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Slav-nsk [51]
3 years ago
11

How does the law of definite proportions apply to hydrates?

Chemistry
1 answer:
vladimir1956 [14]3 years ago
4 0
The law of definite proportions would state that a hydrate always contain exactly the same proportion of salt and water by mass.
strictly speaking, the law of definite proportion states that a compound always 
contains exactly the same proportion of elements by mass.
But the law is often applied to groupings of elements in compound.
Hydrates are salt that have a certain amount of  water asa part of their structure.
The water is chemically combined with the compound in a definite ratio.
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"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"
Andru [333]

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

<em>Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical</em>

<em />

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

  • <em>pKa NH₃/NH₄⁺</em>

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = <em>4.74</em>

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

  • <em>Moles NH₃</em>

<em>2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃</em>

  • <em>H-H equation:</em>

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 =  log [0.400 moles] / [NH₄Cl]

0.0087 =  [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

<h3>245.66g of NH₄Cl is the mass we need to add to obtain the desire pH</h3>

<em />

3 0
4 years ago
Two processes are described below:
UNO [17]

Answer:

i would say D i just did this but i kinda forgot so sorry if im wrong or A

Explanation:

4 0
3 years ago
Read 2 more answers
The valences of metal x,y and z are 1,2 and 3 respectively. What are the formulae of their;a) hydroxides, b) sulphates, c) hydro
Rina8888 [55]

Answer:

See answer below

Explanation:

AS we know that the valence for those metals X, Y, and Z are 1, 2 and 3, we can determine the formula of each compound.

1. Hydroxides.

An hydroxide is formed when an oxyde of a metal reacts with water. When this happens, the general molecular formula is:

Meₐ(OH)ₙ

Where:

a: valence or charge of the hydroxide (Which is -1)

n: valence of the metal.

Following this, the formula for X, Y and Z would be:

XOH

Y(OH)₂

Z(OH)₃

2. Sulphates

Sulphates follow a similar rule of hydroxide in the general molecular formula, but instead of having a charge of -1, it has a charge of -2 so:

Mₐ(SO₄)ₙ

So, following the rule:

X₂SO₄

Y₂(SO₄)₂ ------> YSO₄

Z₂(SO₄)₃

3. Hydrogens

Following the same rule as the previous, hydrogens works with a charge of -1, so:

MₐHₙ

Then:

XH

YH₂

ZH₃

4. Carbonates.

This follows the same rule as sulphates, with the same charge so:

Mₐ(CO₃)ₙ

Then:

X₂CO₃

YCO₃

Z₂(CO₃)₃

5. Nitrates

Follow the same rule as the hydroxides, with the same charge of -1.

Mₐ(NO₃)ₙ

Then:

XNO₃

Y(NO₃)₂

Z(NO₃)₂

6. Phosphates

In the case of phosphates, these have a charge of -3 so:

Mₐ(PO₄)ₙ

Then:

X₃PO₄

Y₃(PO₄)₂

Z₃(PO₄)₃ ----> ZPO₄

Hope this helps

6 0
3 years ago
The equilibrium constant (Kp) for the decomposition of phosphorus pentachloride (PCl5) to phosphorus trichloride (PCl3) and mole
Oksanka [162]

Answer: The partial pressure of Cl_2 is 1.86 atm

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_c

The given balanced equilibrium reaction is,

                            PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)

Pressure at eqm.     0.973 atm                 0.548atm      x atm

The expression for equilibrium constant for this reaction will be,

K_c=\frac{(p_{PCl_3}\times (p_{Cl_2})}{(p_{PCl_5})}

Now put all the given values in this expression, we get :

1.05=\frac{(0.548)\times (x)}{(0.973)}

By solving the term 'x', we get :

x = 1.86 atm

Thus, the partial pressure of Cl_2 is 1.86 atm

4 0
3 years ago
Give two examples of steroids
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3 0
3 years ago
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