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3 years ago

What is the energy transformation taking place as a moving roller coaster slowly climbs a steep hill?

Chemistry

1 answer:

Delvig [45]3 years ago

7 0

Kinetic energy to potential energy. Hope this helps.

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Complete and balance the reaction in acidic solution. equation: ZnS + NO_{3}^{-} -> Zn^{2+} + S + NO ZnS+NO−3⟶Zn2++S+NO Which

kotykmax [81]

Answer:

Balanced reaction: $3ZnS+2NO_{3}^{-}+8H^{+}\rightarrow 3Zn^{2+}+3S+2NO+4H_{2}O$

S is oxidized and N is reduced.

$NO_{3}^{-}$ is the oxidizing agent and ZnS is the reducing agent.

Explanation:

Reaction: $ZnS+NO_{3}^{-}\rightarrow Zn^{2+}+S+NO$

$\Rightarrow$ Oxidation: $ZnS\rightarrow Zn^{2+}+S$

Balance charge: $ZnS-2e^{-}\rightarrow Zn^{2+}+S$ ...............(1)

$\Rightarrow$ Reduction: $NO_{3}^{-}\rightarrow NO$

Balance H and O in acidic medium : $NO_{3}^{-}+4H^{+}\rightarrow NO+2H_{2}O$

Balance charge: $NO_{3}^{-}+4H^{+}+3e^{-}\rightarrow NO+2H_{2}O$ ...............(2)

[ $3\times$ Equation-(1)] + [ $2\times$ Equation-(2)]:

$3ZnS+2NO_{3}^{-}+8H^{+}\rightarrow 3Zn^{2+}+3S+2NO+4H_{2}O$

Oxidation number of S increases from (-2) to (0) for the conversion of ZnS to S. Therefore S is oxidized.

Oxidation number of N decreases from (+5) to (+2) for the conversion of $NO_{3}^{-}$ to NO. Therefore N is reduced.

$NO_{3}^{-}$ consumes electron from ZnS. Therefore $NO_{3}^{-}$ is the oxidizing agent and ZnS is the reducing agent.

7 0

4 years ago

The net force on an object that is accelerating at a rate of 4 m/sec is 255 N. What is the mass of the object in kg?

aleksklad [387]

Answer:

Given - acc = 4 m/ s^2

Force = 255 N

To find = mass

Solution -

Using the formula,

$f = m \times a$

255 = Mass × 4

255 /4 = Mass

63.75 = Mass

1 N = kg m/s^2

63.75 kg = Mass

7 0

3 years ago

Given that Kp = 3.5 x 10-4 for the reaction 2 CO(g) <=> C(graphite) + CO2(g), what is the partial pressure of CO2(g) at eq

spin [16.1K]

Answer: The partial pressure of carbon dioxide at equilibrium is 0.0056 atm

Explanation:

The given chemical equation follows:

$2CO(g)\rightleftharpoons C\text{ (graphite)}+CO_2(g)$

Initial: 4.00

At eqllm: 4.00-2x x x

The expression of $K_p$ for above reaction follows:

$K_p=\frac{p_{CO_2}}{(p_{CO})^2}$

The partial pressure of pure solids and liquids are taken as 1 in the equilibrium constant expression.

We are given:

$K_p=3.5\times 10^{-4}$

Putting values in above expression, we get:

$3.5\times 10^{-4}=\frac{x}{(4-2x)^2}\\\\x=0.0056,718.28$

Neglecting the value of x = 718.28 because equilibrium pressure cannot be greater than initial pressure

Partial pressure of $CO_2$ = 0.0056 atm

Hence, the partial pressure of carbon dioxide at equilibrium is 0.0056 atm

4 0

3 years ago

NO LINKSSS PLEASE!!!

Temka [501]

Answer: a

Explanation:

are u a male or female

7 0

3 years ago

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What is the Octet rule?

trasher [3.6K]

8 electrons makes a full shell in an atom

6 0

4 years ago

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