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3 years ago

How many atoms are in KOH?

Chemistry

1 answer:

Nata [24]3 years ago

3 0

Answer:

KOH contains only one K, so a mole of KOH contains one mole of K. How many atoms of K are in 1 mole of KOH? There are 6.022 × 1023 atoms of potassium in every mole of potassium. Since one mole of KOH contains one mole of K, the answer is 6.022×1023 atoms of K.

Explanation:

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What happens when you touch two hot bowls made of different materials? The wooden and metal bowl are both at the same temperatur

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3 years ago

What mass of salt would you need to add to 1.00kg of water to achieve a freezing point of -5 degrees C

Serga [27]

Answer:

The type of salt to be added to the water is not known from the question but no worries, I will try to give you the step by step procedure to answer any type of question similar to this.

To answer this question, we should know some facts.

1. the molar freezing point depression constant of water (Kf) = 1.86 K kg/mol

2. the molar mass of the salt if NaCl = 58.5 g/mol ; KCl = 74.5 g/mol

3. since the salt can dissociate if NaCl or KCl into two ions, the Van't Hoff factor ( i )= 2

Note that: the change in freezing point, molarity, deepression constant and van't Hoff factor are related by this formula;

ΔTf = i Kf m

So lets take NaCl as the salt:

Molar mass = 58.5 g/mol

Van't Hoff factor = 2

1. calculate the number of moles

So we can calculate the molarity of the salt NaCl from the formula;

m = ΔTf / i Kf

m = 5 / 2 * 1.86

m = 5 / 3.72

m = 1.344 mol/kg

2. calculate the number of moles of the salt required

Next is to multiply the molarity by the mass of water. Density of water = 1kg/L

number of moles = 1.344 mol/kg * 1 Kg/L * 1 kg water

number of moles = 1.344 moles.

3. calculate the mass of the salt.

numner of moles = mass / molar mass

mass = number of moles * molar mass

mass = 1.344 * 58.5

mass = 78.624 g of NaCl salt.

You can follow these steps to solve for the type of salt you are given in the question.

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4 years ago

Consider the reaction: A (aq) ⇌ B (aq) at 253 K where the initial concentration of A = 1.00 M and the initial concentration of B

Ugo [173]

Answer:

The maximum amount of work that can be done by this system is -2.71 kJ/mol

Explanation:

Maximum amount of work denoted change in gibbs free energy $(\Delta G)$ during the reaction.

Equilibrium concentration of B = 0.357 M

So equilibrium concentration of A = (1-0.357) M = 0.643 M

So equilibrium constant at 253 K, $K_{eq}= \frac{[B]}{[A]}$

[A] and [B] represent equilibrium concentrations

$K_{eq}=\frac{0.357}{0.643}=0.555$

When concentration of A = 0.867 M then B = (1-0.867) M = 0.133 M

So reaction quotient at this situation, $Q=\frac{0.133}{0.867}=0.153$

We know, $\Delta G=RTln(\frac{Q}{K_{eq}})$

where R is gas constant and T is temperature in kelvin

Here R is 8.314 J/(mol.K), T is 253 K, Q is 0.153 and $K_{eq}$ is 0.555

So, $\Delta G=8.314\times 253\times ln(\frac{0.153}{0.555})mol/K$

= -2710 J/mol

= -2.71 kJ/mol

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3 years ago