Which of the following statements is NOT true about practice?

stich3 [128]

Answer:

$\large \boxed{\text{-851.4 kJ/mol}}$

Explanation:

2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s); ΔᵣH = ?

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

$\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})$

2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s)

ΔfH°/kJ·mol⁻¹: 0 -824.3 -1675.7 0

$\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & [1(-1675.7) + 2(0)] - [2(0) - 1(-824.3)]\\& = & -1675.7 + 824.3\\& = & \textbf{-851.4 kJ/mol}\\\end{array}\\\text{The enthalpy change is } \large \boxed{\textbf{-851.4 kJ/mol}}$

7 0

3 years ago

H2CO3(aq) + H200 H30 (aq) + HCO3 (aq).

timofeeve [1]

Answer:

$K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}$

Explanation:

Several rules should be followed to write any equilibrium expression properly. In the context of this problem, we're dealing with an aqueous equilibrium:

an equilibrium constant is, first of all, a fraction;
in the numerator of the fraction, we have a product of the concentrations of our products (right-hand side of the equation);
in the denominator of the fraction, we have a product of the concentrations of our reactants (left-hand side o the equation);
each concentration should be raised to the power of the coefficient in the balanced chemical equation;
only aqueous species and gases are included in the equilibrium constant, solids and liquids are omitted.

Following the guidelines, we will omit liquid water and we will include all the other species in the constant. Each coefficient in the balanced equation is '1', so no powers required. Multiply the concentrations of the two products and divide by the concentration of carbonic acid:

$K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}$

4 0

3 years ago

A sample of 53.0 of carbon dioxide was obtained by heating 1.31 g of calcium carbont. What is the percent yield for this reactio

MaRussiya [10]

Answer:

92.04%

Explanation:

Given:

Mass of CO₂ obtained = 53.0 grams

Mass of calcium carbonate heated = 1.31 grams

Now,

the molar mass of the calcium carbonate = 100.08 grams

The number of moles heated in the problem = Mass / Molar mass

= (1.31 grams) / (100.08 grams/moles)

= 0.013088 moles

now,

1 mol of calcium carbonate yields 1 mol of CO₂

thus,

0.013088 moles of calcium carbonate will yield = 0.013088 mol of CO₂

now,

Theoretical mass of 0.013088 moles of CO₂ will be

= Number of moles × Molar mass of CO₂

= 0.013088 × 44 = 0.5758 grams

Thus, the percent yield for this reaction = $\frac{\textup{Actual yield}}{\textup{Theoretical yield}}\times100$

or

the percent yield for this reaction = $\frac{0.53}{0.5758}\times100$

or

the percent yield for this reaction = 92.04%

6 0

2 years ago

What does the host mean when he says that the acid has "donated a proton" to the water?

weqwewe [10]

Answer:

Option B

Explanation:

As Brønsted-Lowry theory states, acids are the ones that can donate protons.

When a proton is donated, it is released to become medium more acidic.

HCl is a strong acid.

HCl (l) + H₂O (l) → H₃O⁺ (aq) + Cl⁻(aq)

These always reffers to strong acid where the dissociation is 100% completed.

In a weak acid, dissociation is not 100% complete, that's why we have an equilibrium.

HA (l) + H₂O (l) ⇄ H₃O⁺ (aq) + A⁻(aq) Ka

5 0

2 years ago

If I had 3.50 x 10 24molecules of Cl2 gas, how many grams is this?

Zinaida [17]

Answer:

412 g Cl₂

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Table
Moles
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 3.50 × 10²⁴ molecules Cl₂

[Solve] grams Cl₂

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of Cl₂ - 2(35.45) = 70.9 g/mol

Step 3: Convert

[DA] Set up: $\displaystyle 3.50 \cdot 10^{24} \ molecules \ Cl_2(\frac{1 \ mol \ Cl_2}{6.022 \cdot 10^{23} \ molecules \ Cl_2})(\frac{70.9 \ g \ Cl_2}{1 \ mol \ Cl_2})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 412.072 \ g \ Cl_2$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

412.072 g Cl₂ ≈ 412 g Cl₂

5 0

2 years ago