A 2.00-mol sample of hydrogen gas is heated at constant pressure from 294 K to 414 K. (a) Calculate the energy transferred to the gas by heat. kJ (b) Calculate the increase in its internal energy. kJ (c) Calculate the work done on the gas. kJ

Explanation:

the reaction is indeed an oxidation reduction reaction

In the original solution you have the mixture of a weak acid (Hypochlorous acid) and its conjugate base (Sodium hypochlorite). That is a buffer.

The barium hydroxide will react with hypochlorous acid. If this reaction cause the complete reaction of hypochlorous acid, the buffer break its capacity and the pH change in several units. In this case:

The addition of barium hydroxide will raise the pH slightly because the buffer still working.

The initial moles of those species are:

Hypochlorous acid:

[tex]1L * \frac{0.370mol}{1L} = 0.370 moles[/tex]

Sodium hypochlorite:

[tex]1L * \frac{0.493mol}{1L} = 0.493 moles[/tex]

Now, a strong acid as barium hydroxide (Ba(OH)₂) reacts with a weak acid as hypochlorous acid (HClO) as follows:

Ba(OH)₂ + 2HClO → Ba(ClO)₂ + 2H₂O

For a complete reaction of 0.092 moles of barium hydroxide are required:

[tex]0.092 moles Ba(OH)_2*\frac{2mol HClO}{1molBa(OH)_2} = 0.184 moles HClO[/tex]

As there are 0.370 moles, the moles of HClO after the reaction are:

0.370 moles - 0.184 moles = 0.186 moles of HClO will remain

As you still have hypochlorite and hypochlorous acid you still have a buffer.

Thus, the pH will raise slightly because the amount of acid is decreasing and slightly because the buffer can keep the pH.

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Answer:

[tex]K=K_1*K_2\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}[/tex]

Explanation:

Hello there!

In this case, for the given chemical reaction, it turns out firstly necessary to write the equilibrium expression for both reactions 1 and 2:

[tex]K_1=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \\\\K_2=\frac{[CO_2][H_2]}{[CO][H_2O]}[/tex]

Now, when we combine them to get the overall expression, we infer these two are multiplied to get:

[tex]K=K_1*K_2\\\\K=\frac{[CO][H_2]^3}{[CH_4][H_2O]} *\frac{[CO_2][H_2]}{[CO][H_2O]}\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}[/tex]

Regards!

Answer:

answer in picture

Explanation:

Finding Nemo??? sorry I really need these points

Answer:

2-nonyne

explanation:

First consider the type of bonds, there are tripple bonds of carbon to carbon, in position 2 from the right.

hence it us alkyn.

There are 9 carbons.

Complete Question

A sample of aluminum, which has a specific heat capacity of 0.897 JB loc ! is put into a calorimeter (see sketch at right) that contains 200.0 g of water. The aluminum sample starts off at 85.6 °C and the temperature of the water starts off at 16.0 °C. When the temperature of the water stops changing it's 20.1 °C. The pressure remains constant at 1 atm. Calculate the mass of the aluminum sample.

Answer:

[tex]M=58g[/tex]

Explanation:

From the question we are told that:

Heat Capacity [tex]H=0.897[/tex]

Mass of water [tex]M=200g[/tex]

Initial Temperature of Aluminium [tex]T_a=85.6[/tex]

Initial Temperature of Water [tex]T_{w1}=16.0[/tex]

Final Temperature of Water  [tex]T_{w2}=16.0[/tex]

Generally

Heat loss=Heat Gain

Therefore

[tex]M*0.897*(85.6-20.1) =200*4.184*(20.1-16)[/tex]

[tex]M=58g[/tex]

Answer:

0

Explanation:

answer from gauth math

Answer:

22.55 mmHg (0.03 atm)

Explanation:

According to Raoult's law, the vapor pressure (Psolution) of a solution is given by:

Psolution = Xsolvent x Psolvent

Where Xsolvent is the mole fraction of the solvent in the solution and Psolvent is the vapor pressure of the pure solvent.

From the data, we have: Psolvent = 25.76 mmHg

We have to calculate Xsolvent, which is equal to the moles of solvent divided into the total number of moles.

The solution is composed of the solute (glucose) dissolved in the solvent (water). So, the total number of moles is calculated from the moles of solute and solvent.

To calculate the moles of solute (glucose), we divide the mass of glucose into its molar mass:

moles of glucose = mass/molar mass = 550 g/(180.2 g/mol) = 3.05 mol

The same for the moles of solvent (water). The mass of water is obtained from the product of the volume and density:

mass of water = volume x density = 1020.0 mL x 1.00 g/mL = 1020.0 g

molar mass H₂O = (1 g/mol x 2) + 16 g/mol = 18 g/mol

moles of water = mass water/molar masss = 1020.0 g/(18 g/mol) = 56.67 mol

Now, we can calculate Xsolvent:

Xsolvent = moles of water/total moles

total moles = moles glucose + moles water = 3.05 mol + 56.67 mol = 59.72 mol

⇒ Xsolvent = 56.67 mol/(59.72 mol) = 0.9489

Finally, we calculate the vapor pressure of the solution:

Psolution = 0.9489 x 23.76 mmHg = 22.55 mmHg

22.55 mmHg x 1 atm/760 mmHg = 0.03 atm

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Answer:

a) E2

b) SN2

c) SN2

Explanation:

A substitution reaction involves replacement of an atom or group in a molecule by another atom or group. An elimination reaction is the loss of two atoms from the same molecule leading to the formation of a multiple bond in the molecule.

We must note that primary alkyl halides never undergo SN1/E1 reactions. However, the presence of a strong bulky base such as tert BuO- , E2 reactions predominate. In the presence of strong bases such as OH^- and good nucleophiles such as I^-, SN2 mechanism predominates.

Answer:

23592U+10n→14454Xe+9038Sr+210n

Explanation:

a nuclear reaction for the neutron-induced fission of U−235 to form Xe−144 and Sr−90.

Answer:

C . Kinetic Molecular Theory

Answer:

B. Complex B.

Explanation:

Complex B will have binding site with high affinity for reduced plastocyanin due to greater redox potential. The high number of redox potential will will transport electron chain in cyanobacteria.

Redox potential is the measure of the electron gain or loss to the electrode. For reduced plastocyanin complex B will have the highest affinity.

Electron affinity is the energy released when the atom gets attached to the atom or other molecule. The high number of redox potential increases the electron transport in the cell.

The greater the redox potential more will be its tendency to show electron affinity. To bond with reduced species, the oxidized species must have greater redox potential.

Therefore, option B. complex b will have the highest affinity.

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Answer:

11.76 g/cm^3

Explanation:

Mass of empty flask and stopper = 64.232g

Mass of flask filled with water = 153.617 g

Mass of water = 153.617 g - 64.232g = 89.385 g

Mass of flask, stopper and metal = 143.557 g

Mass of metal = 143.557 g - 64.232g = 79.325 g

Mass of water, flask, stopper and metal = 226.196 g

Mass of water = 226.196 g - 143.557 g = 82.639 g

Since mass of water =volume of water

Volume occupied by metal = 89.385 cm^3 - 82.639 cm^3 = 6.746 cm^3

Density of metal = mass/volume = 79.325 g/6.746 cm^3

= 11.76 g/cm^3

Answer:

Explanation:

Answer:

6.02×10^23 I hope it helps you

Explanation:

I hope it helps you

Answer:

Part A

HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)

Part B

ClCH2CH2CO2H(aq) + NaOH(aq) ------> ClCH2CH2CO2Na(aq) + H2O(l)

Explanation:

The reaction between an alkanoic acid and a base is a neutralization reaction. The reaction occurs as follows;

RCOOH + NaOH ----> RCOONa + H2O

We have to note the fact that the net ionic reaction still remains;

H^+(aq) + OH^-(aq) ---> H2O(l)

In both cases, the reaction can occur and they actually do occur as written.

Answer:

D. beta particle

Explanation:

Number of protons increases from 38 to 39 indicating beta decay (only one proton up from parent isotope to daughter isotope) Also atomic mass (on top of an isotope), 90 stays the same as beta particle is very small.

Answer:

it is 11.55 and ik because I just had that question

18.0 grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride.

Let's consider the following balanced equation.

4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g)

The molar ratio of C₈H₄N₂ to Cu(C₃₂H₁₆N₈) is 4:1. The moles of Cu(C₃₂H₁₆N₈) produced from 0.125 moles of C₈H₄N₂ are:

[tex]0.125 mol C_8H_4N_2 \times \frac{1molCu(C_{32}H_{16}N_8)}{4mol C_8H_4N_2} = 0.0313 molCu(C_{32}H_{16}N_8)[/tex]

The molar mass of Cu(C₃₂H₁₆N₈) is 576.07 g/mol. The mass corresponding to 0.0313 moles of Cu(C₃₂H₁₆N₈) is:

[tex]0.0313 moles \times \frac{576.07g}{mol} = 18.0 g[/tex]

18.0 grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride.

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Answer:

1. 17.5 g of CO₂

2. The limiting reactant is carbon (graphite), and its formula is C(graphite)

3. 3.7 g of O₂

Explanation:

First, we have to write the chemical equation for the reaction. For this, we have to know the chemical formula of each reactant and product:

Thus, we write the chemical equation:

C(graphite) + O₂(g) → CO₂(g)

The equation is already balanced because it has the same number of C and O atoms on both sides. Thus, we can see that 1 mol of C(graphite) reacts with 1 mol of O₂ and produce 1 mol of CO₂ (mole-to-mole reaction).

Now we convert the grams of reactants to moles by using the molecular weight (Mw) of each compound:

Mw(C) = 12 g/mol

moles of C(graphite) = 4.77g/(12 g/mol) = 0.3975 mol

Mw(O₂) = 16 g/mol x 2 = 32 g/mol

moles of O₂ = 16.4 g/(32 g/mol) = 0.5125 mol

Now, we can compare the stoichiometric ratio (given by the moles of reactants in the equation) with the actual ratio (given by the mass of reactants we have):

stoichiometric ratio ⇒ 1 mol C(graphite)/mol O₂

actual ratio ⇒ 0.3975 mol C(graphite)/0.5125 mol O₂

We can see that we need 0.3975 moles of O₂ to react with C(graphite) and we have more moles (0.5125 mol) so the excess reactant is O₂. Thus, the limiting reactant is C(graphite).

The amount of product (CO₂) that is formed is calculated from the amount of limiting reactant. We can see in the chemical equation that 1 mol of CO₂ is produced from 1 mol of C(graphite) ⇒ stoichiometric ratio = 1 mol CO₂/mol C(graphite).

Thus, we multiply the moles of C(graphite) we have by the stoichiometric ratio to calculate the moles of CO₂ produced:

moles of CO₂ = 0.3975 mol C(graphite) x 1 mol CO₂/mol C(graphite) = 0.3975 mol CO₂

Now, we convert the moles of CO₂ to mass by using the Mw:

Mw(CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol

mass of CO₂ = 0.3975 mol CO₂ x 44 g/mol CO₂ = 17.5 g

Therefore, the maximum amount of carbon dioxide (CO₂) formed is 17.5 g.

Since this is a mole-to-mole reaction, the moles of excess reactant that remains after the reaction is complete is calculated as the difference between the moles of excess reactant and limiting reactant:

remaining moles of O₂ = 0.5125 mol - 0.3975 mol = 0.115 mol O₂

Finally, we convert the moles of O₂ to mass with the Mw (32 g/mol) :

mass of O₂ = 0.115 mol O₂ x 32 g/mol = 3.68 g

Therefore, the mass of the excess reagent that remains after the reaction is complete is 3.7 g.

Explanation:

The required concentration of [tex]HNO_3[/tex] M1 =0.222 M.

The required volume of [tex]HNO_3[/tex] is V1 =225 mL.

The standard solution of [tex]HNO_3[/tex] is M2 =16 M.

The volume of standard solution required can be calculated as shown below:

Since the number of moles of solute does not change on dilution.

The number of moles [tex]n=molarity * volume[/tex]

[tex]M_1.V_1=M_2.V_2[/tex]

[tex]V2=\frac{M_1.V_1}{M_2} \\=0.222M x 225 mL / 16 M\\=3.12 mL[/tex]

Hence, 3.12 mL of 16 m nitric acid is required to prepare 0.222 M and 225 mL of nitric acid.

Answer:

a. A beta particle has a negative charge. d. A beta particle is a high-energy electron.

Explanation:

Identify the correct descriptions of beta particles.

a. A beta particle has a negative charge. YES. A beta particle is originated in the following nuclear reaction: ¹₀n ⇒ ¹₁H + ⁰₋₁e (beta particle.)

b. A beta particle contains neutrons. NO. It is a electron originated in the nucleus.

c. A beta particle is less massive than a gamma ray. NO. Gamma rays don't have mass while a beta particle has a mass which is half of one thousandth of the mass of a proton.

d. A beta particle is a high-energy electron. YES. Beta particles are nuclear originated hig-energy electrons.

The correct option is d.: 1.5 grams of glucose is produced from 2.20 g of CO₂.

To find the mass of glucose produced, first you must know the balanced reaction. For this, the Law of Conservation of Matter is followed.

The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

So, in this case, the balanced reaction is:

6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the amounts of moles of each reactant and product participate in the reaction:

So, you know that 2.20 g of CO₂ react, whose molar weight is 44.01 g/mole. By definition of molar mass, 1 mole of CO₂ has 44.01 g. So, the number of moles that 2.20 grams of the compound represent is calculated as:

[tex]moles of CO_{2} =2.20 grams*\frac{1 mole}{44.01 grams}[/tex]

moles of CO₂= 0.05 moles

Now you must follow the following rule of three: if by stoichiometry of the reaction 6 moles of CO₂ produce 1 mole of C₆H₁₂O₆, 0.05 moles of CO₂ produce how many moles of C₆H₁₂O₆?

[tex]moles of C_{6} H_{12} O_{6} =\frac{0.05moles of CO_{2} *1 mole of C_{6} H_{12} O_{6}}{6moles of CO_{2}}[/tex]

moles of C₆H₁₂O₆= 8.33*10⁻³

Being the molar mass of glucose 180.18 g/mole, the mass that 8.33*10⁻³ moles of the compound represent is calculated as:

[tex]mass of glucose =8.33*10^{-3} moles*\frac{180.18 grams}{1 mole}[/tex]

mass of glucose= 1.5 grams

In summary, the correct option is d.:  1.5 grams of glucose is produced from 2.20 g of CO₂.

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Answer:

the answer is C

Explanation:

a molecule can be made up of two atoms of the same kind, as when two oxygen atoms bind together to make an oxygen molecule

I'd really appreciate a brainleast

Answer:

This is an acid-base reaction (neutralization): Ba(OH) 2 is a base, H 2SO 4 is an acid. This is a precipitation reaction: BaSO 4 is the formed precipitate.

Answer:

an if/then statement

Explanation:

Explanation:

P1V1/T1 = P2V2/T2

=((390.89×313.9)/76.6)×((718.3×V2)/154.2)

= (122700.371/76.7) × ((718.3×V2)/154.2)

make V2 the subject of the formula...

V2 =(122700.371×154.2)/(76.6×718.3)

V2 =18920397.21/55021.78

V2 = 343.87

Answer:

False

Explanation:

The first reaction is;

NO(g) + 1/2O2(g) ---->NO2(g)

K= [NO2]/[NO] [ O2]^1/2

The second reaction is;

2NO(g) + O2(g) ---->2NO2(g)

K'= [NO2]^2/[NO]^2 [O2]

It now follows that;

K'= K^2

Hence the statement in the question is false

After the double replacement reaction from mixing 50.0g of sulfuric acid and 40.0 grams of barium chloride is complete, 31.16 grams of sulfuric acid and 0 grams of barium chloride remain.

The equation of the reaction between sulfuric acid and barium chloride is

BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

From this equation of reaction, it means 1 mole of barium chloride will completely react with 1 mole of sulfuric acid.

From the question, we have 50.0g of sulfuric acid and 40.0 grams of barium chloride.

First, we will determine the number of moles of the sulfuric acid and barium chloride present.

Number of moles is given by the formula

Number of moles = Mass / Molar mass

For sulfuric acid

Mass = 50.0 g

Molar mass = 98.079 g/mol

∴ Number of moles = 50.0 / 98.079

Numbers of moles of sulfuric = 0.509793 mol

For barium chloride

Mass = 40.0 grams

Molar mass of barium chloride = 208.23 g/mol

∴ Number of moles = 40.0 / 208.23

Number of moles of barium chloride = 0.192095 mol

Since the number of moles of sulfuric acid is more than that of barium chloride, then the limiting reagent is barium chloride and the excess reagent is sulfuric acid

NOTE: A limiting reagent is the reactant that is completely used up in a reaction, and it determines when the reaction stops.

Hence, barium chloride will be used up during the reaction (that is, 0 grams will remain after the reaction is complete).

For the mass of sulfuric acid that will remain,

First, we will determine the number of mole that will remain.

Since 1 mole of barium chloride completely reacts with 1 mole of sulfuric acid, then 0.192095 mol of barium chloride will react with 0.192095 mol of sulfuric acid.

∴ The remaining number moles of sulfuric acid = 0.509793 mol - 0.192095 mol

The remaining number moles of sulfuric acid = 0.317698 mol

Then, from

Mass = Number of moles × Molar mass

Mass = 0.317698 mol ×98.079 g/mol

Mass = 31. 1595 gram

Mass ≅ 31.16 grams

∴ 31.16 grams remains after the reaction is complete.

Hence, 31.16 grams of sulfuric acid and 0 grams of barium chloride remain after the double replacement reaction is complete.

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Answer:

% NiCl2 = 24.13%

% water = 78.57%

Explanation:

Mass percentage = mass of solute/mass of solution × 100

According to this question, a solution contains 159 g of NiCl2 in 500 g of water. Hence, mass of the solution is calculated as follows:

Mass of solution = 159g + 500g

Mass of solution = 659g

Therefore;

A) % Mass of NiCl2 in solution = mass of NiCl2/mass of solution × 100

% Mass of NiCl2 in solution = 159/659 × 100

% Mass of NiCl2 in solution = 0.2413 × 100

= 24.13%

B) % Mass of water in solution = mass of water/mass of solution × 100

% Mass of water in solution = 500/659 × 100

% Mass of water in solution = 0.7587 × 100

% Mass of water in solution = 75.87%