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3 years ago

What is the type of compound of copper and oxide?

1 answer:

5 0

Copper(II) oxide or cupric oxide is the inorganic compound with the formula CuO. A black solid, it is one of the two stable oxides of copper, the other being Cu2O or cuprous oxide. As a mineral, it is known as tenorite and paramelaconite.

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Give the expression for the solubility product constant for baf2.

svetoff [14.1K]

Solubility product constant (Ksp) is applied to the saturated ionic solutions which are in equilibrium with its solid form. The solid is partially dissociated into its ions.

For the BaF, the dissociation as follows;
BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)

Hence,
Ksp = [Ba²⁺(aq)] [F⁻(aq)]²

3 0

4 years ago

Please help Asap………………

vladimir2022 [97]

Answer:

3rd

Explanation:

6 0

3 years ago

Read 2 more answers

Draw the product formed when the compound shown below undergoes a reaction with hcl in ch2cl2.

Taya2010 [7]

The product formed when HCl and CH2Cl2 react is CH4
H
H-C- H methane structure
H

HCl react with CH2Cl2 to form methane (CH4) and chlorine gas(Cl2)
that is,

2HCl(g) + CH2Cl2(l) = CH4 (g) +2Cl2 (g)

6 0

4 years ago

Calculate the enthalpy of the reaction

harkovskaia [24]

Answer : The enthalpy of the reaction is, -2552 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

$4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)$ $\Delta H_A=+2035kJ$

(2) $2B(s)+3H_2(g)\rightarrow B_2H_6(g)$ $\Delta H_B=+36kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_C=-285kJ$

(4) $H_2O(l)\rightarrow H_2O(g)$ $\Delta H_D=+44kJ$

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

$\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D$

$\Delta H=-2(+2035kJ)+2(+36kJ)-6(-285kJ)-6(+44)$

$\Delta H=-2552kJ$

Therefore, the enthalpy of the reaction is, -2552 kJ/mole

4 0

3 years ago

Terms that have to be used: Products, Reactants, Balanced, Unbalanced, Atoms

emmainna [20.7K]

Answer:

A Chemical equation is balanced if the reactants side of the chemical reaction has the same number total of atoms of elements as the products side of the chemical equation. While a Chemical equation is unbalanced if the reactants side of the chemical reaction has a different number total of atoms of elements as the products side of the chemical equation

Explanation:

Self explanatory

3 0

3 years ago