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3 years ago

.....k+.....HgCl2 gives u...... Hg+ KCl

Chemistry

1 answer:

worty [1.4K]3 years ago

5 0

Answer:

2K + HgCl2 —> Hg + 2KCl

Explanation:

K + HgCl2 —> Hg + KCl

The above equation is not balanced

To balance the equation, do the following.

First, put 2 in front of KCl, the equation becomes,

K + HgCl2 —> Hg + 2KCl

Now we see that Cl is balanced but K is not. To balance K, put 2 in front of K as shown below:

2K + HgCl2 —> Hg + 2KCl

Now we can that the equation is balanced

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If the rate of a reaction increases by a factor of 64 when the concentration of reactant increases by a factor of 4, what is the

igomit [66]

The reaction is of order three with respect to the reactant.

<h3>Explanation</h3>

The rate of a reaction of order n about a certain reactant is proportion to the concentration of that reactant raised to the n-th power. This is true only if concentrations of any other reactants stay constant in the whole process.

In other words, Rate = constant × [Reactant]ⁿ, Rate ∝ [Reactant]ⁿ. (The symbol "∝" reads "proportional to".)

In this question,

[4 × Reactant]ⁿ ÷ [Reactant]ⁿ = 64.

In other words, 4ⁿ = 64, where n is the order of the reaction with respect to this reactant.

It might take some guesswork to find the value of n. Alternatively, n can be solved directly with a calculator using logarithms. Taking natural log of both sides:

$\ln{4^n} = \ln{64}\\n\; \ln{4} = \ln{64}\\n = \frac{\ln{64}}{\ln{4}} = 3$ .

Evaluating $\ln(64) / \ln(4)$ on Google or on a calculator with support for ln (the natural log) will give the value of n- no guesswork required.

n = 3. Therefore, the reaction is of order three with respect to this reactant.

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3 years ago

Which property altered during a chemical change is not altered during a physical change?

marissa [1.9K]

<h3>During physical changes, the composition od the original substance is not altered, but the properties of the original substance are altered. During a chemical change the composition od the original substance is not altered and the change is irreversible. Melting of butter and wax is an example of chemical changes.</h3>

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2 years ago

Read 2 more answers

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nata0808 [166]

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3 years ago

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At 500 degree C, F_2 gas is stable and does not dissociate, but at 840 degree C, some dissociation occurs: F_2 (g) 2 F(g). A fla

bazaltina [42]

Answer:

2.73 is the equilibrium constant for the dissociation of $F_2$ gas at 840 degree Celsius.

Explanation:

$F_2(g)\rightleftharpoons 2F(g)$

Initial

0.600 atm 0

Equilibrium

(0.600 atm - p) 2p

Total pressure at equilibrium = P = 0.984 atm

P= 0.600 atm - p)+2p=0.984 atm

p = 0.384 atm

Partial pressure of the $F_2$ gas , $p_{f_2}$ = (0.600 atm - 0.384 atm)=0.216 atm

Partial pressure of the $F$ gas, $p_{f}$ = 2(0.384 atm)=0.768 atm

$K_p=\frac{(p_{F})^2}{p_{F_2}}$

$K_p=\frac{(0.768 atm)^2}{0.216 atm}=2.73$

2.73 is the equilibrium constant for the dissociation of $F_2$ gas at 840 degree Celsius.

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4 years ago

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4 years ago