How much energy is needed to vaporize 75.0 g of diethyl ether (c4h10o) at its boiling point (34.6°c), given that δhvap of diethyl ether = 26.5 kj/mol?

The compound MnCL2 called

Manganese cloride

Answer:

Explanation:

Cloruro de manganeso

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

Answer:

James Watson

Explanation:

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

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

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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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

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.

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]

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.

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.

Answer:

17.08 moles of manganese dioxide

Explanation:

From the balanced reaction equation;

Zn + 2 MnO2 + 1 H2O → Zn(OH)2 + Mn2O3 (notice that I did not put 1 as a stoichiometric coefficient. It is expected that any specie written without a coefficient should have a coefficient of 1)

It is clear from the reaction equation that 2 moles of manganese dioxide produced 1 mole of zinc II hydroxide

Hence x moles of manganese dioxide will produce 8.54 moles of zinc II hydroxide

x= 2× 8.54/1

x= 17.08 moles of manganese dioxide

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

The AH for the reaction is 1043 kJ/mol and the reaction is endothermic.

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.

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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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

Answer:

D. Atmospheres

Explanation:

Answer:C

Explanation:

Answer:

Radioactive

Explanation:

Just did this on study island

Answer:

Base

Explanation:

acid + base = salt + water

The compound reacts with an acid to form a salt and water is 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.

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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.

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

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.

The equilibrium constant for the decomposition of water is approximately 1.01 × 10^-13.

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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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

Answer:hope we can be friends

can i please get brainliest

Although phlorizin inhibition of Na+-glucose cotransport occurs within a few seconds, 3H-phlorizin binding to the sodium-coupled glucose transport protein(s) requires several minutes to reach equilibrium (the fast-acting slow-binding paradigm). Using kinetic models of arbitrary dimension that can be reduced to a two-state diagram according to Cha’s formalism, we show that three basic mechanisms of inhibitor binding can be identified whereby the inhibitor binding step either (A) represents, (B) precedes, or (C) follows the rate-limiting step in a binding reaction. We demonstrate that each of mechanisms A–C is associated with a set of unique kinetic properties, and that the time scale over which one may expect to observe mechanism C is conditioned by the turnover number of the catalytic cycle. In contrast, mechanisms A and B may be relevant to either fast-acting or slow-binding inhibitors.

Explanation:

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.

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.

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.

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.

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.

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:

The 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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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:

Answer:

the rectives are just not doing anything so just do sodium

Explanation:

I believe that leading zeros, and tail zeros without a decimal point are not significant