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3 years ago

Which part of an atom is involved in chemical reactions?

Chemistry

2 answers:

algol [13]3 years ago

8 0

Answer:

a nucleus and its particles

Explanation:

torisob [31]3 years ago

5 0

Answer: Valence electrons are involved in a chemical reaction.

Explanation:

A chemical reaction takes place when there is an involvement of valence electrons. Bond formation involves either sharing of valence electrons or transfer of valence electrons. Bond breaking involves the separation of valence electrons.

When valence electrons are shared between the atoms, it leads to the formation of covalent bond.

When transfer of electrons takes place from one atom to another atom, it leads to the formation of ionic bond.

Hence, valence electrons are involved in a chemical reaction.

I think is A

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3 years ago

During decomposition of water,10g of oxygen gas was formed.How much hydrogen gas was formed

statuscvo [17]

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3 years ago

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Bas_tet [7]

Answer:

Gases are easily compressed. We can see evidence of this in Table 1 in Thermal Expansion of Solids and Liquids, where you will note that gases have the largest coefficients of volume expansion. The large coefficients mean that gases expand and contract very rapidly with temperature changes. In addition, you will note that most gases expand at the same rate, or have the same β. This raises the question as to why gases should all act in nearly the same way, when liquids and solids have widely varying expansion rates.

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3 years ago

At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---> +H2(g) + Br2 (g). If the initial partial pressures o

Damm [24]

Answer : The partial pressure of $H_2$ at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of $HBr$ = $1.0\times 10^{-2}atm$

The partial pressure of $H_2$ = $2.0\times 10^{-4}atm$

The partial pressure of $Br_2$ = $2.0\times 10^{-4}atm$

$K_p=4.2\times 10^{-9}$

The balanced equilibrium reaction is,

$2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)$

Initial pressure 1.0×10⁻² 2.0×10⁻⁴ 2.0×10⁻⁴

At eqm. (1.0×10⁻²-2p) (2.0×10⁻⁴+p) (2.0×10⁻⁴+p)

The expression of equilibrium constant $K_p$ for the reaction will be:

$K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}$

Now put all the values in this expression, we get :

$4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}$

$p=-1.99\times 10^{-4}$

The partial pressure of $H_2$ at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of $H_2$ at equilibrium is, 1.0 × 10⁻⁶

4 0

3 years ago

6. What are two uses for nonmetals?

lesantik [10]

Uses of nonmetals in our daily life: Oxygen which is 21% by volume helps in the respiration process.

Nonmetals used in fertilizers: Fertilizers contain nitrogen.

Nonmetals used in crackers: Sulphur and phosphorus are used in fireworks.

5 0

3 years ago