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3 years ago

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

Chemistry

1 answer:

Sav [38]3 years ago

5 0

Answer:

1 mol Al; 1 mol O2

Explanation:ol Al; 10 mol O2

15.4 mol Al; 10.7 mol O2

or the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

4Al(s) + 302(g) —> 2Al2O3(s)

Express your answer aor the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

4Al(s) + 302(g) —> 2Al2O3(s)

Express your answer as a chemical formula.

1 mol Al; 1 mol O2

4 mol Al; 2.5 mol O2

12 mol Al; 10 mol O2

15.4 mol Al; 10.7 mol O2

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Determine the mass of 9.2 x 10^18 molecules of dinitrogen tetroxide.

steposvetlana [31]

Answer: 1.4x10-3 g N2O4

Explanation: First convert molecules of N2O4 to moles using Avogadro's Number. Then convert moles to mass using the molar mass of N2O4.

9.2x10^18 molecules N2O4 x 1 mole N2O4 / 6.022x10²³ molecules N2O4

= 1.53x10-5 moles N2O4

1.53x10-5 moles N2O4 x 92 g N2O4/ 1 mole N2O4

= 1.4x10-3 g N2O4

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3 years ago

Word elements can be used to show chemical reactions. When hydrogen burns in oxygen. Water is produced. The word equation is hyd

frez [133]

Iron + sulphur - iron sulphide

8 0

3 years ago

Read 2 more answers

La densidad del óxido de magnesio. MgO, es de 3.581 g/cm3 El MgO, es de 3.581 g/cm3 El MgO cristaliza con ordenamiento cúbico co

Jobisdone [24]

Answer:

$a=4.213cm$

$r=1.490x10^{-8}cm$

Explanation:

Hola.

En este caso, para calcular la longitud (a) de una cara de celda unitaria, consideramos la siguiente ecuación:

$\rho =\frac{#at*M}{a^3N_A}$

En la que consideramos el número de átomos por celda (4 para FCC), la masa molar (40.3 g/mol para MgO) y el número de avogadro para obtener:

$3.581g/mol = \frac{4atom/celda*40.3g/mol}{a^3*6.02x10^{23}atom/mol}$

Despejando para a, obtenemos:

$a^3 = \frac{4atom/celda*40.3g/mol}{3.581g/cm^3*6.02x10^{23}atom/mol}\\\\a=\sqrt[3]{7.478cm^3} \\\\a=4.213cm$

Finalmente, el radio lo calculamos como:

$r=\frac{\sqrt{2}*a}{4}=\frac{\sqrt{2}*4.213x10^{-8}cm}{4}\\\\r=1.490x10^{-8}cm$

¡Saludos!

6 0

3 years ago

Cal is titrating 50.8 mL of 0.319 M HBr with 0.337 M Ba(OH)2. How many mL of Ba(OH)2 does Cal need to add to reach the equivalen

alexgriva [62]

Answer:

$V_{base}=24.04mL$

Explanation:

Hello!

In this case, according to the chemical reaction by which HBr reacts with Ba(OH)2:

$2HBr+Ba(OH)_2\rightarrow BaBr_2+2H_2O$

We can see there is a 2:1 mole ratio between the acid and the base; thus, at the equivalent point we can write:

$2M_{base}V_{base}=M_{acid}V_{acid}$

Therefore, for is to compute the volume of the used base, we proceed as shown below:

$V_{base}=\frac{M_{acid}V_{acid}}{2M_{base}}$

And we plug in to obtain:

$V_{base}=\frac{0.319M*50.8mL}{2*0.337M}\\\\V_{base}=24.04mL$

Best regards!

8 0

3 years ago

Consider the chemical formula for one unit of aluminum sulfate, Al2(SO4)3. How many total atoms would be present in 22 units of

EastWind [94]

Answer:

1.32×10²⁵ atoms of sulfate are contained in 22 units of it

Explanation:

1 unit = 1 mol

Al₂(SO₄)₃ → Aluminum sulfate

As 1 unit = 1 mol, 1 unit has 6.02×10²³ atoms of aluminum sulfate.

Let's make a rule of three:

1 unit of Al₂(SO₄)₃ contains 02×10²³ atoms

Then, 22 units of Al₂(SO₄)₃ must contain (22 . 6.02×10²³) / 1 = 1.32×10²⁵ atoms

3 0

3 years ago