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katen-ka-za [31]

Why not search it up and figure it out and then write it how you would?

6 0

3 years ago

WILL MARK AS BRAINIEST PLEASE HELP 20 POINTS

Alisiya [41]

The subscript is the amount of atoms in each molecule and the coefficient is the amount of molecules. there are 4 Hydrogen, 2 Sulfur, and 8 Oxygen in this particular substance.

6 0

3 years ago

as a gas occupied a volume of 250cc at 250at 700mm pressure temperature,25degree Celsius what additional pressure is required?re

Allisa [31]

Additional pressure required : 175 mm

<h3>Further explanation</h3>

Boyle's Law

At a constant temperature, the gas volume is inversely proportional to the pressure applied

$\rm p_1V_1=p_2.V_2\\\\\dfrac{p_1}{p_2}=\dfrac{V_2}{V_1}$

V₁=250 ml

P₁=700 mmHg

$\tt V_2=\dfrac{4}{5}\times 250=200~ml$

$\tt P_2=\dfrac{P_1.V_1}{V_2}\\\\P_2=\dfrac{700\times 250}{200}=875~mmHg$

Additional pressure :

$\tt 875-700=175~mm$

4 0

3 years ago

Order the follow processes from (1) the least work done by the system to (5) the most work done by one mole of an ideal gas at 2

quester [9]

Answer : The order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

Explanation :

<u>The formula used for isothermally irreversible expansion is :</u>

$w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)$

where,

w = work done

$p_{ext}$ = external pressure

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

<u>The expression used for work done in reversible isothermal expansion will be,</u>

$w=-nRT\ln (\frac{V_2}{V_1})$

where,

w = work done = ?

n = number of moles of gas = 1 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = $25^oC=273+25=298K$

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

First we have to determine the work done for the following process.

(1) An isothermal expansion from 1 L to 10 L at an external pressure of 2.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(2.5atm)\times (10-1)L$

$w=-22.5L.atm=-22.5\times 101.3J=-2279.25J$

(2) A free isothermal expansion from 1 L to 100 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{100L}{1L})$

$w=-11409.6J$

(3) A reversible isothermal expansion from 0.5 L to 4 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{4L}{0.5L})$

$w=-5151.97J$

(4) A reversible isothermal expansion from 0.5 L to 40 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{40L}{0.5L})$

$w=-10856.8J$

(5) An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(0.5atm)\times (100-1)L$

$w=-49.5L.atm=-49.5\times 101.3J=-5014.35J$

Thus, the order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

8 0

3 years ago

Brainliest Available! Thank you in advance!

liraira [26]

Answer:

3.0×10⁻¹³ M

Explanation:

The solubility product Ksp is the product of the concentrations of the ions involved. This relation can be used to find the solubility of interest.

<h3>Equation</h3>

The power of each concentration in the equation for Ksp is the coefficient of the species in the balanced equation.

Ksp = [Al₃⁺³]×[OH⁻]³

<h3>Solving for [Al₃⁺³]</h3>

The initial concentration [OH⁻] is that in water, 10⁻⁷ M. The reaction equation tells us there are 3 OH ions for each Al₃ ion. If x is the concentration [Al₃⁺³], then the reaction increases the concentration [OH⁻] by 3x.

This means the solubility product equation is ...

Ksp = x(10⁻⁷ +3x)³

For the given Ksp = 3×10⁻³⁴, we can estimate the value of x will be less than 10⁻⁸. This means the sum will be dominated by the 10⁻⁷ term, and we can figure x from ...

3.0×10⁻³⁴ = x(10⁻⁷)³

Then x = [Al₃⁺³] will be ...

$[\text{Al}_3^{\,+3}]=\dfrac{3.0\times10^{-34}}{10^{-21}}\approx \boxed{3.0\times10^{-13}\qquad\text{moles per liter}}$

We note this value is significantly less than 10⁻⁷, so our assumption that it could be neglected in the original Ksp equation is substantiated.

__

<em>Additional comment</em>

The attachment shows the solution of the 4th-degree Ksp equation in x. The only positive real root (on the bottom line) rounds to 3.0×10^-13.

6 0

2 years ago

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