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1 year ago

For the reaction of sodium bromide with chlorine gas

Chemistry

1 answer:

lina2011 [118]1 year ago

7 0

The reaction of sodium bromide with chlorine gas is Cl₂(aq) + 2Na + 2Br? 2Na + 2Cl⁻ + Br²(aq).

<h3>What is sodium bromide?</h3>

Sodium bromide is an inorganic compound, white, crystalline with high melting point.

The reaction between halogens is redox reaction

Oxidation – 2Br⁻ ? Br₂ + 2e⁻ loss of electron.

Reduction – Cl₂ + 2e⁻ ? 2Cl⁻ gains of electron.

Thus, the correct option is Cl₂(aq) + 2Na + 2Br? 2Na + 2Cl⁻ + Br²(aq).

Learn more about sodium bromide

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Deep sea divers use a mixture of helium and oxygen to breathe. Assume that a diver is going to a depth of 150 feet where the tot

Svetlanka [38]

Answer:

4.525% is the percentage by volume of oxygen in the gas mixture.

Explanation:

Total pressure of the mixture = p = 4.42 atm

Partial pressure of the oxygen = $p_1=0.20 atm$

Partial pressure of the helium = $p_2$

$p_1=p\times \chi_1$ (Dalton law of partial pressure)

$0.20 atm=4.42 atm\times \chi_1$

$\chi_1=\frac{0.20 atm}{4.42 atm}=0.04525$

$\chi_2=1-\chi_1=1-0.04525=0.95475$

$chi_1+chi_2=1$

$n_1=0.04525 mol,n_2=0.95475 mol$

According Avogadro law:

$Moles\propto Volume$ (At temperature and pressure)

Volume occupied by oxygen gas = $V_1$

Total moles of gases = n = 1 mol

Total Volume of the gases = V

$\frac{n_1}{V_1}=\frac{n}{V}$

$\frac{V_1}{V}=\frac{n_1}{n}=\frac{0.04525 mol}{1 mol}$

Percent by volume of oxygen in the gas mixture:

$\frac{V_1}{V}\times 100=\frac{0.04525 mol}{1 mol}\times 100=4.525\%$

6 0

3 years ago

What is the associated deBroglie wavelength of a H2 molecule moving on one direction with kinetic energy of (3/2 kT) at 30 K

andrey2020 [161]

Answer: The de-Broglie's wavelength of a hydrogen molecule is $3.26\AA$

Explanation:

Kinetic energy is the measure of temperature of the system.

The equation used to calculate kinetic energy of a particle follows:

$E=\frac{3}{2}kT$

where,

E = kinetic energy of the particles = ?

k = Boltzmann constant = $1.38\times 10^{-23}J/K$

T = temperature of the particle = 30 K

Putting values in above equation, we get:

$E=\frac{3}{2}\times 1.38\times 10^{-23}J/K\times 30K\\\\E=6.21\times 10^{-22}J$

Calculating the mass of 1 molecule of hydrogen gas:

Conversion factor used: 1 kg = 1000 g

1 mole of hydrogen gas has a mass of 2 grams or $2\times 10^{-3}kg$

According to mole concept:

$6.022\times 10^{23}$ number of molecules occupy 1 mole of a gas.

As, $6.022\times 10^{23}$ number of hydrogen molecules has a mass of $2\times 10^{-3}kg$

So, 1 molecule of hydrogen will have a mass of = $\frac{2\times 10^{-3}kg}{6.022\times 10^{23}}\times 1=3.32\times 10^{-27}kg$

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

$\lambda=\frac{h}{\sqrt{2mE_k}}$

where,

$\lambda$ = De-Broglie's wavelength = ?

h = Planck's constant = $6.624\times 10^{-34}Js$

m = mass of 1 hydrogen molecule = $3.32\times 10^{-27}kg$

$E_k$ = kinetic energy of the particle = $6.21\times 10^{-22}J$

Putting values in above equation, we get:

$\lambda=\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 3.32\times 10^{-27}kg\times 6.21\times 10^{-22}J}}$

$\lambda=3.26\times 10^{-10}m=3.26\AA$ (Conversion factor: $1\AA=10^{-10}m$ )

Hence, the de-Broglie's wavelength of a hydrogen molecule is $3.26\AA$

3 0

3 years ago

What is the equilibrium membrane potential due to na ions if the extracellular concentration of na ions is 142 mm and the intrac

Ganezh [65]

The equilibrium membrane potential is 41.9 mV.

To calculate the membrane potential, we use the Nernst Equation:

V_Na = (RT)/(zF) ln{[Na]_o/[Na]_ i}

where

• V_Na = the equilibrium membrane potential due to the sodium ions

• R = the universal gas constant [8.314 J·K^(-1)mol^(-1)]

• T = the Kelvin temperature

• z = the charge on the ion (+1)

• F = the Faraday constant [96 485 C·mol^(-1) = 96 485 J·V^(-1)mol^(-1)]

• [Na]_o = the concentration of Na^(+) outside the cell

• [Na]_i = the concentration of Na^(+) inside the cell

∴ V_Na =

[8.314 J·K^(-1)mol^(-1) × 293.15 K]/[1 × 96 485 J·V^(-1)mol^(-1)] ln(142 mM/27 mM) = 0.025 26 V × ln5.26 = 1.66× 25.26 mV = 41.9 mV

4 0

3 years ago

For the reaction shown, calculate how many moles of NH3 form when 16.72 moles of reactant completely reacts:

Agata [3.3K]

Answer : The moles of $NH_3$ formed are, 22.3 moles.

Explanation : Given,

Moles of $N_2H_4$ = 16.72 mol

The given chemical reaction is:

$3N_2H_4(l)\rightarrow 4NH_3(g)+2N_2(g)$

From the balanced chemical reaction, we conclude that:

As, 3 moles of $N_2H_4$ react to give 4 moles of $NH_3$

So, 16.72 moles of $N_2H_4$ react to give $\frac{4}{3}\times 16.72=22.3$ moles of $NH_3$

Therefore, the moles of $NH_3$ formed are, 22.3 moles.

8 0

3 years ago

With what compound will nh3 experience only dispersion intermolecular forces? with what compound will nh3 experience only disper

Burka [1]

Answer is: ammonia experience only dispersion intermolecular forces with BF₃ (boron trifluoride) because BF₃ is only nonpolar molecule (vectors of dipole moments cansel each other, dipole moment is zero).
The London dispersion force (intermolecular force) is a temporary attractive force between molecules.

6 0

2 years ago

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