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3 years ago

14

Write any four limitations of a balanced chemical equation.

1 answer:

Valentin [98]3 years ago

3 0

Answer:

1. The physical states of the reactants and products.

2. The concentration of the reactants and products.

3. The conditions such as temperature, pressure or catalyst etc which affect the reaction.

4. The heat changes accompanying the reaction i.e. whether heat is evolved or absorbed during the reaction.

Explanation:

Hope it helps! ^w^

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What has to happen in order for dew to form on grass?

jeka57 [31]

It has to be moist in the grass

4 0

3 years ago

Rango carefully mixed 15.6 g of sodium with chlorine. The reaction produced 39.7 g of sodium chloride. How many grams of chlorin

Evgen [1.6K]

Answer:

24.1g of chlorine & 15.6g of sodium

Explanation:

according to the law of conservation of mass

Na+Cl=NaCl only

5 0

3 years ago

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

8 0

3 years ago

Titanium has an HCP crystal structure, a c/a ratio of 1.669, an atomic weight of 47.87 g/mol, and a density of 4.51 g/cm3. Compu

IrinaK [193]

Answer : The atomic radius for Ti is, $1.45\times 10^{-8}cm$

Explanation :

Atomic weight = 47.87 g/mole

Avogadro's number $(N_{A})=6.022\times 10^{23} mol^{-1}$

First we have to calculate the volume of HCP crystal structure.

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times V}$ .............(1)

where,

$\rho$ = density = $4.51g/cm^3$

Z = number of atom in unit cell (for HCP = 6)

M = atomic mass = 47.87 g/mole

$(N_{A})$ = Avogadro's number

V = volume of HCP crystal structure = ?

Now put all the values in above formula (1), we get

$4.51g/cm^3=\frac{6\times (47.87g/mol)}{(6.022\times 10^{23}mol^{-1}) \times V}$

$V=1.06\times 10^{-22}cm^3$

Now we have to calculate the atomic radius for Ti.

Formula used :

$V=6R^2c\sqrt{3}$

Given:

c/a ratio = 1.669 that means, c = 1.669 a

Now put (c = 1.669 a) and (a = 2R) in this formula, we get:

$V=6R^2\times (1.669a)\sqrt{3}$

$V=6R^2\times (1.669\times 2R)\sqrt{3}$

$V=(1.669)\times (12\sqrt{3})R^3$

Now put all the given values in this formula, we get:

$1.06\times 10^{-22}cm^3=(1.669)\times (12\sqrt{3})R^3$

$R=1.45\times 10^{-8}cm$

Therefore, the atomic radius for Ti is, $1.45\times 10^{-8}cm$

3 0

3 years ago

Write the complete balanced equation for the reaction between CaCO3 and KCI

EleoNora [17]

CaCO3 + 2KCL ⇒ CaCl2 + K2CO3

It is balanced as so based on the charges given on the periodic table and polyatomic ions.

Calcium has the charge of 2 but CO3 also shares the same charge, thus cancelling that out.

Potassium has a charge of 1 while Chlorine also shares a charge of 1, also cancelling it out.

Thus, if it performs a double replacement reaction, they would take these charges to the new elements that do not cancel out their charges.

Therefore, we need the coefficient of 2 in front of Potassium Chloride in order to balance the equation as on the products side of the equation, Potassium and Chlorine both have a subscript of 2.

Hope this helps!

5 0

3 years ago