that energy is absorbed from the surroundings during the reaction.
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To find the ΔH for the decomposition of sodium bicarbonate, we subtract the sum of the enthalpies of formation of the reactants from those of the products, which results in -95 kJ. This indicates the reaction is exothermic.
Explanation:To determine the enthalpy change (ΔH) for the decomposition of sodium bicarbonate (baking soda), we use the given enthalpies of formation for the reactants and products in the reaction:
2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(l)
The enthalpy (ΔH) for the reaction is calculated as follows:
ΔH = { ΔHf(products) - ΔHf(reactants) }
For this reaction, the enthalpy change is:
ΔH = [(-1311 kJ/mol for Na2CO3) + (-394 kJ/mol for CO2) + (-286 kJ/mol for H2O)] - [2 x (-948 kJ/mol for NaHCO3)]
ΔH = (-1311 - 394 - 286) - (2 x -948)
ΔH = -1991 + 1896
ΔH = -95 kJ
Since the enthalpy change is negative, we can classify this reaction as exothermic. An exothermic reaction is one that releases heat to the surroundings, as evidenced by the negative sign of ΔH.
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How many grams of silver chloride are produced from 5.0g of silver nitrate reacting with an excess of
barium chloride?
How many grams of barium chloride would actually be necessary to complete the reaction of the silver nitrate?
Final answer:
From 5.0 grams of silver nitrate reacting with barium chloride, 4.21 grams of silver chloride are produced. The exact amount of barium chloride is not calculated because it is in excess.
Explanation:
To determine how many grams of silver chloride are produced from the reaction of 5.0g of silver nitrate (AgNO3) with an excess of barium chloride (BaCl2), we use the stoichiometry of the balanced chemical equation. The balanced equation is:
AgNO3 (aq) + BaCl2 (aq) → AgCl (s) + Ba(NO3)2 (aq)
To calculate the mol product, we follow the equation:
(mol product) = (mol reactant) × (stoichiometric mole ratio)
For AgNO3 with a molar mass of 169.88 g/mol, we calculate the moles of reactant:
moles of AgNO3 = mass of AgNO3 / molar mass of AgNO3
= 5.0 g / 169.88 g/mol
= 0.0294 moles of AgNO3
For silver chloride (AgCl) with a molar mass of 143.32 g/mol, we calculate the mass of AgCl produced:
mass of AgCl = moles of AgCl x molar mass of AgCl
= 0.0294 moles × 143.32 g/mol
= 4.2094 grams of AgCl (rounded to 4.21 grams)
Regarding the amount of barium chloride needed, since it is in excess and not limiting the reaction, the exact amount needed is not calculated. However, if one wanted to calculate it, we would use the stoichiometry of the reaction based on the amount of silver nitrate, ensuring that barium chloride is in excess.
In a study of the formation of NOx air pollution, a chamber heated to 2200°C was filled with air (0.790 atm N₂, 0.210 atm O₂). What are the equilibrium partial pressures of N₂, O₂, and NO if [tex]K_p[/tex] = 0.0460 for the following reaction:
[tex]N_2(g)+O_2(g) \rightleftharpoons 2NO(g)[/tex]
Answer:
N₂ = 0.7515atm
O₂ = 0.1715atm
NO = 0.0770atm
Explanation:
For the reaction:
N₂(g) + O₂(g) ⇄ 2NO(g)
Where Kp is defined as:
[tex]Kp = \frac{P_{NO}^2}{P_{N_2}P_{O_2}}}[/tex]
Pressures in equilibrium are:
N₂ = 0.790atm - X
O₂ = 0.210atm - X
NO = 2X
Replacing in Kp:
0.0460 = [2X]² / [0.790atm - X] [0.210atm - X]
0.0460 = 4X² / 0.1659 - X + X²
0.0460X² - 0.0460X + 7.6314x10⁻³ = 4X²
-3.954X² - 0.0460X + 7.6314x10⁻³ = 0
Solving for X:
X = - 0.050 → False answer. There is no negative concentrations.
X = 0.0385 atm → Right answer.
Replacing for pressures in equilibrium:
N₂ = 0.790atm - X = 0.7515atm
O₂ = 0.210atm - X = 0.1715atm
NO = 2X = 0.0770atm
Answer:
partial pressure N2 = 0.7515 atm
partial pressure O2 = 0.1715 atm
partial pressure NO = 0.077 atm
Explanation:
Step 1: Data given
Temperature = 2200 °C
Pressure of N2 = 0.790 atm
Pressure of O2 = 0.210 atm
Kp = 0.0460
Step 2: The balanced equation
N2(g) + O2(g) ⇆ 2NO(g)
Step 3: The pressure at equilibrium
pN2 = 0.790 - X atm
pO2 = 0.210 - X atm
pNO = 2X
Step 4: Define Kp and the partial pressures
Kp = (pNO)² / (pO2 * pN2)
0.0460 = 4X² / (0.210 - X)(0.790 - X)
X = 0.0385
pN2 = 0.790 - 0.0385 = 0.7515 atm
pO2 = 0.210 - 0.0385 = 0.1715 atm
pNO = 2*0.0385 = 0.077 atm
A balloon is floating around outside your window. The temperature outside is -1 ∘C , and the air pressure is 0.700 atm . Your neighbor, who released the balloon, tells you that he filled it with 3.80 moles of gas. What is the volume of gas inside this balloon?
Answer:
[tex]\large \boxed{\text{121 L}}[/tex]
Explanation:
We can use the Ideal Gas Law.
pV = nRT
Data:
p = 0.700 atm
n = 3.80 mol
T = -1 °C
Calculations:
1. Convert the temperature to kelvins
T = (-1 + 273.15) K= 272.15 K
2. Calculate the volume
[tex]\begin{array}{rcl}pV &=& nRT\\\text{0.700 atm} \times V & = & \text{3.80 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{272.15 K}\\0.700V & = & \text{84.86 L}\\V & = & \textbf{121 L} \\\end{array}\\\text{The volume of the balloon is $\large \boxed{\textbf{121 L}}$}[/tex]
A catalyst:
A. Increases the energy of the activated complex
B. decreases the ∆H of the reaction
C. Decreases the energy of the reactants
D. Decreases the activation energy of the reaction
Answer:
D
Explanation:
well, just need to remember
Answer:
Catalyst decreases activation energy
Explanation:
Consider the attached diagram and note the annotation => top of transition diagram for catalyzed reaction is lower than uncatalyzed reaction.
Calculate the concentration of a solution prepared by adding 15.00 mL of 2.00 × 10 − 3 M K M n O 4 from a buret into a 50.00 mL volumetric flask, which is then filled to the 50.00 mL graduation mark with distilled water.
Answer:
The concentration of the solution is 6 [tex]10^{-4}[/tex] mol/dm3.
Explanation:
We have the relationship between concentration and volume to be:
[tex]c_{1} v_{1} = c_{2} v_{2}[/tex] where c1v1 and c2v2 are initial and final concentrations and volumes respectively.
NOTE: (volume should be in litres if concentration is in Molar, M. But seeing as you have 2 × 10-3, I assume you have converted from mL to L already.)
15mL × 0.002mol/dm3 = 50mL × [tex]x[/tex] mol/dm3
∴ [tex]x[/tex] = [tex]\frac{0.03}{50}[/tex]
= 0.0006 ≅ 6 ×[tex]10^{-4}[/tex]mol/dm3
Based on the data provided, the final concentration of the solution is 6 * 10⁻⁴ mol/dm3
What is concentration of a solution?Concentration of a solution is the amount of solute dissolved in a given volume of solution in litres.
The formula: C1V1 = C2V2 is used to calculate the concentration of solutions after dilution where
C1 and C2 are initial and final concentrations and V1 and V2 are the final and initial volumes respectively.Using C1V1 = C2V2:
15 × 0.002= 50 × C2
C2 = 6 * 10⁻⁴ mol/dm3
Therefore, the final concentration of the solution is 6 * 10⁻⁴ mol/dm3.
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I need help with my chemistry homework
Answer:
false, false, true
Calculate the mass of carbon dioxide produced by the combustion of 571.3 g of octane.
Answer:
The answer to your question is 1795.5 grams of CO₂
Explanation:
Data
mass of carbon dioxide = ?
mass of Octane = 571.3 g
Process
1.- Write the balanced chemical reaction
C₈H₁₆ + 12O₂ ⇒ 8CO₂ + 8H₂O
2.- Calculate the molar mass of octane and carbon dioxide
Octane = (12 x 8) + (16 x 1) =96 + 16 = 112 g
Carbon dioxide = 8[ 12 + 2(16)] = 352 g
3.- Use proportions to find the mass of carbon dioxide
112 g of Octane ---------------- 352 g of carbon dioxide
571.3 g of Octane ------------- x
x = (571.3 x 352) / 112
x = 201097.6 / 112
x = 1795.5 grams of CO₂
The oxidation of copper(I) oxide, Cu 2 O ( s ) , to copper(II) oxide, CuO ( s ) , is an exothermic process. 2 Cu 2 O ( s ) + O 2 ( g ) ⟶ 4 CuO ( s ) Δ H ∘ rxn = − 292.0 kJ mol Calculate the energy released as heat when 9.94 g Cu 2 O ( s ) undergo oxidation at constant pressure.
Answer:
The energy released as heat when 9.94 g Cu 2 O ( s ) undergo oxidation at constant pressure is -10.142 kJ
Explanation:
Here we have
2Cu₂O ( s ) + O₂ ( g ) ⟶ 4 CuO ( s ) Δ H ∘ rxn = − 292.0 kJ mol
In the above reaction, 2 Moles of Cu₂O (copper (I) oxide) react with one mole of O₂ to produce 4 moles of CuO, with the release of − 292.0 kJ/mol of energy
Therefore,
1 Moles of Cu₂O (copper (I) oxide) react with 0.5 mole of O₂ to produce 2 moles of CuO, with the release of − 146.0 kJ of energy
We have 9.94 g of Cu₂O with molar mass given as 143.09 g/mol
Hence the number of moles in 9.94 g of Cu₂O is given as
9.94/143.09 = 6.95 × 10⁻² moles of Cu₂O
6.95 × 10⁻² moles of Cu₂O will therefore produce 6.95 × 10⁻² × − 146.0 kJ mol or -10.142 kJ.
Final answer:
The energy released as heat when 9.94 g of copper(I) oxide undergoes oxidation is calculated by determining the moles of Cu2O from its mass and using the stoichiometry of the reaction along with its enthalpy change to find the energy change. The result is approximately -10.14 kJ.
Explanation:
The problem involves calculating the energy released as heat during the oxidation of copper(I) oxide (Cu2O) to copper(II) oxide (CuO), given the mass of copper(I) oxide and the enthalpy change of the reaction. The reaction is 2 Cu2O(s) + O2(g) → 4 CuO(s) with an enthalpy change (ΔH°rxn) of -292.0 kJ/mol.
To find the energy released, we first need to find the number of moles of Cu2O. The molar mass of Cu2O is 143.09 g/mol (2*63.55 + 15.99). Using the given mass of Cu2O (9.94 g), we have:
moles of Cu2O = mass (g) / molar mass (g/mol) = 9.94 g / 143.09 g/mol = 0.0694 mol
The reaction stoichiometry shows that 2 moles of Cu2O yield -292.0 kJ, so we need to calculate the energy for 0.0694 moles:
Energy released = 0.0694 mol * (-292.0 kJ/mol) / 2 = -10.14 kJ
Therefore, the energy released as heat when 9.94 g of Cu2O undergoes oxidation at constant pressure is approximately -10.14 kJ.
Given: Pb(s) + PbO2(s) + 2H2SO4(l) → 2PbSO4(s) + 2H2O(l); ΔH° = –509.2 kJ SO3(g) + H2O(l) → H2SO4(l); ΔH° = –130. kJ determine ΔH° for the following thermochemical equation. Pb(s) + PbO2(s) + 2SO3(g) → 2PbSO4(s) Group of answer choices
ΔH° = -769.2 kJ is required.
Explanation:
Given equations are:
Pb(s) + PbO₂(s) + 2 H₂SO₄(l) → 2 PbSO₄(s) + 2H₂O(l); ΔH° = –509.2 kJ ------1
SO₃(g) + H₂O(l) → H₂SO₄(l); ΔH° = –130. kJ ----2
From the above 2 equations, by adding or subtracting or multiplying or dividing the required amount to get the final equation by means of Hess's law.
Multiplying eq. 2 by 2 we will get,
2 SO₃(g) + 2 H₂O(l) → 2 H₂SO₄(l) ; ΔH° = -260 kJ ----3
Adding it to eq. 1 we will get,
Pb(s) + PbO₂(s) + 2 H₂SO₄(l) → 2 PbSO₄(s) + 2H₂O(l); ΔH° = –509.2 kJ ------1
2 H₂SO₄ and 2 H₂O gets cancelled since they are on opposite sides, and the ΔH° values are added to get the ΔH° value of the required equation as,
Pb(s) + PbO₂(s) + 2 SO₃(g) → 2 PbSO₄(s)
ΔH° = -509.2 - 260 = -769.2 kJ
Is a volumetric flask calibrated to contain or to dispense?
Volumetric flask are calibrated to contain.
A volumetric flask is calibrated to contain a specific volume of solution when filled to its calibration mark, while pipettes are designed to deliver precise volumes of liquids.
Explanation:A volumetric flask is calibrated 'to contain' (T. C.) a specific volume of solution rather than to dispense it. This means when the flask is filled to its calibration mark, it is accurate to a specific volume such as 10.00 mL ± 0.02 mL for a 10-mL volumetric flask or 250.0 mL ± 0.12 mL for a 250-mL volumetric flask. Unlike volumetric flasks, pipettes like volumetric and graduated pipettes are designed to deliver a known volume of liquid, either a single volume in the case of volumetric pipettes or variable volumes for graduated pipettes. To achieve accurate measurements, it is essential that both pipets and volumetric flasks are clean because any residue can affect the volume of liquids either delivered or contained.
How many moles of C5H12 are there in 362.8 grams of the compound
Answer: 5.039moles
Explanation:
No. of moles=mass/molar mass
= 362.7/72= 5.039moles
Molar mass of C5H12= 72
Final answer:
To find the number of moles of C5H12 in 362.8 grams, divide the mass by the molar mass of C5H12 (72.0 g/mol), resulting in 5.04 moles.
Explanation:
To calculate how many moles of C5H12 are in 362.8 grams of the compound, you first need to determine the molar mass of C5H12. The molar mass of a compound is calculated by summing the molar masses of all atoms in the molecule. For C5H12, the calculation is as follows:
Carbon (C) has a molar mass of 12.0 g/mol, and there are 5 carbon atoms, which contributes 5 * 12.0 = 60.0 g/mol.Hydrogen (H) has a molar mass of 1.0 g/mol, and there are 12 hydrogen atoms, which contributes 12 * 1.0 = 12.0 g/mol.
Therefore, the molar mass of C5H12 is 60.0 g/mol (carbon) + 12.0 g/mol (hydrogen) = 72.0 g/mol.
To find the number of moles, divide the mass of the compound by its molar mass:
Moles of C5H12 = Mass of C5H12 / Molar mass of C5H12 = 362.8 g / 72.0 g/mol = 5.04 moles.
So, there are 5.04 moles of C5H12 in 362.8 grams of the compound.
PLEASE HELP DUE TODAY 55 POINTS
The volume of a gas is 27.5 mL at 22.0°C and 0.974 atm. What will the volume be at 15.0°C
and 0.993 atm?
(5 Points)
22.1 mL
26.3 mL
2.2 mL
Answer : The final volume of gas will be, 26.3 mL
Explanation :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 0.974 atm
[tex]P_2[/tex] = final pressure of gas = 0.993 atm
[tex]V_1[/tex] = initial volume of gas = 27.5 mL
[tex]V_2[/tex] = final volume of gas = ?
[tex]T_1[/tex] = initial temperature of gas = [tex]22.0^oC=273+22.0=295K[/tex]
[tex]T_2[/tex] = final temperature of gas = [tex]15.0^oC=273+15.0=288K[/tex]
Now put all the given values in the above equation, we get:
[tex]\frac{0.974 atm\times 27.5 mL}{295K}=\frac{0.993 atm\times V_2}{288K}[/tex]
[tex]V_2=26.3mL[/tex]
Therefore, the final volume of gas will be, 26.3 mL
Answer:
volume of gas will be, 26.3 mL
Explanation:
2 Al + 3 H2SO4 --> Al2(SO4)3 + 3 H2 If you have 7.6 moles of Al, then how many moles of H2SO4 will be needed to react completely with it?
Answer:
11.4 moles of H₂SO₄ are needed to completely react the 7.6 moles of Al
Explanation:
The equation indicates that 2 moles of aluminum react to 3 moles of sulfuric acid in order to produce 1 mol of aluminum sulfate and 3 moles of hydrogen gas.
The reaction is: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂
This question can be solved with an easy rule of three. Ratio in the reaciton is 2:3, so we propose:
2 moles of Al react with 3 moles of sulfuric acid
Then, 7.6 moles of Al will react with 11.4 moles of H₂SO₄
How many atoms are there in 1.50 moles carbon?
Answer:
9.03*10^23 atoms of C
Explanation:
1 mol of any substance contains 6.02*10^23 particle of this substance.
1 mol C --- 6.02*10^23 atoms of C
1.50 mol C --- x atoms of C
x = 1.50*6.02*10^23 = 9.03*10^23 atoms of C
who is the first scientist that discovered DNA?
Answer:
James Watson
Explanation:
Which units of pressure are needed if you are going to use 0.0821 as your ideal
gas constant?
A.psi
B.Pascals
C.torr
D.atmospheres
E.mmHg
Answer:
D. Atmospheres
Explanation:
Calculate the equilibrium constant for the decomposition of water 2h2o(l) 2h2(g) + o2(g) at 25°c, given that g°f (h2o(l)) = –237.2 kj/mol.
The equilibrium constant for the decomposition of water is approximately 1.01 × 10^-13.
Explanation:The equilibrium constant (Kc) for the decomposition of water can be calculated using the equation: Kc = [H2]2[O2]/[H2O]2.
Given that ΔG°f(H2O(l)) = -237.2 kJ/mol, we can use the equation ΔG° = -RTlnK to find the equilibrium constant. R is the ideal gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin (25 °C + 273.15 = 298.15 K). Plugging in the values, we can solve for K.
ΔG° = -RTlnK
-237.2 kJ/mol = -(8.314 J/(mol·K) × 298.15 K) × lnK
lnK = -237.2 kJ/mol ÷ (8.314 J/(mol·K) × 298.15 K)
lnK ≈ -29.155
K ≈ e-29.155
K ≈ 1.01 × 10-13
Therefore, the equilibrium constant for the decomposition of water at 25 °C is approximately 1.01 × 10-13.
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What is the maximum number of moles of H2O that can be produced from the reaction of 5.6 mol H2 and 4.7 mol O2?
H2 + O2
What is the limiting reactant?
What is the excess reactant?
Answer:
Limiting reactant is H2.
Excess reactant is O2.
Maximum 5.6 mol of H2O can be produced.
Explanation:
2H2 + O2 ----> 2 H2O
from reaction 2 mol 1 mol
given 5.6mol 4.7mol
calculated 5.6mol 2.8 mol
We can see that for 5.6 mol H2 only 2.8 mol O2 needed, but we have 4.7 mol O2 given, so we have excess of O2.
Then limiting reactant is H2.
Excess reactant is O2.
2H2 + O2 ----> 2 H2O
from reaction 2 mol 2 mol
given 5.6 mol x mol = 5.6 mol
Final answer:
The maximum number of moles of H₂O that can be produced from 5.6 mol H₂ and 4.7 mol O₂ is 5.6 mol H₂O, with hydrogen (H₂) as the limiting reactant and oxygen (O₂) as the excess reactant.
Explanation:
To answer the question of the maximum number of moles of H₂O that can be produced from the reaction of 5.6 mol H₂ and 4.7 mol O₂, we must first look at the balanced chemical equation for the reaction which is 2H₂ + O₂ ightarrow 2H₂O. From this equation, we can see that every 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.
Now, let's see if we have enough of each reactant:
Hydrogen: 5.6 moles H₂ is available
Oxygen: 4.7 moles O₂ is available
According to the stoichiometry of the equation, oxygen will run out first since 4.7 moles of oxygen can react completely with (4.7 times 2) = 9.4 moles of hydrogen. But only 5.6 moles of hydrogen are available, which is less than 9.4 moles, so actually, hydrogen will limit the reaction.
Thus, the limiting reactant is hydrogen (H₂) and the excess reactant is oxygen (O₂). The maximum number o2f moles of water that can be produced is therefore equal to the moles of hydrogen available, which is 5.6 moles of H₂, resulting in 5.6 moles of H₂O being produced.
"how many liters of a 0.2 m naoh solution are needed in order to have 0.4 moles of naoh?"
Answer:
2
Explanation:
well since the question has given you the concentration of the solution 0.2mol/L and the wanted amount (moles) of naoh(0.4mol) you are able put this into the formula n=cV; where n is the moles of naoh the solution, c is the concentraton of the solution and V is the volume of the solution in litres.
therefore the for the solution to have 0.4 moles of naoh you put it into the formula, giving you:
0.4 = 0.2V
V = 2
V = 2 litres
Final answer:
To obtain 0.4 moles of NaOH from a 0.2 M NaOH solution, you would need to measure out 2 liters of the solution.
Explanation:
To calculate the volume of a 0.2 M NaOH solution needed to have 0.4 moles of NaOH, we can use the molarity equation, which is:
Molarity (M) = Moles of solute / Volume of solution in liters (L)
We can rearrange this equation to solve for the volume:
Volume of solution (L) = Moles of solute / Molarity (M)
Substituting the given values:
Volume of solution (L) = 0.4 moles NaOH / 0.2 M NaOH
Volume of solution (L) = 2 liters
Therefore, you would need 2 liters of a 0.2 M NaOH solution to have 0.4 moles of NaOH.
A kettle has a store of 400J of thermal energy in the heating filament. 345J is transferred by heating to thermal energy store of water. How much energy is dissipated as sound?
Answer:
55J
Explanation:
∆E = 400-345 = 55J
Answer:
Energy dissipated as sound is 55 J
Explanation:
Here we have the principle of conservation of energy which states that energy can neither be created nor destroyed but can be transformed from one form to another
Where there is an initial 400 J of thermal energy in the heating element of the electric kettle we have;
Total available energy = 400 J
The energy (heat) transferred to the water is given as 345 J
The heat dissipated as sound of the kettle during heating is then found as follows;
Total available energy = Heat transferred to water + Energy dissipated as sound
400 = 345 + Energy dissipated as sound
∴ Energy dissipated as sound = 400 - 345 = 55 J.
3.00 L of Ch4 is known to contain 0.650 moles at a certain temperature and pressure if the volume of a chamber of CH4 increased from 7.00L to 8.20 L how many moles CH4 were added assume temperature and pressure stayed constant
Final answer:
When the volume of a chamber containing CH₄ increased from 7.00L to 8.20L, 0.260 moles of CH₄ were added, assuming constant temperature and pressure.
Explanation:
To determine how many moles of CH₄ (methane) were added when the volume of the chamber increased from 7.00L to 8.20L, we can use the molar volume concept under the assumption that temperature and pressure remain constant, therefore following Avogadro's Law. According to Avogadro's Law, equal volumes of gases at the same temperature and pressure contain an equal number of moles. You can calculate the number of moles in the new volume by setting up a proportion based on the known conditions (3.00 L contains 0.650 moles) and then solving for the number of moles in the new volume of 8.20 L.
First, determine the number of moles in the 7.00 L chamber:
moles in 3.00 L / 3.00 L = moles in 7.00 L / 7.00 L
0.650 moles / 3.00 L = x moles / 7.00 L
x = (0.650 moles / 3.00 L) × 7.00 L
x = 1.517 moles in 7.00 L
Now, calculate the number of moles in the increased volume of 8.20 L:
moles in 3.00 L / 3.00 L = moles in 8.20 L / 8.20 L
0.650 moles / 3.00 L = y moles / 8.20 L
y = (0.650 moles / 3.00 L) × 8.20 L
y = 1.777 moles in 8.20 L
The number of moles added is the difference between the moles in 8.20 L and the moles in 7.00 L:
moles added = 1.777 moles - 1.517 moles
moles added = 0.260 moles
Therefore, when the volume of the CH₄ chamber increased from 7.00L to 8.20L, 0.260 moles of CH₄ were added.
Which compound reacts with an acid to form a salt and water?
Answer:
Base
Explanation:
acid + base = salt + water
The compound reacts with an acid to form a salt and water is base.
What is a base?A base is a substances that is slippery to touch,corrosive and sour taste which react with acid to give salt and water. It turns red lithmus paper to blue. It is the degree of hydroxide ion in a solution.
The compound reacts with an acid to form a salt and water is base.
Learn more about base below.
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At constant temperature, if a gas occupies 312 mL at a pressure of 1.60 atm, what pressure is necessary for this gas to occupy a volume of 500 mL?
Answer:
P₂ = 1.0 atm
Explanation:
Boyles Law problem => P ∝ 1/V at constant temperature (T).
Empirical equation
P ∝ 1/V => P = k(1/V) => k = P·V => for comparing two different case conditions, k₁ = k₂ => P₁V₁ = P₂V₂
Given
P₁ = 1.6 atm
V₁ = 312 ml
P₂ = ?
V₂ = 500 ml
P₁V₁ = P₂V₂ => P₂ = P₁V₁/V₂ =1.6 atm x 312 ml / 500ml = 1.0 atm
What types of elements are useful for dating materials?
A.
metals
B.
nonmetals
C.
radioactive
D.
noble gases
Answer:C
Explanation:
Answer:
Radioactive
Explanation:
Just did this on study island
What’s the oxidation number for NH3?
Answer: 0
Explanation:
The oxidation number for nitrogen in NH3 (ammonia) is -3, determined by setting up the equation x + 3(+1) = 0, where x represents the oxidation number of nitrogen.
Explanation:The oxidation number for NH3 (ammonia) can be found by considering the usual oxidation states of nitrogen and hydrogen. Hydrogen typically has an oxidation number of +1, except when it forms hydride compounds with metals. Since ammonia consists of one nitrogen atom and three hydrogen atoms, and since each hydrogen has an oxidation number of +1, the total oxidation number contributed by the hydrogen atoms is +3 (3 x +1).
To find the oxidation number of nitrogen in NH3, let the oxidation number be represented as x. The sum of the oxidation numbers in a neutral compound is zero. Therefore, if we have x as the oxidation number of nitrogen and +3 from the hydrogen atoms, we can set up the equation x + 3(+1) = 0 to solve for x. Simplifying, we get x = -3. Thus, the oxidation number of nitrogen in NH3 is -3.
Following this approach, we can understand how oxidation numbers reflect the degree of electron transfer between atoms in a chemical compound or during a chemical reaction, such as NH3 reacting with O2 to form N2 and H2O. The reaction 4 NH3 + 3 O2 → 2 N2 + 6 H2O illustrates this electron transfer process and the involvement of oxidation states in balancing chemical equations.
Which statement describes all solids?
They contain loosely packed atoms.
They have a definite shape and volume.
They have a smooth, rigid surface.
They flow with resistance.
Option B is correct. The statement that describes all solids is that they have a definite shape and volume
States of matter are one of the ways in which matter exists. Matter can exist as a solid, liquid, gas, and plasma.
The states of matter have different characteristics. Some of the properties of solids are:
Solids are known for their definite shape.Higher densityStrong intermolecular forceThe liquid contains loosely packed atoms eliminating the first option. Based on the explanations above, we can conclude that solids have a definite shape and volume.
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Al poner en contacto vinagre y bicarbonato se desprende un gas, dióxido de carbono, y se forma una sustancia llamada acetato de sodio y agua, cuáles son los reactivos de la reaccion
Answer:
the rectives are just not doing anything so just do sodium
Explanation:
Significant figures chemistry
I believe that leading zeros, and tail zeros without a decimal point are not significant
If a gas at 25oC occupies 3.60 L at a pressure of 2.50 atm, what will be its volume at a pressure of 1.00 atm if the temperature does NOT change?
Answer:
9L
Explanation:
Given parameters:
Initial volume V₁ = 3.6L
Initial pressure P₁ = 2.5atm
Final pressure P₂ = 1atm
Unknown:
Final volume V₂ = ?
Condition: constant temperature = 25°C
Solution:
This problem compares the volume and pressure of a gas at constant temperature.
This is highly synonymous to the postulate of Boyle's law. It states that "the volume of a fixed mass of gas is inversely proportional to the pressure provided that temperature is constant".
Mathematically;
P₁V₁ = P₂V₂
where P and V are pressure and volume
1 and 2 are initial and final states
Input the parameters and solve for V₂;
2.5 x 3.6 = 1 x V₂
V₂ = 9L