Which of these prevents conduction from occuring?

An inflated balloon is left outside overnight. initially it has a volume of 1.84 L when the temperature is 293.4 K and the press

Mrac [35]

Answer:

ffff

Explanation:

5 0

3 years ago

Iodine-131 half life 8.040 how long will it take for a 40 gram sample of iodine-131 to decay to 0.75 grams

otez555 [7]

Answer:

See explanation below

Explanation:

To get this, we need to apply the general expression for half life decay:

N = N₀e(-λt) (1)

Where:

N and N₀ would be the final and innitial quantities, in this case, masses.

t: time required to decay

λ: factor related to half life

From the above expression we need λ and t. To get λ we use the following expression:

λ = t₁₂/ln2 (2)

And we have the value of half life, so, replacing we have:

λ = 8.04 / ln2 = 11.6

Now, we can replace in (1) and then, solve for t:

0.75 = 40 exp(-11.6t)

0.75 / 40 = exp(-11.6t)

ln(0.01875) = -11.6t

-3.9766 = -11.6t

t = -3.9766 / -11.6

3 0

3 years ago

Why does it have nerves connected to your organs

Kamila [148]

Answer:

The nervous system helps all the parts of the body to communicate with each other. It also reacts to changes both outside and inside the body. The nervous system uses both electrical and chemical means to send and receive messages

Explanation:

8 0

2 years ago

Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe

baherus [9]

Answer:

The equilibrium constant is $K =0.02867$

The equilibrium pressure for oxygen gas is $P_{O_2} = 0.09367\ atm$

Explanation:

From the question we are told that

The equation of the chemical reaction is

$M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}$

The Gibbs free energy for $M_3 O_4_{(s)}$ is $\Delta G^o_{1} = -8.80 \ kJ/mol$

The Gibbs free energy for $M{(s)}$ is $\Delta G^o_{2} = 0 \ kJ/mol$

The Gibbs free energy for $O_2{(s)}$ is $\Delta G^o_{3} = 0 \ kJ/mol$

The Gibbs free energy of the reaction is mathematically represented as

$\Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r$

$\Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)$

Substituting values

From the balanced equation

$\Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]$

$\Delta G^o_{re} = 8.80 kJ/mol =8800J/mol$

The Gibbs free energy of the reaction can also be represented mathematically as

$\Delta G^o_{re} = -RTln K$

Where R is the gas constant with a value of $R = 8.314 J/mol \cdot K$

T is the temperature with a given value of $T = 298 K$

K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction is mathematically represented as

$K_p =[ P_{O_2}]^{\frac{3}{2} }$

Where $[ P_{O_2}]$ is the equilibrium pressure of oxygen

The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the second equation of the Gibbs free energy of the reaction

$K = e^{- \frac{\Delta G^o_{re}}{RT} }$

Substituting values

$K= e^{\frac{8800}{8.314 * 298} }$

$K =0.02867$

Now substituting this into the equation above to obtain the equilibrium of oxygen

$0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}$

multiplying through by $1 ^{\frac{2}{3} }$

$P_{O_2} = [0.02867]^{\frac{2}{3} }$

$P_{O_2} = 0.09367\ atm$

3 0

3 years ago

A whale is swimming in the ocean. It communicates with another whale by sending out a sound. Which of the following best

ludmilkaskok [199]

Answer:

A.

The sound is transmitted through water until it reaches the other whale.

Explanation:

When one whale makes a sound, the sound is transmitted through water until it reaches the other whale. The sound causes a ripple-like effect in the water, producing waves that travel in the water. The waves are heard in all directions. This is why humans can hear the sounds animals make underwater.

6 0

3 years ago