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3 years ago

Dispersal of matter and energy can be directly related to spontaneity if the: Select the correct answer below

Chemistry

1 answer:

n200080 [17]3 years ago

7 0

Answer:

d. there are no conditions for this correlation.

Explanation: Dispersal of matter is a term used to describe the various processes and activities involved in the transfer of materials from one location to another.

The spontaneity of a process tries to explain that a process will continue to occur when an initial force has been applied, it doesn't require the continous input of energy for it to continue to take place.

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A powder contains FeSO4⋅7H2O (molar mass=278.01 g/mol), among other components. A 3.930 g sample of the powder was dissolved in

fredd [130]

Answer:

The mass of FeSO4*7H2O in the sample is 1.21 grams

Explanation:

Step 1: Calculate moles of Fe2O3

moles of Fe2O3 = mass of Fe2O3 / Molar mass of Fe2O3

moles of Fe2O3 = 0.348 grams / 159.69 g/mole = 0.00218 moles

Step 2: Calculate moles of Fe

4 Fe + 3O2 → 2Fe2O3

For 4 moles of Fe consumed there is 2 moles of Fe2O3 produced

This means it has a ratio 2:1

So 0.00218 moles of Fe2O3 produced , there is 2*0.00218 = 0.00436 moles of Fe consumed

Step 3: Calculate moles of FeSO4*7H2O

Fe + H2SO4 + 7H2O → FeSO4*7H20 + H2

For 1 mole of Fe consumed there is 1 mole of FeSO4*7H2O produced

This means for 0.00436 moles there is 0.00436 moles of Fe2SO4*H2O produced

Step 4: Calculate the mass of FeSO4*7H2O in the sample

mass of FeSO4*7H2O = 0.00436 moles * 278.01 g/mole = 1.212 g

The mass of FeSO4*7H2O in the sample is 1.21 grams

8 0

3 years ago

The following reaction was carried out in a 3.25 L reaction vessel at 1100 K:

kipiarov [429]

Answer : The value of reaction quotient Q is, 0.498

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

$C(s)+H_2O(g)\rightarrow CO(g)+H_2(g)$

The expression for reaction quotient will be :

$Q=\frac{[CO][H_2]}{[H_2O]}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now we have to calculate the concentration of $H_2O,CO\text{ and }H_2$

$\text{Concentration of }H_2O=\frac{\text{Moles of }H_2O}{\text{Volume of solution}}=\frac{13.3mol}{3.25L}=4.09M$

and,

$\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{3.40mol}{3.25L}=1.05M$

and,

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{6.30mol}{3.25L}=1.94M$

Now put all the given values in the above expression, we get:

$Q=\frac{(1.05)\times (1.94)}{(4.09)}=0.498$

Thus, the value of reaction quotient Q is, 0.498

3 0

3 years ago

1.Chlorobenzene, C6H5Cl, is used in the production of chemicals such as aspirin and dyes. One way that chlorobenzene is prepared

miv72 [106K]

Answer:

$m_{C_6H_5Cl}=65.7gC_6H_5Cl$

Explanation:

Hello!

In this case, according to the given balanced chemical reaction:

$C_6H_6 (l) + Cl_2 (g) \rightarrow C_6H_5Cl (s) + HCl (g)$

We can see there is 1:1 between benzene and chlorobenzene as the relavant product; thus, since the molar mass of benzene is 78.11 g/mol and that of chlorobenzene is 112.55 g/mol, the theoretical yield for this reaction turns out:

$m_{C_6H_5Cl}=45.6gC_6H_6*\frac{1molC_6H_6}{78.11gC_6H_6 }*\frac{1molC_6H_5Cl}{1molC_6H_6}*\frac{112.55gC_6H_5Cl}{1molC_6H_5Cl} \\\\m_{C_6H_5Cl}=65.7gC_6H_5Cl$

Best regards!

8 0

3 years ago

A scientist evaluates its livestock and determines which livestock it is

Rina8888 [55]

Any animals look for steangth mostly because the offspring can probobly onther trait is looks. The male will try to impess the female.

4 0

3 years ago

An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +

Maurinko [17]

Answer:

The simplified expression for the fraction is $\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$

Explanation:

From the given information:

O3* → O3 (1) fluorescence

O + O2 (2) decomposition

O3* + M → O3 + M (3) deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

$\text {X} = \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) } } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{ {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3 \times cM) }$

$\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$ since cM is the concentration of the inert molecule

7 0

3 years ago