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4 years ago

6

when an atomic ion or molecule of a recant switches places with an unpaired recant the type of reaction is called

1 answer:

siniylev [52]4 years ago

4 0

I think it's single replacement reaction.

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Please help me on this exsam question ASAP!!! Please

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3 years ago

How do oil spills affect marine ecosystems?

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3 years ago

How can I get the 3 sig figs of this number 187515.4207

zimovet [89]

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1.88 x 10^5

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6 0

3 years ago

Read 2 more answers

A mixture of 0.307 M Cl 2 , 0.465 M F 2 , and 0.706 M ClF is enclosed in a vessel and heated to 2500 K . Cl 2 ( g ) + F 2 ( g )

monitta

<u>Answer:</u> The equilibrium concentration of chlorine gas, fluorine gas and ClF gas is 0.159 M, 0.317 M and 1.002 M respectively.

<u>Explanation:</u>

We are given:

Initial concentration of chlorine gas = 0.307 M

Initial concentration of fluorine gas = 0.465 M

Initial concentration of ClF gas = 0.706 M

The given chemical equation follows:

$Cl_2(g)+F_2(g)\rightleftharpoons 2ClF(g)$

<u>Initial:</u> 0.307 0.465 0.706

<u>At eqllm:</u> 0.307-x 0.465-x 0.706+2x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[ClF]^2}{[Cl_2][F_2]}$

We are given:

$K_c=20.0$

Putting values in above equation, we get:

$20.0=\frac{(0.706+2x)^2}{(0.307-x)(0.465-x)}\\\\x=0.148,0.993$

Neglecting the value of x = 0.993 because the equilibrium concentrations of chlorine and fluorine gases will become negative, which is not possible

So, equilibrium concentration of chlorine gas = (0.307 - x) = [0.307 - 0.148] = 0.159 M

Equilibrium concentration of fluorine gas = (0.465 - x) = [0.465 - 0.148] = 0.317 M

Equilibrium concentration of ClF gas = (0.706 + 2x) = [0.706 + 2(0.148)] = 1.002 M

Hence, the equilibrium concentration of chlorine gas, fluorine gas and ClF gas is 0.159 M, 0.317 M and 1.002 M respectively.

5 0

3 years ago

How many molecules are in 23 moles of oxygen (O2)? Round the answer to one digit past the decimal point. If the answer is less t

melamori03 [73]

To Find :

How many molecules are in 23 moles of oxygen (O₂).

Solution :

We know, 1 mole of any atom/molecule contains 6.022 × 10²³ numbers of atoms/molecules.

So, number of molecules in 23 moles of O₂ are :

$n= 23\times 6.022\times 10^{23}\\\\n =138.506\times 10^{23}\\\\n = 1.385 \times 10^{25}\ molecules$

Hence, this is the required solution.

6 0

3 years ago