Which substance is an electrolyte? A) O2 B) Xe C) C3H8 D) KNO3

1 answer:

6 0

Answer:
KNO₃ is an electrolyte.

Explanation:
Electrolyte is any specie which when dissolved in water produces a cation and anoin. Electrolytes has the ability to conduct electricity. In given options O₂, Xe and C₃H₈ are non electrolytes because they cannot dissociate into cation and an anion due to their covalent bonds or monoatomic nature. The dissociation of Potassium Nitrate is as follow,

KNO₃ → K⁺ ₍aq₎ + NO₃⁻ ₍aq₎

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Please help me with this I will give out extra points with the brain thing if you help me with this science question 1/7 image b

oksian1 [2.3K]

Answer:

D) The baby will continue to move forward as the car slows and be push into the padded car seat.

Explanation:

Newton's first law states that an object tends to be in a state of rest or motion unless and until an unbalanced external force acts on it. This means a body would continue to be in a state of motion unless external force stops it. A body in the state of rest will remain at rest unless an external force moves it.

An infant car seat is made to face the rear of the car. This is because, in case of a front-end collision, the car would come to a sudden stop but the bodies inside the car are in a state of motion and a sudden halt will cause the body to push in the opposite direction.

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7 0

3 years ago

Read 2 more answers

One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is: