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3 years ago

Calculate the number of moles in 4.22x1022 molecules of H,O

Chemistry

1 answer:

Arturiano [62]3 years ago

7 0

Answer:

0.07 moles of water

Explanation:

Given data:

Number of molecules of water = 4.22×10²²

Number of moles of water = ?

Solution:

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules of water

4.22×10²² molecules × 1 mol /6.022 × 10²³ molecules

0.7 ×10⁻¹ mol

0.07 moles of water

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A mixture of methane and carbon dioxide gases contains methane at a partial pressure of 431 mm Hg and carbon dioxide at a

KatRina [158]

Answer:

XCH₄ = 0.461

XCO₂ = 0.539

Explanation:

Step 1: Given data

Partial pressure of methane (pCH₄): 431 mmHg

Partial pressure of carbon dioxide (pCO₂): 504 mmHg

Step 2: Calculate the total pressure in the container

We will sum both partial pressures.

P = pCH₄ + pCO₂

P = 431 mmHg + 504 mmHg = 935 mmHg

Step 3: Calculate the mole fraction of each gas

We will use the following expression.

Xi = pi / P

XCH₄ = pCH₄/P = 431 mmHg/935 mmHg = 0.461

XCO₂ = pCO₂/P = 504 mmHg/935 mmHg = 0.539

3 0

3 years ago

Theory is often used as a synonym for idea in everyday life.

Rasek [7]

I think B. As an idea is just a way that could be possible

8 0

3 years ago

A piece of an unknown substance weighing 124.0 grams is heated in boiling water to 100.0oc. when the substance is placed in a ca

MakcuM [25]

The heat cause 300g water temperature increase from 20 to 26 celcius. The heat transferred would be: 300g * (26 °C -20 °C) *4.2 joule/gram °C= 7560J

The unknown substance is added to the water, so its final temperature should be the same as the water. The calculation would be:

7560J= 124g * (100-26)* specific heat

specific heat= 7560J / 124g / 74 °C= 0.8238 J/gram °C

3 0

4 years ago

Calculate the solubility of o2 in water at a partial pressure of o2 of 120 torr at 25 ̊c. the henry's law constant for o2 at 25

Vladimir79 [104]

Answer:

1) 2.054 x 10⁻⁴ mol/L.

2) Decreasing the temperature will increase the solubilty of O₂ gas in water.

Explanation:

1) The solubility of O₂ gas in water:

We cam calculate the solubility of O₂ in water using Henry's law: <em>Cgas = K P</em>,

where, Cgas is the solubility if gas,

K is henry's law constant (K for O₂ at 25 ̊C is 1.3 x 10⁻³ mol/l atm),

P is the partial pressure of O₂ (P = 120 torr / 760 = 0.158 atm).

Cgas = K P = (1.3 x 10⁻³ mol/l atm) (0.158 atm) = 2.054 x 10⁻⁴ mol/L.

2) The effect of decreasing temperature on the solubility O₂ gas in water:

Decreasing the temperature will increase the solubilty of O₂ gas in water.

When the temperature increases, the solubility of O₂ gas in water will decrease because the increase in T will increase the kinetic energy of gas particles and increase its motion that will break intermolecular bonds and escape from solution.

Decreasing the temperature will increase the solubility of O₂ gas in water will because the kinetic energy of gas particles will decrease and limit its motion that can not break the intermolecular bonds and increase the solubility of O₂ gas.

6 0

3 years ago

Read 2 more answers

Draw the bridged bromonium ion that is formed as an intermediate during the bromination of this alkene. include hydrogen atoms,

gogolik [260]

<h2>Answer</h2>

Bromination:

Any reaction or process in which bromine (and no other elements) are introduced into a molecule.

Bromonium Ion:

The bromonium ion is formed when alkenes react with bromine. When the π cloud of the alkene (acting as a nucleophile) approaches the bromine molecule (acting as an electrophile), the σ-bond electrons of Br2 are pushed away, resulting in the departure of the bromide anion.(2)

Mechanism:

Step 1:

In the first step of the reaction, a bromine molecule approaches the electron-rich alkene carbon–carbon double bond. The bromine atom closer to the bond takes on a partial positive charge as its electrons are repelled by the electrons of the double bond. The atom is electrophilic at this time and is attacked by the pi electrons of the alkene [carbon–carbon double bond]. It forms for an instant a single sigma bond to both of the carbon atoms involved (2). The bonding of bromine is special in this intermediate, due to its relatively large size compared to carbon, the bromide ion is capable of interacting with both carbons which once shared the π-bond, making a three-membered ring. The bromide ion acquires a positive formal charge. At this moment the halogen ion is called a "bromonium ion".

Step 2:

When the first bromine atom attacks the carbon–carbon π-bond, it leaves behind one of its electrons with the other bromine that it was bonded to in Br2. That other atom is now a negative bromide anion and is attracted to the slight positive charge on the carbon atoms. It is blocked from nucleophilic attack on one side of the carbon chain by the first bromine atom and can only attack from the other side. As it attacks and forms a bond with one of the carbons, the bond between the first bromine atom and the other carbon atoms breaks, leaving each carbon atom with a halogen substituent.

In this way the two halogens add in an anti addition fashion, and when the alkene is part of a cycle the dibromide adopts the trans configuration.

6 0

3 years ago