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4 years ago

Logs in a fire, and charged batteries are examples of?

Chemistry

1 answer:

vlabodo [156]4 years ago

7 0

Answer:

Logs in a fire is an example of something burning, also changing its state of form. I think thats chemical change; going from wood to ashes. And charged batteries means they haven't been used yet or they were just charged, resulting in physical change.

Explanation:

I just learned this in science.

I hope this helps!

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The same reaction is begun with an initial concentration of 0.05 M O3 and 0.02 M NO. Under these conditions, the reaction reache

Ivahew [28]

The balanced equation of the reaction is:

O3(g) + NO (g) → O2 (g) + NO2 (g)

Then the ratios of reaction is 1 mol O3 : 1 mol NO : 1 mol O2 : 1 mol NO2

If you have initially 0.05 M of O3 and 0.02 M of NO, the reaction will end when all the NO is consumed.

The by the stoichiometry 0.02 mol of O3 will be consumed in 8 seconds.

And the rate of reaction is change in concetration divided by the time.

The change in concentration in O3 is 0.02 M

Then, the rate respect O3 is 0.02 M / 8 seconds = 0.0025 M/s

8 0

3 years ago

How did the surface area of the metal strip expose to the solution affect the

frez [133]

Answer:

See explanation below

Explanation:

In an electrochemical cell, electricity is obtained by the gradual deterioration of the anode.

Hence, surface area of the metal will affect the length of time within which the electrochemical cell works.

The greater the surface area of the metal, the longer the electrochemical cell can function and the greater the quantity of electricity produced, hence the answer above.

3 0

3 years ago

The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?

anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

$K_a\times K_b=K_w$

$K_b$ for sodium formate is $K_b=\frac {K_w}{K_a}$

Given that:

$K_a$ of formic acid = $1.8\times 10^{-4}$

And, $K_w=10^{-14}$

So,

$K_b=\frac {10^{-14}}{1.8\times 10^{-4}}$

$K_b=5.5556\times 10^{-11}$

Concentration = 0.35 M

HCOONa ⇒ Na⁺ + HCOO⁻

Consider the ICE take for the formate ion as:

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻

At t=0 0.35 - -

At t =equilibrium (0.35-x) x x

The expression for dissociation constant of sodium formate is:

$K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}$

$5.5556\times 10^{-11}=\frac {x^2}{0.35-x}$

Solving for x, we get:

x = 0.44×10⁻⁵ M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

pH = 14 - 4.64 = 9.36

5 0

3 years ago

Group the following electron configurations in pairs that would represent similar chemical properties at their atoms;

marishachu [46]

Answer: Pairs are: (a) and (d), (b) and (f), (c) and (e)

Explanation:

In a periodic table, elements are arranged in 18 vertical columns known as groups and 7 horizontal rows known as periods.

Elements arranged in a group show similar chemical properties because of the presence of same number of valence electrons.

Valence electrons are defined as the electrons which are present in the outermost shell of an atom. Outermost shell has the highest value of 'n' that is principal quantum number.

For the given options:

For a:

The given electronic configuration is: $1s^22s^22p^63s^2$

The number of valence electrons in the given configuration are 2

For b:

The given electronic configuration is: $1s^22s^22p^63s^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

For c:

The given electronic configuration is: $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For d:

The given electronic configuration is: $1s^22s^2$

The number of valence electrons in the given configuration are 2

For e:

The given electronic configuration is: $1s^22s^22p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For f:

The given electronic configuration is: $1s^22s^22p^63s^23p^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

Electronic configuration of (a) and (d) will form a pair, (b) and (f) will form a pair, (c) and (e) will form a pair and will have similar chemical properties.

Hence, the pairs are: (a) and (d), (b) and (f), (c) and (e)

4 0

3 years ago

Consider the following reaction: 2{\rm{ N}}_2 {\rm{O(}}g)\; \rightarrow \;2{\rm{ N}}_2 (g)\; + \;{\rm{O}}_2 (g)

fredd [130]

Answer:

a. 5.9 × 10⁻³ M/s

b. 0.012 M/s

Explanation:

Let's consider the following reaction.

2 N₂O(g) → 2 N₂(g) + O₂(g)

Time (t): 12.0 s

Δn(O₂): 1.7 × 10⁻² mol

Volume (V): 0.240 L

We can find the average rate of the reaction over this time interval using the following expression.

r = Δn(O₂) / V × t

r = 1.7 × 10⁻² mol / 0.240 L × 12.0 s

r = 5.9 × 10⁻³ M/s

b. The molar ratio of N₂O to O₂ is 2:1. The rate of change of N₂O is:

5.9 × 10⁻³ mol O₂/L.s × (2 mol N₂O/1 mol O₂) = 0.012 M/s

4 0

3 years ago