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3 years ago

Can 750 mL of water dissolve 0.60 mol of gold(III) chloride (AuCl3)?

Chemistry

1 answer:

nydimaria [60]3 years ago

3 0

Answer:

Yes

Explanation:

1. Mass of 0.60 mol of AuCl₃

$\text{Mass} = \text{0.60 mol} \times \dfrac{\text{303.33 g}}{\text{1 mol}} = \text{184 g}$

2. Mass of AuCl₃ in 750 mL

The solubility of AuCl₃ is 68 g/100 mL.

In 750 mL of water, you can dissolve

$\text{Mass of AuCl}_{3} = \text{750 mL} \times \dfrac{\text{68 g}}{\text{100 mL}} = \text{510 g AlCl}_{3}$

∴ Yes, 750 mL of water can dissolve 0.60 mol of AuCl₃.

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What is the oxidation number for CO2

anzhelika [568]

Answer:

Explanation:

The oxidation number of C in carbon dioxide (CO2) is (rules 1 & 2): 0 + (2 x 2) = +4 [Check (rule 3): +4 + 2(-2) = 0] The oxidation number of C in methane (CH4) is (rules 1 & 2): 0 – (4 x1) = -4 [Check (rule 3): -4 + 4(-1) = 0].

7 0

3 years ago

Calculate the moles of calcium oxalate formed from 5 grams of potassium oxalate (molar mass 184.24 g/mol) the mole mole ratio fr

Nataly_w [17]

5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.

Calcium oxalate (CaC₂O₄) is obtained by the reaction of 5 g of potassium oxalate (K₂C₂O₄).

We can calculate the moles of CaC₂O₄ obtained considering the following relationships.

The molar mass of K₂C₂O₄ is 184.24 g/mol.
The mole ratio of K₂C₂O₄ to CaC₂O₄ is 2:1.

$5 g K_2C_2O_4 \times \frac{1molK_2C_2O_4}{184.24gK_2C_2O_4} \times \frac{1molCaC_2O_4}{2molK_2C_2O_4} = 0.03molCaC_2O_4$

5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.

Learn more: brainly.com/question/15288923

7 0

3 years ago

One way to represent this equilibrium is: 2 Al(s) 3 Br2(l)2 AlBr3(s) We could also write this reaction three other ways, listed

myrzilka [38]

The question is incomplete, the complete question is;

Aluminum metal and bromine liquid (red) react violently to make aluminum bromide (white powder). One way to represent this equilibrium is:

Al(s) + 3/2 Br2(l)AlBr3(s)

We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above.

1) 2 AlBr3(s) 2 Al(s) + 3 Br2(l)

2) 2 Al(s) + 3 Br2(l) 2 AlBr3(s)

3) AlBr3(s) Al(s) + 3/2 Br2(l)

Answer:

See explanation

Explanation:

We have that; Al(s) + 3/2 Br2(l)AlBr3(s)

So;

Al(s) + 3/2 Br₂(l) = AlBr₃(s)

K = [ AlBr₃] / [ Al] [ Br₂]³/²

K² = [ AlBr₃]² / [ Al ] ² [ Br₂]³

Now;

1) 2 AlBr₃ = 2 Al(s) + 3 Br₂(l) =

K₁ = [ Al ] ² [ Br₂]³ / [ AlBr₃]²

K₁ = ( 1 / K² ) = K⁻²

For the second reaction;

2 ) 2 Al(s) + 3 Br₂(l) = 2 AlBr₃(s)

K₂ = [ AlBr₃ ]² / [ Al ]² [ Br₂ ]³

K₂ = K²

For the third reaction;

3 )

AlBr₃(s) = Al(s) + 3/2 Br₂(l)

K₃ = [ Al ] [ Br₂ ] ³/² / [ AlBr₃ ]

= ( 1 / K ) = K⁻¹

7 0

3 years ago

Why is it important to know your variables when conducting an investigatory problem ?

Juli2301 [7.4K]

You can harm ur self if you dont know variables . Bcs the values of variables should limit.

5 0

4 years ago

Sydney was organizing a box of resistors by resistance range. She knows that there are several factors that go into determining

kati45 [8]

The resistance of the wire is dependent on temperature, diameter and nature of the material.

<h3>What is resistance?</h3>

The term resistance refers to the opposition offered to the flow of current in a circuit. The resistance is obtained form Ohm's law which states that;

V = IR

Where;

V = voltage

I = current

R = resistance

Hence, the resistance is dependent on;

temperature
diameter
material

Learn more about resistance: brainly.com/question/15067823

7 0

2 years ago