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3 years ago

If you had excess aluminum, how many moles of aluminum chloride could be produced from 28.0 g of chlorine gas, Cl2?

Chemistry

1 answer:

Mice21 [21]3 years ago

6 0

Answer:

Yes chemistry. Try to add then multiply the top. Get the moles and you will find it.

Explanation:

Try to add then multiply the moles in the equation

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Students are designing an experiment to test the law of conservation of mass using the following materials: 10 g baking soda, 30

GaryK [48]

Answer:

1. Students need to measure masses of the items.

2. Put baking soda and vinegar in a plastic bag and close it.

3. Mix the two, allowing for a reaction to occur.

4. Figure the mass of the plastic bag while the two components are inside.

5. The combined mass should be equal to what each weighed on their own.

Explanation:

6 0

2 years ago

What mass of CO is needed to react completely with55.0 g of Fe2O3(s)+CO(g) yield Fe(s)+CO2(g)?

Katarina [22]

Answer:

28.9 g

Explanation:

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

Gather all the information in one place with molar masses above the formulas and masses below them.

$M_{r}$ : 159.69 28.01

Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂

Mass/g: 55.0

1. Use the molar mass of Fe₂O₃ to calculate the moles of Fe₂O₃.

$\text{Moles of Fe$_{2}$O$_{3}$} =\text{55.0 g Fe$_{2}$O$_{3}$} \times \frac{\text{1 mol Fe$_{2}$O$_{3}$}}{\text{159.69 g Fe$_{2}$O$_{3}$}}= \text{0.3444 mol Fe$_{2}$O$_{3}$}$

2. Use the molar ratio of CO:Fe₂O₃ to calculate the moles of CO.

$\text{Moles of CO} = \text{0.3444 mol Fe$_{2}$O$_{3}$} \times \frac{\text{3 mol CO}}{\text{1 mol Fe$_{2}$O$_{3}$}}= \text{1.033 mol CO}$

3.Use the molar mass of CO to calculate the mass of CO.

$\text{Mass of CO} = \text{1.033 mol CO} \times \frac{\text{28.01 g CO} }{\text{1 mol CO}}= \textbf{28.9 g CO}$

3 0

3 years ago

Read 2 more answers

Calculate the energy (in J/atom) for vacancy formation in silver, given that the equilibrium number of vacancies at 800 C is 3.6

MAXImum [283]

Answer:

the energy vacancies for formation in silver is $\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

Explanation:

Given that:

the equilibrium number of vacancies at 800 °C

i.e T = 800°C is 3.6 x 10¹⁷ cm3

Atomic weight of sliver = 107.9 g/mol

Density of silver = 9.5 g/cm³

Let's first determine the number of atoms in silver

Let silver be represented by N

SO;

$N = \dfrac{N_A* \rho _{Ag}}{A_{Ag}}$

where ;

$N_A =$ avogadro's number = $6.023*10^{23} \ atoms/mol$

$\rho _{Ag}$ = Density of silver = 9.5 g/cm³

$A_{Ag}$ = Atomic weight of sliver = 107.9 g/mol

$N = \dfrac{(6.023*10^{23} \ atoms/mol)*( 9.5 \ g/cm^3)}{(107.9 \ g/mol)}$

N = 5.30 × 10²⁸ atoms/m³

However;

The equation for equilibrium number of vacancies can be represented by the equation:

$N_v = N \ e^{^{-\dfrac{Q_v}{KT}}$

From above; Considering the natural logarithm on both sides; we have:

$In \ N_v =In N - \dfrac{Q_v}{KT}$

Making $Q_v$ the subject of the formula; we have:

${Q_v = - {KT} In( \dfrac{ \ N_v }{ N})$

where;

K = Boltzmann constant = 8.62 × 10⁻⁵ eV/atom .K

Temperature T = 800 °C = (800+ 273) K = 1073 K

$Q _v =-( 8.62*10^{-5} \ eV/atom.K * 1073 \ K) \ In( \dfrac{3.6*10^{17}}{5.3 0*10^{28}})$

$\mathbf{Q_v = 2.38 \ eV/atom}$

Where;

1 eV = 1.602176565 × 10⁻¹⁹ J

Then

$Q_v = (2.38 \ * 1.602176565 * 10^{-19} ) J/atom }$

$\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

Thus, the energy vacancies for formation in silver is $\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

8 0

3 years ago

Of the elements: b, c, f, li, and na. the element with the highest ionization energy is

padilas [110]

Carbon has the highest ionization energy as its energy 1086KJ\Mol and the rest are between 500 and 800.

6 0

3 years ago

Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?

s2008m [1.1K]

Answer:

65.08 g.

Explanation:

For the reaction, the balanced equation is:

2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

n = mass/molar mass = (36.2 g)/(133.34 g/mol) = 0.2715 mol.

Using cross multiplication:

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass = (0.4072 mol)(159.808 g/mol ) = 65.08 g.

4 0

3 years ago