The most common sulfur ion has a charge of

What is an isotope? Explain why isotopes having different number of neutrons are still considered to be the same element on the

Mazyrski [523]

Isotopes of any given factor all incorporate the equal variety of protons, so they have the identical atomic wide variety (for example, the atomic wide variety of helium is usually 2). Isotopes of a given factor include exceptional numbers of neutrons, therefore, special isotopes have special mass numbers.

3 0

3 years ago

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PLEASE ANSWER ASAP

marta [7]

Its an inelastic collision because the force from the bat causes it to bounce back. It is also an elastic force because catching the ball call for the energy of the ball to be deformed and restored into the mitt.

8 0

3 years ago

N204(0) + 2NO2(g)

user100 [1]

setup 1 : to the right

setup 2 : equilibrium

setup 3 : to the left

<h3>Further explanation</h3>

The reaction quotient (Q) : determine a reaction has reached equilibrium

For reaction :

aA+bB⇔cC+dD

$\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}$

Comparing Q with K( the equilibrium constant) :

K is the product of ions in an equilibrium saturated state

Q is the product of the ion ions from the reacting substance

Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)

Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium

Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)

Keq = 6.16 x 10⁻³

Q for reaction N₂O₄(0) ⇒ 2NO₂(g)

$\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}$

Setup 1 :

$\tt Q=\dfrac{0.0064^2}{0.098}=0.000418=4.18\times 10^{-4}$

Q<K⇒The reaction moved to the right (products)

Setup 2 :

$\tt Q=\dfrac{0.0304^2}{0.15}=0.00616=6.16\times 10^{-3}$

Q=K⇒the system at equilibrium

Setup 3 :

$\tt Q=\dfrac{0.230^2}{0.420}=0.126$

Q>K⇒The reaction moved to the left (reactants)

8 0

2 years ago

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Is the process represented below an example of a physical or a chemical

Basile [38]

It is a physical change because only the states as being changes, not the actual bonds in the compound.

6 0

2 years ago

15.0 moles of gas are in a 8.00 L tank at 22.3 ∘C∘C . Calculate the difference in pressure between methane and an ideal gas unde

leva [86]

Answer:

$\Delta P=4.10atm$

Explanation:

Hello!

In this case, since the ideal gas equation is used under the assumption of no interaction between molecules and perfectly sphere-shaped molecules but the van der Waals equation actually includes those effects, we can compute each pressure as shown below, considering the temperature in kelvins (22.3+273.15=295.45K):

$P^{ideal}=\frac{nRT}{V}=\frac{15.0mol*0.08206\frac{atm*L}{mol*K}*295.45K}{8.00L}=45.5atm$

Next, since the VdW equation requires the molar volume, we proceed as shown below:

$v=\frac{8.00L}{15.0mol}=0.533\frac{L}{mol}$

Now, we use its definition:

$P^{VdW}=\frac{RT}{v-b} -\frac{a}{v^2}$

Thus, by plugging in we obtain:

$P^{VdW}=\frac{0.08206\frac{atm*L}{mol*K}*295.45K}{0.533mol/L-0.0430L/mol} -\frac{2.300L^2*atm/mol^2}{(0.533L/mol)^2}\\\\P^{VdW}=49.44atm-8.09atm\\\\P^{VdW}=41.4atm$

Thus, the pressure difference is:

$\Delta P=45.5atm-41.4atm\\\\\Delta P=4.10atm$

Best regards!

6 0

3 years ago