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3 years ago

What laws did dalton propose

Chemistry

1 answer:

Zina [86]3 years ago

6 0

Answer:

Dalton based his theory on two laws: the law of conservation of mass and the law of constant composition. The law of conservation of mass says that matter is not created or destroyed in a closed system.

Explanation:

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Give the oxidation state of the metal species in each complex. ru(cn)(co)4 -

kogti [31]

The given complex ion is as follow,

[Ru (CN) (CO)₄]⁻

Where;
[ ] = Coordination Sphere

Ru = Central Metal Atom = <span>Ruthenium

CN = Cyanide Ligand

CO = Carbonyl Ligand

The charge on Ru is calculated as follow,

Ru + (CN) + (CO)</span>₄ = -1
Where;
-1 = overall charge on sphere

0 = Charge on neutral CO

-1 = Charge on CN

So, Putting values,

Ru + (-1) + (0)₄ = -1

Ru - 1 + 0 = -1

Ru - 1 = -1

Ru = -1 + 1

Ru = 0
Result:
<span>Oxidation state of the metal species in each complex [Ru(CN)(CO)</span>₄]⁻ is zero.

3 0

4 years ago

Read 2 more answers

Predidt the direction and the relative value of the dipole moment of the following bond: C-Cl

WINSTONCH [101]

Answer:

D. C → Cl electronegativity difference > 0.5

Explanation:

The electro negativity of an atom in a compound refers to its ability to attract the electrons of a bond towards itself.

On the Pauling's scale, carbon has an electro negativity value of 2.55 while that of chlorine is 3.16. The difference in electro negativity between the both atoms is about 0.61.

The dipole is aslways directed towards the more electronegative atom. Hence the direction is ; C → Cl

5 0

3 years ago

PLEASE HELP!!!!!!!

dsp73

Answer: 1. $2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu$

2. 3 moles of $CuCl_2$ : 2 moles of $Al$

3. 0.33 moles of $CuCl_2$ : 0.92 moles of $Al$

4. $CuCl_2$ is the limiting reagent and $Al$ is the excess reagent.

5. Theoretical yield of $AlCl_3$ is 29.3 g

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Al=\frac{25.0g}{27g/mol}=0.92moles$

$\text{Moles of} CuCl_2=\frac{45.0g}{134g/mol}=0.33moles$

The balanced chemical equation is:

$2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu$

According to stoichiometry :

3 moles of $CuCl_2$ require = 2 moles of $Al$

Thus 0.33 moles of $CuCl_2$ will require= $\frac{2}{3}\times 0.33=0.22moles$ of $Al$

Thus $CuCl_2$ is the limiting reagent as it limits the formation of product and $Al$ is the excess reagent.

As 3 moles of $CuCl_2$ give = 2 moles of $AlCl_3$

Thus 0.33 moles of $CuCl_2$ give = $\frac{2}{3}\times 0.33=0.22moles$ of $AlCl_3$

Theoretical yield of $AlCl_3=moles\times {\text {Molar mass}}=0.22moles\times 133.34g/mol=29.3$

Thus 29.3 g of aluminium chloride is formed.

5 0

3 years ago

Determine the molar mass of a 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm. group of answer choices

lilavasa [31]

Considering the ideal gas law and the definition of molar mass, the molar mass of the sample of gas is 9.17 $\frac{g}{mol}$ .

<h3>Ideal gas law</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that occupies.
T is its temperature.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
n is the number of moles of the gas.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Molar mass of the sample of gas</h3>

In this case you know:

P= 0.980 arm
V= 1.20 L
T= 287 K
R= 0.082 $\frac{atmL}{molK}$
n= ?

Replacing in the ideal gas law:

0.980 atm× 1.20 L= n× 0.082 $\frac{atmL}{molK}$ × 287 K

Solving:

(0.980 atm× 1.20 L)÷ (0.082 $\frac{atmL}{molK}$ × 287 K)= n

<u><em>0.04997 moles= n</em></u>

On the other hand, you know that the<u><em> mass of the sample of gas</em></u> is <u><em>0.458 grams</em></u>. Replacing in the definition of molar mass:

$molar mass=\frac{0.458 grams}{0.04997 moles}$