What is Industrial chemistry?

(5.50 x 10^6) x (2.16x 10^-2)

frutty [35]

Answer:1.188

⋅

10

5

1.188 Times 10 to the fifth power

Explanation:

6 0

3 years ago

PLEASE HELP! Find out whether the chemical equation is balanced or unbalanced.

nordsb [41]

One chemical reaction is called the Haber process, a method for preparing ammonia by reacting nitrogen gas with hydrogen gas:

This equation shows you what happens in the reaction, but it doesn’t show you how much of each element you need to produce the ammonia. To find out how much of each element you need, you have to balance the equation — make sure that the number of atoms on the left side of the equation equals the number of atoms on the right.

You know the reactants and the product for this reaction, and you can’t change them. You can’t change the compounds, and you can’t change the subscripts, because that would change the compounds.

So the only thing you can do to balance the equation is add coefficients, whole numbers in front of the compounds or elements in the equation. Coefficients tell you how many atoms or molecules you have.

For example, if you write the following, it means you have two water molecules:

Each water molecule is composed of two hydrogen atoms and one oxygen atom. So with two water molecules (represented above), you have a total of 4 hydrogen atoms and 2 oxygen atoms.

You can balance equations by using a method called balancing by inspection. You take each atom in turn and balance it by adding appropriate coefficients to one side or the other.

With that in mind, take another look at the equation for preparing ammonia: HOPE THIS HELPS

4 0

3 years ago

. Helium is stored at 293 K and 500 kPa in a 1-cm-thick, 2-m-inner-diamater spherical container made of fused silica. The area w

Ivahew [28]

Answer:

(a) 45.17×10^-14 kg/s

(b) since the amount of helium escaped due to diffusion is insignificant, the final pressure drop in the tank remains the same as the initial pressure 500kPa.

Explanation:

Helium gas at temperature T=293k

Helium gas at pressure P= 500kPa

The inner diameter of spherical tank is $D_1 = 2m$

The inner radius of spherical tank is : $r_1 = \frac{D_1}{2}$

= $\frac{2}{2}$

=1m

Thickness of the container r = 1cm =0.01m

Outer radius of the spherical tank is ;

t = $r_2 - r_1$

$-r_2 = -t - r-1$

multiplying through with (-) we have ;

$r_2 = t + r_1$

$r_2 = 1 + 0.01m$

$r_2 = 1.01m$

From table of binary diffusion coefficients of solids, the diffusion coefficients of helium in silica is noted as

$D_A_B =4.0 ×10^-14 \frac{m^2}{s}$

From table molar mass and gas constant, the molecular weight of helium is:

M = 4.003kg/kmol

The solubility of helium in fused silica is determined from Table of Solubility of selected gases and soilids.

$S_He$ = 0.00045 kmol/m³. bar

Considering total molar concentration as constant, the molar concentration of helium inside the container is determined as

$C_B_I = S_H_e×P$

= 0.00045kmol/m³. bar × (5)

$C_B_I = 2.25×10^-3 kmol/m³$

From one dimensional mass transfer through spherical layers is expressed as:

$N_di_f_f= 4πr_1 r_2 D_A_B \frac{C_B_I - C_B_2}{r_2 - r_1}$

substituting all the values in the above relation, we have;

$M_di_f_f= 4π(1) (1.01) (4.0×10^-14) \frac{2.25 × 10^-3 -0}{1.01-1}$

$M_di_f_f=11.42×10^-14kmol/s$

(a) The mass flow rate is expressed as

$M_di_f_f = MN_diff$

$M_di_f_f=4.003×11.42×10^-14$

$M_di_f_f=45.71×10^-14kg/s$

(b) The pressure drop in the tank after a week;

For one week the mass flow rate of helium is

$N_di_ff =11.42×10^-14kmol/s$

$N_di_ff= 11.42×10^-14×7×24×3600 kmol/week$

$N_di_f_f=6.9×10^-8kmol/week$

The volume of the spherical tank is $V=\frac{4}{3} πr_1^3$

$V=\frac{4}{3}π×1^3$

V = 4.189m³

The initial mass of helium in the sphere is determined from the ideal gas equation:

PV=NRT

where R is the universal gas constant and its value is R = 8.314 KJ/Kmol.k

N= PV/RT

N= 500 × 4.189/ 8.314 × 293

N= 0.86kmol

The number of moles of helium gas remaining in the tank after one week is:

$N_di_f_f-final = 0.86 - 6.9 × 10^-8 kmol/week$

$N_di_f_f-final ≅ 0.86$

therefore, since the amount of helium escaped due to diffusion is insignificant, the final pressure drop in the tank remains the same as the initial pressure 500kPa.

3 0

3 years ago

If sea water has a density of 1.025 kg/L and 3.5% of the mass in sea water is salt, determine the mass of salt dissolved in 500

Alja [10]

68 kg. There are 58 kg salt in 500 gal seawater.

Step 1. Convert gallons to litres

1 US gal = 3.79 L (1 Imp gal = 4.55 L)

Step 2. Find the volume of the seawater

Volume = 500 gal × (3.79 L/1 gal) = 1895 L

Step 3. Find the mass of the seawater

Mass = 1895 L × (1.025 kg/1 L) = 1942 kg

Step 4. Find the mass of the salt

Mass of salt = 1942 kg seawater × (3.5 kg salt/100 kg seawater) = 68 kg salt

6 0

4 years ago

Please help me, please

viva [34]

C of course is the correct

3 0

3 years ago

What is Industrial chemistry?​

What is Industrial chemistry?