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3 years ago

Describe how chemical energy is related to chemical change

1 answer:

7 0

During a chemical change, chemical energy may be changed to other forms of energy. Other forms of energy may also be changed to chemical energy.

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Please help me. I need help

Doss [256]

Answer:

they both collided is one i guess

Explanation:

5 0

3 years ago

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3. 2 HCl (aq)

tekilochka [14]

Explanation:

balance the equation ...HCl,Ba(OH) and BaCl2 are aqueous

know groups of the elements H 1+ ,Cl -,Ba 2+,OH-

write complete ionic eqution and eliminate spectator ions those appearing on both sides Cl and Ba

6 0

3 years ago

Which applies to the collision theory?

sweet [91]

A) Particles need to collide in order to react

4 0

3 years ago

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I just started learning about kinetic molecular theory, and I’m not sure how to answer the question circled below

lions [1.4K]

Answer : The value of 'R' is $0.0821\text{ L atm }mol^{-1}K^{-1}$

Solution : Given,

At STP conditions,

Pressure = 1 atm

Temperature = 273 K

Number of moles = 1 mole

Volume = 22.4 L

Formula used : $R=\frac{PV}{nT}$

where,

R = Gas constant

P = pressure of gas

T = temperature of gas

V = volume of gas

n = number of moles of gas

Now put all the given values in this formula, we get the values of 'R'.

$R=\frac{(1atm)\times (22.4L)}{(1mole)\times (273K)}$

$R=0.0821\text{ L atm }mol^{-1}K^{-1}$

Therefore, the value of 'R' is $0.0821\text{ L atm }mol^{-1}K^{-1}$ .

7 0

3 years ago

Just as the depletion of stratospheric ozone today threatens life on Earth today, its accumulation was one of the crucial proces

inessss [21]

Answer:

(a) rate = -(1/3) Δ[O₂]/Δt = +(1/2) Δ[O₃]/Δt

(b) Δ[O₃]/Δt = 1.07x10⁻⁵ mol/Ls

Explanation:

By definition, the reaction rate for a chemical reaction can be expressed by the decrease in the concentration of reactants or the increase in the concentration of products:

aX → bY (1)

$rate= -\frac{1}{a} \frac{\Delta[X]}{ \Delta t} = +\frac{1}{b} \frac{\Delta[Y]}{ \Delta t}$

where, a and b are the coefficients of de reactant X and product Y, respectively.

(a) Based on the definition above, we can express the rate of reaction (2) as follows:

3O₂(g) → 2O₃(g) (2)

$rate = -\frac{1}{3} \frac{\Delta[O_{2}]}{\Delta t} = +\frac{1}{2} \frac{ \Delta[O_{3}]}{ \Delta t}$ (3)

(b) From the rate of disappearance of O₂ in equation (3), we can find the rate of appearance of O₃:

$rate = +\frac{1}{2} \frac{\Delta[O_{3}]}{ \Delta t} = -\frac{1}{3} \frac{\Delta[O_{2}]}{ \Delta t}$

$+\frac{\Delta[O_{3}]}{ \Delta t} = -\frac{2}{3} \frac{\Delta[-1.61 \cdot 10^{-5}]}{ \Delta t}$

$\frac{\Delta[O_{3}]}{ \Delta t} = 1.07 \cdot 10^{-5} \frac{mol}{Ls}$

So the rate of appearance of O₃ is 1.07x10⁻⁵ mol/Ls.

Have a nice day!

6 0

4 years ago