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3 years ago

Emperical formula for Al³⁺

Chemistry

1 answer:

ad-work [718]3 years ago

4 0

Answer:

Al2O3

Explanation:

If we combine two Al3+ ions and three O2- ions, and use subscripts to indicate the number of each ion in the formula, we obtain Al2O3 as the empirical formula

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The relative atomic mass of an element is the mass of its atoms compared with that of which element?

mariarad [96]

In calculating the relative atomic mass of an element with isotopes, the relative mass and proportion of each is taken into account. For example, naturally occurring chlorine consists of atoms of relative isotopic masses 35 (75%) and 37 (25%). Its relative atomic mass is 35.5.

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4 years ago

What’s the force of a pitching machine on a baseball?

worty [1.4K]

Answer:

The Machine exerts a force of 9N.

Explanation:

Initial velocity (u) = 0 m/s

Final velocity (v) = 30 m/s

Time (t) = 0.5 s

Acceleration (a) = (v-u)/t

= (30-0)/0.5 m/s²

= 30/0.5 m/s²

= 60 m/s²

Mass = 0.15 kg

Force = Mass*Acceleration

= 0.15*60 N

= 9 N

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2 years ago

Evidence shows that interstellar bodies are moving far away from earth some faster than others which best suggest that the unive

igor_vitrenko [27]

The answer is B

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3 years ago

Read 2 more answers

Aluminum and cloride undergoes a synthesis reaction if 8molnof Al reacts with 10mol of cl, what is the maximum amount of AlCl3 c

Inessa [10]

Answer: The mass of $AlCl_3$ produced is 889.38 g

Explanation:

We are given:

Moles of Al = 8 mol

Moles of $Cl_2$ = 10 mol

For the given chemical reaction:

$2Al+3Cl_2\rightarrow 2AlCl_3$

By stoichiometry of the reaction:

If 3 moles of chlorine gas reacts with 2 moles of Al

So, 10 moles of chlorine gas will react with = $\frac{2}{3}\times 10=6.67mol$ of Al

As the given amount of Al is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, chlorine gas is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 3 moles of $Cl_2$ produces 2 mole of $AlCl_3$

So, 10 moles of $Cl_2$ will produce = $\frac{2}{3}\times 10=6.67mol$ of $AlCl_3$

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

We know, molar mass of $AlCl_3$ = 133.34 g/mol

Putting values in above equation, we get:

$\text{Mass of }AlCl_3=(6.67mol\times 133.34g/mol)=889.38g$

Hence, the mass of $AlCl_3$ produced is 889.38 g

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3 years ago

Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold

Sonja [21]

Answer:

The Number of gold atoms are = $N_{Au}=1.81718*10^{22}\ atoms/cm^3$

Explanation:

The formula we are going to use is:

$N_{Au}=\frac{N_A*C_{Au}}{{\frac{C_{Au}A_{Au} }{\rho_{Au}}}+\frac{A_{Au}}{\rho_{Ag}}(100-C_{Au})}$

Where:

$N_{Au}$ are number of gold atoms.

$N_A$ is Avogadro Number.

$C_{Au}$ is the amount of gold.

$A_{Au}$ is the atomic weight of gold.

$\rho_{Au}$ is the density of gold.

$\rho_{Ag}$ is the density of silver.

$C_{Ag}$ is the amount of silver.

$N_{Au}=\frac{6.023*10^{23}*45\%wt}{\frac{45\%wt*196.97}{19.32}+\frac{196.97}{10.49}(100-45\%wt)}\\ N_{Au}=1.81718*10^{22}\ atoms/cm^3$

The Number of gold atoms are = $N_{Au}=1.81718*10^{22}\ atoms/cm^3$

5 0

4 years ago