All categories

3 years ago

How many atoms are in 3.6 moles of titanium?

Chemistry

2 answers:

kaheart [24]3 years ago

6 0

Answer:

69420

Explanation:

Ber [7]3 years ago

6 0

Answer:

We need to calculate number atoms, which can be determined as follows: We know that 1 mol contains 6.02214076×1023 number of atoms. i.e. is Avogadro's constant. Therefore the answer to the above question is that the number of atoms in 3.5 mol of Arsenic is 2.11×104 number of atoms.

Explanation:

You might be interested in

Given the following thermodynamic data, calculate the lattice energy of LiCl:

tiny-mole [99]

Answer:

$\boxed{\text{-862 kJ/mol}}$

Explanation:

One way to calculate the lattice energy is to use Hess's Law.

The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s); U = ?

We must generate this reaction rom the equations given.

(1) Li(s) + ½Cl₂ (g) ⟶ LiCl(s); ΔHf° = -409 kJ·mol⁻¹

(2) Li(s) ⟶ Li(g); ΔHsub = 161 kJ·mol⁻¹

(3) Cl₂(g) ⟶ 2Cl(g) BE = 243 kJ·mol⁻¹

(4) Li(g) ⟶Li⁺(g) +e⁻ IE₁ = 520 kJ·mol⁻¹

(5) Cl(g) + e⁻ ⟶ Cl⁻(g) EA₁ = -349 kJ·mol⁻¹

Now, we put these equations together to get the lattice energy.

E/kJ

(5) Li⁺(g) +e⁻ ⟶ Li(g) 520

(6) Li(g) ⟶ Li(s) -161

(7) Li(s) + ½Cl₂(g) ⟶ LiCl(s) -409

(8) Cl(g) ⟶ ½Cl₂(g) -121.5

(9) Cl⁻(g) ⟶ Cl(g) + e⁻ +349

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s) -862

The lattice energy of LiCl is $\boxed{\textbf{-862 kJ/mol}}$ .

3 0

3 years ago

Elimination of the pharmaceutical IV antibiotic gentamicin follows first-order kinetics. If the half-life of gentamicin is 1.5 h

Genrish500 [490]

Explanation:

The given data is:

The half-life of gentamicin is 1.5 hrs.

The reaction follows first-order kinetics.

The initial concentration of the reactants is 8.4 x 10-5 M.

The concentration of reactant after 8 hrs can be calculated as shown below:

The formula of the half-life of the first-order reaction is:

$k=\frac{0.693}{t_1_/_2}$

Where k = rate constant

t1/2=half-life

So, the rate constant k value is:

$k=\frac{0.693}{1.5 hrs}$

The expression for the rate constant is :

$k=\frac{2.303}{t} log \frac{initial concentration}{concentration after time "t"}$

Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:

$\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6$

Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.

5 0

3 years ago

Why are animal vaccinated?

juin [17]

Answer:

to go against dangerous viral and bacteria

Explanation:

5 0

3 years ago

Read 2 more answers

If distance, d, is proportional to time, t, and at time t =2.00 seconds, the distance is 6.20 m, find the distance when t = 3.50

Allushta [10]

At 3.5s the distance would be 10.85

Find 1/4th of 2.00 and add that to 6.2

5 0

3 years ago

Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

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

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

4 years ago