Identify the element that cannot participate in nuclear fission reactions. (Hint: Think about the size of the atom.)

What are the four categories commonly used to classify chemicals and chemical agents in the graphic communications industry

Oduvanchick [21]

Answer:

Organic

Plate making

Ink mists

Gas, fumes and dust

Explanation:

the four categories commonly used to classify chemicals and chemical agents in the graphic communications industry are Organic ,Plate making ,Ink mists, Gas, fumes and dust.

Organic refer to those chemicals gotten from living matter and are natural.

Plate making refers to how plates are make and it carry image in the printing process.

Ink mist gotten from ink is use in graphic designs for painting and color indication.

5 0

2 years ago

For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [

sveta [45]

Explanation:

The given data is as follows.

$E^{o}$ = 0.483, $[Co^{3+}]$ = 0.173 M,

$[Co^{2+}]$ = 0.433 M, $[Cl^{-}]$ = 0.306 M,

$P_{Cl_{2}}$ = 9.0 atm

According to the ideal gas equation, PV = nRT

or, P = $\frac{n}{V}RT$

Also, we know that

Density = $\frac{mass}{volume}$

So, P = MRT

and, M = $\frac{P}{RT}$

= $\frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}$

= $\frac{9.0}{24.436}$

= 0.368 mol/L

Now, we will calculate the cell potential as follows.

E = $E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}$

= $0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}$

= $0.483 - 0.02955 log \frac{0.0689}{0.0162}$

= $0.483 - 0.02955 \times 0.628$

= 0.483 - 0.0185

= 0.4645 V

Thus, we can conclude that the cell potential of given cell at $25^{o}C$ is 0.4645 V.

4 0

3 years ago

carbon dioxide at 25 degrees celsius and 101.3 kpa has a density of 1.799 kg/m^3 determine gas constant

dusya [7]

Answer:

Explanation:

Temperature of gas in absolute scale T = 25 + 273 = 298 .

pressure of gas P = 1.013 x 10⁵ N / m²

density D = 1.799 kg / m³

= 1799 g / m³

From gas formula

PV / T = n R Where P is pressure , V is volume and T is absolute temperature , n is no of moles

P / T = n R / V

P / T = m R /M V where m is mass of gas and M is molecular weight .

m / V = D ( density )

P / T = DR/ M

PM / DT = R

Putting the values

1.013 x 10⁵ x 44 / (1799 x 298)

R = 8.314.09 J / K mole

5 0

2 years ago

7. What determines what type of element an atom is?

frozen [14]

The amount of protons

3 0

2 years ago

If the molecular weight of a semiconductor is 27.9 grams/mole and the diamond lattice constant is 0.503 nm, what is the density

adelina 88 [10]

Explanation:

The given data is as follows.

Mass = 27.9 g/mol

As we know that according to Avogadro's number there are $6.023 \times 10^{26}$ atom present in 1 mole. Therefore, weight of 1 atom will be as follows.