CHEMISTRY COLLEGE

Which compound contains covalent bonds?A - BeO
B- NaCl
C- CaS
D- MgF2
E - SO2

Answers

Answer 1

Answer:

Answer:

Explanation:

Thus is because both the elements are non metals.

Answer 2

Answer:

Answer:

SO2

Explanation:

SO2 is covalent since it is formed by the sharing of electrons between sulphur and oxygen also both the sulphur and oxegen atoms are non metal.

and we know that non-metal non-metal always bond together via covelent bonding

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Iron has density of 7.9g/cm3. What is the mass of a cube with the length of one side equal to 1.64x10squared cm3?

Answers

Given data:

Density of iron (Fe) = 7.9 g/cm3

Length of one side of the iron cube = 1.64 * 10^2 cm

Now, the volume (V) of a cube in which the length of the side is 'a' cm is given as:

V = a^3

Volume of iron cube = (1.64 *10^2 cm)^3 = 4.41 * 10^6 cm3

The density (D) of an object of mass (m) and volume (V) is given as:

D = m/V

or, m = D*V

Therefore, mass of iron cube = 7.9 g/cm3 * 4.41 * 10^6 cm3

= 34.84 *10^6 g

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 in a CSTR. When pure A is fed to a 10 dm 3 PFR at 300 K and a volumetric flow rate of 5 dm 3 /s, the conversion is 80%. When a mixture of 50% A and 50% inert (I) is fed to a 10 dm 3 CSTR at 320 K and a volumetric flow rate of 5 dm 3 /s, the conversion is also 80%. What is the activation energy in cal/mol

Answers

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.

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

Rearrange the formula is as follows.

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

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)

$k=(5m^(3)/s)/(10dm^(3))[(1+1)ln (1)/(1-0.8)-1 * 0.8] = 1.2s^(-1)$

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.

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

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)

$(10dm^(3))/(5dm^(3))=(0.8(1+0.5(0.8)))/(k(1-0.8))=2.8s^(-1)$

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

$k(T_(2))=k(T_(1))exp[(E)/(R)((1)/(T_(1))-(1)/(T_(2)))]$

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

Rearrange the above formula is as follows.

$E= R *((T_(1)T_(2))/(T_(1)-T_(2)))ln(k(T_(2)))/(k(T_(1)))$

Substitute the all values.

$=1.987cal/molK((300K *320K)/(320K *300K))ln (2.8)/(1.2)=8.081 *10^(3)cal\,mol^(-1)$

$=8.1\,kcal\,mol^(-1)$

Therefore, the activation energy is $=8.1\,kcal\,mol^(-1)$

Calculate the number of C atoms in 9.837 x 1024 molecules of CO2.

Please help

Answers

Answer:

Explanation:

1 molecule contains 1 carbon atom.

9.837 * 10^24 molecules contains 9.837 * 10^24 atom of carbon.

It's a 1 to 1 ratio.

A certain substance X condenses at a temperature of 120.7 degree C. But if a 500, g sample of X is prepared with 55.4 g of urea (NH_2)_2 CO) dissolved in it, the sample is found to have a condensation point of 125.2 degree C instead. Calculate the molal boiling point elevation constant K_b of X. Round your answer to 2 significant digits.

Answers

Answer: The molal boiling point elevation constant $k_b$ of X is $2.4^0C/m$

Explanation:

Formula used for Elevation in boiling point :

$\Delta T_b=k_b* m$

or,

$T_b-T^o_b=i* k_b* (w_2* 1000)/(M_2* w_1)$

where,

$T_b-T^o_b =(125.2-120.7)^0C=4.5^0C$

$k_b$ = boiling point constant = ?

m = molality

$w_2$ = mass of solute (urea) = 55.4 g

$w_1$ = mass of solvent X = 500 g

$M_2$ = molar mass of solute (urea) = 60 g/mol

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

$4.5^oC=k_b* (55.4g* 1000)/(60* 500g)$

$k_b=2.4^0C/m$

Thus the molal boiling point elevation constant $k_b$ of X is $2.4^0C/m$

Isn't this false? For the industrial production of indigo carmine, a blue food colouring additive, a synthetic process with an E-factor of 17.4 produces less waste than a synthetic process with an E-factor of 3.0.

The answer I got was False, is this correct?

Answers

Answer: yes it is false

Explanation:

The statement is false. A synthetic process with a lower E-factor produces less waste than a process with a higher E-factor.

The E-factor is a measure of the waste generated during a manufacturing process. It is calculated by dividing the total mass of waste produced by the mass of the desired product. A lower E-factor indicates that less waste is generated per unit of product.

In this case, the synthetic process with an E-factor of 3.0 produces less waste than the process with an E-factor of 17.4. This means that the process with an E-factor of 3.0 is more efficient in terms of waste reduction.

Give an example of a substance that is MORE dense in its solid state when compared to its liquid state.