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2 years ago

How many grams of S are in 475 g of SO2?

Chemistry

1 answer:

enot [183]2 years ago

6 0

There are 237. 5 g of Sulfur,S in 475 g of SO2?

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<h3>Calculation of grams of Sulfur</h3>

From the question, we can say that

The molar mass of sulfur = 32 g/mol

The molar mass of oxygen = 16 g/mol

Therefore,

The molar mass for SO2 = 32 + (16 × 2) g/mol = 64 g/mol

Now,

If 1 mole of SO2 contains 1 mole of S

Then 64 g of SO2, will contain 32g of S;

Such that

475 g of SO2 will give { $\frac{ 32g of S) ( 475 g of SO2)}{64 of SO2}$ } = 237. 5 g of Sulfur.

Learn more about molar mass here :brainly.com/question/18291695

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3 years ago

The enthalpy change for converting 1.00 mol of ice at -50.0 ∘c to water at 60.0∘c is ________ kj. the specific heats of ice, wat

guajiro [1.7K]

First, we have to get:

1- The heat required to increase T of ice from -50 to 0 °C:

according to q formula:

q1 = m*C*ΔT

when m is the mass of ice = mol * molar mass

= 1 mol * 18 mol/g

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and C is the specific heat capacity of ice = 2.09 J/g-K

and ΔT change in temperature = 0- (-50) = 50°C

by substitution:

∴q1 = 18 g * 2.09 J/g-K *50°C

= 1881 J = 1.881 KJ

2- the heat required to melt this mass of ice is :

q2 = n*ΔHfus

when n is the number of moles of ice = 1 mol

and ΔHfus = 6.01 KJ/mol

by substitution:

q2 = 1 mol * 6.01 KJ/mol

= 6.01 KJ

3- the heat required to increase the water temperature from 0°C to 60 °C is:

q3 = m*C*ΔT

when m is the mass of water = 18 g

C is the specific heat capacity of water = 4.18 J/g-K

ΔT is the change of Temperature of water = 60°C - 0°C = 60°C

by substitution:

∴q3 = 18 g * 4.18 J/g-K * 60°C

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3 years ago

For the generic equilibrium HA(aq) ⇌ H+(aq) + A−(aq), which of these statements is true? For the generic equilibrium , which of

timama [110]

<u>Answer:</u> The correct statement is if you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase.

<u>Explanation:</u>

Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect generally decreases the solubility of a solute.

Equilibrium reaction of HA and KA follows the equation:

$HA\rightleftharpoons H^{+}(aq.)+A^{-}(aq.)$

$KA\rightleftharpoons K^+(aq.)+A^{-}(aq.)$

According to Le-Chateliers principle, if there is any change in the variables of the reaction, the equilibrium will shift in the direction in order to minimize the effect.

In the equilibrium reactions, $A^-$ ion is getting increased on the product side, so the equilibrium will shift in the direction to minimize this effect, which is in the direction of HA.

Thus, the addition of KA will shift the equilibrium in the left direction.

Equilibrium constant depends on the temperature of the system. It does not have any effect on any change of pH.

pH is defined as the negative logarithm of hydrogen ions present in the solution

If the solution has high hydrogen ion concentration, then the pH will be low.
If the solution has low hydrogen ion concentration, then the pH will be high.

As, the equilibrium is shifting in the left direction, that means concentration of $H^+$ ions are getting decreases. This will increase the pH of the solution.

Hence, the correct statement is if you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase.

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