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3 years ago

How is the movement of particles in diffusion determined?

1 answer:

3 0

The particles move from an area of high concentration, to an area of low concentration, until the distribution of particles is equal

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If the ph of a solution is decreased from ph 8 to ph 6 it means that the

saw5 [17]

Explanation:

Relation between pH and concentration of hydrogen ions is as follows.

pH = $-log [H^{+}]$

So, it means that an increase in the value of pH will show that there occurs a decrease in concentration of hydrogen ions.

Therefore, the solution becomes basic in nature.

On the other hand, a decrease in the value of pH will show that there occurs an increase in the concentration of hydrogen ions.

Therefore, the solution becomes more acidic in nature.

Hence, if the pH of a solution is decreased from pH 8 to pH 6 it means that the concentration of hydrogen ions has increased in the solution.

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3 years ago

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How to turn form moles to grams

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Turning things to gram so need to convert to the metric system

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What is responsible for the whistling sound that often accompanies firework?

Naya [18.7K]

The wistle is coming from all the chemicals including the fire you use to light the firework and a chemcial reaction is occuring which is the sound that is made because of this

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3 years ago

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Elza [17]

Answer: There are $21.08\times 10^{23}$ molecules in 63.00 g of $H_2O$

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{63.00g}{18g/mol}=3.5moles$

1 mole of $H_2O$ contains = $6.023\times 10^{23}$ molecules

Thus 3.5 moles of $H_2O$ contains = $\frac{6.023\times 10^{23}}{1}\times 3.5=21.08\times 10^{23}$ molecules.

There are $21.08\times 10^{23}$ molecules in 63.00 g of $H_2O$

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3 years ago

Consider the reaction: N2(g) + O2(g) ⇄ 2NO(g) Kc = 0.10 at 2000oC Starting with initial concentrations of 0.040 mol/L of N2 and

IrinaVladis [17]

Answer:

0.011 mol/L

Explanation:

This can be solved with something called an ICE table.

I = initial

C = change

E = equilibrium

Initially, there is 0.04 M of N₂, 0.04 M of O₂, and 0 M of NO.

x amount of N₂ reacts. Since the stoichiometry is 1:1, x amount of O₂ also reacts. This produces 2x of NO.

After the reaction, there is 0.04-x of N₂, 0.04-x of O₂, and 2x of NO.

Here it is in table form:

$\left[\begin{array}{cccc}&N2&O2&NO\\I&0.04&0.04&0\\C&-x&-x&+2x\\E&0.04-x&0.04-x&2x\end{array}\right]$

Now we can use the equilibrium constant:

Kc = [NO]² / ( [N₂] [O₂] )

Substituting:

0.10 = (2x)² / ( (0.04 - x) (0.04 - x) )

Solving:

0.10 = (2x)² / (0.04 - x)²

√0.10 = 2x / (0.04 - x)

(√0.10) (0.04 - x) = 2x

(√0.10)(0.04) - (√0.10)x = 2x

(√0.10)(0.04) = 2x + (√0.10)x

(√0.10)(0.04) = (2 + √0.10)x

x = (√0.10)(0.04) / (2 + √0.10)

x = 0.0055

At equilibrium, the concentration of NO is 2x. So the answer is:

[NO] = 2x

[NO] = 0.011

The equilibrium concentration of NO is 0.011 mol/L.

3 0

3 years ago