How many grams of ammonia would be produced from the reaction of 2.64 moles of hydrogen gas reacting with excess nitrogen?

Gemiola [76]

Answer:

the physical and chemical properties of the products are different from the reactants

Explanation:

A chemical reaction involves the chemical combination of two or more elements/compounds called Reactants to give one or more different elements/compounds called Products. A chemical reaction occurs in such a way that the atoms of the reactants are restructured to form product(s) that is/are entirely different from the reactants.

In a chemical reaction, the physical and chemical properties of the products differ from that of the reactants since different chemical compounds/elements are formed as products. The physical properties of a substance, which include colour, melting and boiling point etc. will differ in the reactants and products formed. Also, the chemical structure and identity of the reactants will be changed to give rise to a different chemical property in the products.

7 0

3 years ago

Lab Report: Paper Chromatography

vampirchik [111]

Answer:

the answer is "I HOPE IT HELPS"

7 0

3 years ago

The rate constant for the reaction 3A equals 4b is 6.00×10 how long will it take the concentration of a to drop from 0.75 to 0.2

Lelu [443]

This question is incomplete, the complete question is;

The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.

how long will it take the concentration of A to drop from 0.75 to 0.25M ?

from the unit of the rate constant we know it is a second reaction order

OPTIONS

a) 2.2×10^−3 min

b) 5.5×10^−3 min

c) 180 min

d) 440 min

e) 5.0×10^2 min

Answer:

it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

Explanation:

Given that;

Rate constant K = 6.00 × 10⁻³ L.mol⁻¹min⁻¹

3A → 4B

given that it is a second reaction order;

k = 1/t [ 1/A - 1/A₀]

kt = [ 1/A - 1/A₀]

t = [ 1/A - 1/A₀] / k

K is the rate constant(6.00 × 10⁻³)

A₀ is initial concentration( 0.75 )

A is final concentration(0.25)

t is time required = ?

so we substitute our values into the equation

t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)

t = 2.6666 / (6.00 × 10⁻³)

t = 444.34 ≈ 440 min {significant figures}

Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

6 0

3 years ago

The energy of a photon needed to cause ejection of an electron from a photoemissive metal is expressed as the sum of the binding

slega [8]

Answer:

$Binding\ energy=43.43\times 10^{-20}\ J$

Explanation:

Using the expression for the photoelectric effect as:

$E=h\nu_0+\frac {1}{2}\times m\times v^2$

Also, $E=\frac {h\times c}{\lambda}$

$\nu_0=\frac {c}{\lambda_0}$

Applying the equation as:

$\frac {h\times c}{\lambda}=\frac {hc}{\lambda_0}+\frac {1}{2}\times m\times v^2$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

$\lambda$ is the wavelength of the light being bombarded

Given, $\lambda=4.00\times 10^{-7}\ m$

$\frac {hc}{\lambda_0}$ is the binding energy or threshold energy

$\frac {1}{2}\times m\times v^2$ is the kinetic energy of the electron emitted. = $6.26\times 10^{-20}\ J$

Thus, applying values as:

$\frac{h\times c}{\lambda}=Binding\ Energy+Kinetic\ Energy$

$\frac{6.626\times 10^{-34}\times 3\times 10^8}{4.00\times 10^{-7}}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$\frac{19.878}{10^{19}\times \:4}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$49.69\times 10^{-20}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$Binding\ energy=43.43\times 10^{-20}\ J$

5 0

3 years ago

30g of naoh is dissolved in 1.5 liter solution the active mass of naoh is?

Pepsi [2]

Answer:

0.5mol/L

Explanation:

First, let us calculate the number of mole NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the question = 30g

Number of mole = Mass /Molar Mass

Number of mole = 30/40 = 0.75mol

Volume = 1.5L

Active mass = mole/Volume

Active mass = 0.75mol/1.5L

Active mass = 0.5mol/L

7 0

3 years ago