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Help me plzz in chemistry

AlexFokin [52]

a) f2 loses electrons and is oxidised. li gains electrons and is reduced

c)sn2+ loses electrons and is oxidised.Al gains electrons and is reduced

b)br2 loses electrons and is oxidised. I- in KI gains electrons and is reduced

5 0

3 years ago

A hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

kupik [55]

hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )

where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )

where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )

where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.

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6 0

3 years ago

Match these items. 1. moles per liter of solution solute 2. spreading out Keq 3. substance being dissolved diffusion 4. general

Degger [83]

Answer:

1. moles per liter of solution - Molarity

2. spreading out - diffusion

3. substance being dissolved- solute

4. general equilibrium constant - Keq

5. the substance doing the dissolving in a solution - solvent

6. solubility product constant - Ksp

Explanation:

Number of moles of substance being dissolved in one liter of solution is defined as molarity

Solute is the substance which is dissolved and solvent is the substance into which the solute is dissolved. For example salt is solute and water is solvent.

Ksp defines the solubility product constant which indicates the equilibrium between a solid and solution and the number of ions of solid required to achieve this equilibrium.

Keq depicts the scenario where reactant and product concentrations are constant in a chemical reaction.

Hence, the correct match is

1. moles per liter of solution - Molarity

2. spreading out - diffusion

3. substance being dissolved- solute

4. general equilibrium constant - Keq

5. the substance doing the dissolving in a solution - solvent

6. solubility product constant - Ksp

8 0

3 years ago

Balancee por tanteo las siguientes ecuaciones químicas. Escriba el nombre a reactantes y productos. H2O5 + H2O ---> HNO3 Na2O

Elan Coil [88]

Answer:

a. N₂O₅ + H₂O ⇒ 2 HNO₃ (pentóxido de dinitrógeno + agua ⇒ ácido nítrico)

b. Na₂O + H₂O ⇒ 2 NaOH (óxido de sodio + agua ⇒ hidróxido de sodio)

Explanation:

Tenemos que balancear, por el método de tanteo, las siguientes ecuaciones químicas.

a. En la primera reacción, el pentóxido de dinitrógeno reacciona con agua para formar ácido nítrico. Es una reacción de síntesis o combinación.

N₂O₅ + H₂O ⇒ HNO₃

Podremos obtener la ecuación balanceada si multiplicamos HNO₃ por 2.

N₂O₅ + H₂O ⇒ 2 HNO₃

b. En la segunda reacción, óxido de sodio reacciona con agua para formar hidróxido de sodio. Es una reacción de síntesis o combinación.

Na₂O + H₂O ⇒ NaOH

Podremos obtener la ecuación balanceada si multiplicamos NaOH por 2.

Na₂O + H₂O ⇒ 2 NaOH

3 0

3 years ago

NaHCO 3 and NaHSO 4 are acid salts. however aqueous solution of NaHCO 3 is basic while aqueous solution of NaHSO 4 is acidic. gi

posledela

Answer:c

Explanation:

3 0

3 years ago