Which of these facts best illustrates why regulation of alcohol consumption is necessary

Which one of the following does not represent 1.000 mol of the indicated

Katen [24]

Answer:

B. 26.00 g Fe

Atomic mass of Fe (Iron) is 55.845g

3 0

3 years ago

Read 2 more answers

Given the reactions below, (1) Na(s) + H2O(l) → NaOH(s) + 1/2 H2(g) LaTeX: \Delta H^o_{rxn}Δ H r x n o = −146 kJ (2) Na2SO4(s) +

slamgirl [31]

Answer:

ΔrxnH = -580.5 kJ

Explanation:

To solve this question we are going to help ourselves with Hess´s law.

Basically the strategy here is to work in an algebraic way with the three first reactions so as to reprduce the desired equation when we add them together, paying particular attention to place the reactants and products in the order that they are in the desired equation.

Notice that in the 3rd reaction we have 2 mol Na₂O (s) which is a reactant but with a coefficient of one, so we will multiply this equation by 1/2-

The 2nd equation has Na₂SO₄ as a reactant and it is a product in our required equation, therefore we will reverse the 2nd . Note the coefficient is 1 so we do not need to multiply.

This leads to the first equation and since we need to cancel 2 NaOH, we will nedd to multiply by 2 the first one.

Taking 1/2 eq 3 + (-) eq 2 + 2 eq 1 should do it.

Na₂O (s) + H₂ (g) ⇒ 2 Na (s) + H₂O(l) ΔrxnHº = 259 / 2 kJ 1/2 eq3

+ 2NaOH(s) + SO₃(g) ⇒ Na₂SO₄ (s) + H₂O (l) ΔrxnHº = -418 kJ - eq 2

+ 2Na (s) + 2 H₂O (l) ⇒ 2 NaOH (s) + H₂ (g) ΔrxnHº = -146 x 2 2 eq 1

Na₂O (s) + SO₃ (g) ⇒ Na₂SO₄ (s) ΔrxnHº = 259/2 + (-418) + (-146) x 2 kJ

ΔrxnH = -580.5 kJ

4 0

3 years ago

Be sure to answer all parts. Consider the following balanced redox reaction (do not include state of matter in your answers): 2C

timurjin [86]

Answer:

The specie which is oxidized is:- $CrO_2^-$

The specie which is reduced is:- $ClO^-$

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

For the given chemical reaction:

$2CrO_2^- + 6ClO^- + 2H_2O\rightarrow 2CrO_4^{2-} + 3Cl_2 + 4OH^-$

The half cell reactions for the above reaction follows:

Oxidation half reaction: $CrO_2^- + 2H_2O + 4OH^-\rightarrow CrO_4^{2-} + 4H_2O + 3e^-$

Reduction half reaction: $2ClO^- + 4H_2O + 2e^-\rightarrow Cl_2 + 2H_2O + 4OH^-$

Thus, the specie which is oxidized is:- $CrO_2^-$

The specie which is reduced is:- $ClO^-$

8 0

4 years ago

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

Answer: The freezing point of solution is 2.6°C

Explanation:

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$

Hence, the freezing point of solution is 2.6°C

8 0

3 years ago

A car engine produces 604 kJ of mechanical energy from fuel capable of producing 2416 kJ of energy. The efficiency of the engine

Anna71 [15]

I think it’s 40% but I’m not sure

6 0

3 years ago