Seasonal variation in Earth's climate are caused by

Place the butane lighter in the sink or tub and let it rest there until needed. why do we soak the lighter in the water bath?

Genrish500 [490]

Answer:

To regulate the gas pressure in the lighter tank and avoid build-up of pressure.

Explanation:

First, you need to understand the properties of this organic molecule. Butane. C₄H₁₀ is a colorless, odorless, but HIGHLY FLAMMABLE liquefied gas. The liquid is flammable at 25⁰C whilst the vapor is flammable at 15⁰C. As you can see, this is an extremely flammable gas. It has a high vapor pressure (tendency by liquids to escape as gas molecules)
Any external heat source induces a pressure build up that might cause the gas to explode when there is an open flame.
Combining the two points above, a thermo-regulated water bath that has a lower temperature (below 15⁰C) will be need to prevent the pressure build up and ensure that any leakage will not have a high vapor pressure.

6 0

3 years ago

Identify the balanced chemical equation that represents a combustion reaction. 2C 4H 10 + 13O 2 ⟶ 10H 2O + 8CO 2 C 4H 10 + 5O 2

kati45 [8]

Answer:

2C₄H₁₀ + 13O₂ ⟶ 10H₂O + 8CO₂

Explanation:

2C₄H₁₀ + 13O₂ ⟶ 10H₂O + 8CO₂ . . . . . balanced combustion reaction

C₄H₁₀ + 5O₂ ⟶ 5H₂O + 4CO₂ . . . . not balanced

The two reactions involving Ca are not combustion reactions.

7 0

3 years ago

For the first order process: AB The half-life of A is 62.1 seconds. If a sample of A initially has 250.0 g, what mass (in g) of

vodomira [7]

Answer: The mass of sample A after given time is 99.05 g.

Explanation:

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

We are given:

$t_{1/2}=62.1s$

Putting values in above equation, we get:

$k=\frac{0.693}{62.1}=0.011s^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $0.011s^{-1}$

t = time taken for decay process = 84.2 s

$[A_o]$ = initial amount of the reactant = 250 g

[A] = amount left after decay process = ?

Putting values in above equation, we get:

$0.011s^{-1}=\frac{2.303}{84.2s}\log\frac{250}{[A]}$

$[A]=99.05g$

Hence, the mass of sample A after given time is 99.05 g.

8 0

3 years ago

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at 298 K is given below. 2 H2(g) + O2(g) → 2

DENIUS [597]

Answer:

(a) ΔG° = -474 kJ/mol; E° = 1.23 V

(b) ΔH° negative; ΔS° negative

(c) Since ΔS is negative, as T increases, ΔG becomes more positive. Therefore, the maximum work obtained will decrease as T increases.

Explanation:

Let's consider the following reaction.

2 H₂(g) + O₂(g) → 2 H₂O(l)

with an equilibrium constant K = 1.34 × 10⁸³

(a) Calculate E° and ΔG° at 298 K for the fuel-cell reaction.

We can calculate the standard Gibbs free energy (ΔG°) using the following expression:

ΔG° = - R × T × lnK

ΔG° = - 8.314 × 10⁻³ kJ . mol⁻¹.K⁻¹ × 298 K × ln 1.34 × 10⁸³ = -474 kJ/mol

To calculate the standard cell potential (E°) we need to write oxidation and reduction half-reactions.

Oxidation: 2 H₂ ⇒ 4 H⁺ + 4 e⁻

Reduction: O₂ + 4 e⁻ ⇒ 2 O²⁻

The moles of electrons (n) involved are 4.

We can calculate E° using the following expression:

$E\°=\frac{0.0591V}{n} .logK\\E\°=\frac{0.0591V}{4} .log1.34 \times 10^{83}=1.23V$

(b) Predict the signs of ΔH° and ΔS° for the fuel-cell reaction. ΔH°: positive negative ΔS°: positive negative

The standard Gibbs free energy is related to the standard enthalpy (ΔH°) and standard entropy (ΔS°) through the following expression:

ΔG° = ΔH° - T.ΔS°

Usually, the major contribution to ΔG° is ΔH°. So, if ΔG° is negative (exergonic), ΔH° is expected to be negative (exothermic).

The entropy is related to the number of moles of gases. There are 3 gaseous moles in the reactants and 0 in the products, so the final state is predicted to be more ordered than the initial state, resulting in a negative ΔS°.

(c) As temperature increases, does the maximum amount of work obtained from the fuel-cell reaction increase, decrease, or remain the same?

The maximum amount of work obtained depends on the standard Gibbs free energy.

wmax = ΔG° = ΔH° - T.ΔS°

Since ΔS is negative, as T increases, ΔG becomes more positive. Therefore, the maximum work obtained will decrease as T increases.

5 0

4 years ago

I need help with the stuff highlighted in the blue

vazorg [7]

Answer:

CO2 + H2O = C6H12O6 + O2

or

6CO2 + 6H2O = C6H12O6 + 6O2 if it need to be balanced

6 0

3 years ago

Read 2 more answers