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Jobisdone [24]
4 years ago
13

Which direction does the decimal move from mega meters to meters

Chemistry
1 answer:
Temka [501]4 years ago
7 0

To solve this problem, let us first understand what does mega meters mean. Now by definition, mega means a factor of one million which is a unit prefix in Metric Systems of units. It is denoted by the unit symbol M.

So for example if we have a mega of something, it means a million of something. So in this case, if we have 1 mega meter, what it means is 1 million meter, which is written as:

1 mega meter = 1,000,000 meter

 

To show the decimal points:

1.0 mega meter = 1,000,000.0 meter

 

So we see that the decimal point move 6 zeroes to the right, therefore the direction is to the right.

 

Answer:

<span>6 places to the right</span>

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Answer:

The oxidation number of an element indicates the hypothetical charge of an atom in a compound. ... When the number of electrons associated with an atom changes, its oxidation number also changes. When an element loses an electron, its oxidation number increases.

Explanation:

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3 0
4 years ago
The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate
Karolina [17]

<u>Answer:</u> The value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 3.45

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta n_g = change in number of moles of gas particles = n_{products}-n_{reactants}=2-1=1

Putting values in above equation, we get:

3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084

The equation used to calculate concentration of a solution is:

\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}

Initial moles of (CH_3)_3CCl(g) = 1.00 mol

Volume of the flask = 5.00 L

So, \text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M

For the given chemical reaction:

                (CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)

Initial:               0.2                    -                        -

At Eqllm:          0.2 - x               x                       x

The expression of K_c for above reaction follows:

K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}

Putting values in above equation, we get:

0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

[(CH_3)_2C=CH]=x=0.094M

[HCl]=x=0.094M

[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M

Hence, the value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

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