All categories

3 years ago

1. What kinds of metals are REALLY reactive? why?

Chemistry

1 answer:

Olin [163]3 years ago

8 0

Answer:

1. Alkali metals (group 1)

2. halogens (Group 17)

3. noble gasses (group 18)

Explanation:

1. alkali metals only have one valence electron meaning that they really want to lose that one valence electron to get a full octet.

2. halogens have 7 valence electrons meaning that they just need to gain 1 to get a full octet.

3. Nobel gasses already have a full octet meaning that they don't want to react. (atoms only react to get a full octet)

I hope this helps. Let me know if anything is unclear.

You might be interested in

Devante buys a helium birthday balloon for his mom at the grocery store, where the temperature is a comfortable 72oF. He takes t

frozen [14]

The balloon would be smaller and not float as well because of the low temperature. The particles in the balloon when in the car will slow down and get closer together because of the low temperature. That caused it to become more dense and smaller in size then before. When in the store, the particulars in the balloon we’re moving faster and more spread apart

5 0

4 years ago

The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed

Oksanka [162]

Answer:

(a) The rate of disappearance of $O_{2}$ is: $4.65*10^{-5}$ M/s

(b) The value of rate constant is: 0.83036 $M^{-2}s^{-1}$

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

$2NO(g)+O_{2}->2NO_{2}$

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = $-\frac{d}{dt}[O_{2}]$ = $\frac{1}{2}\frac{d}{dt}[NO_{2}]$ -----(1)

According to the question, the reaction is second order in NO and first order in $O_{2}$ .

Then we can say that, rate = k $[NO]^{2}[O_{2}]$ -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

$-\frac{d}{dt}[NO]$ = $9.3*10^{-5}$ M/s.

(a) From (1), we can get the rate of disappearance of $O_{2}$ .

Rate of disappearance of $O_{2}$ = $-\frac{d}{dt}[O_{2}]$ = (0.5)*( $9.3*10^{-5}$ ) M/s = $4.65*10^{-5}$ M/s.

(b) The rate of the reaction can be obtained from (1).

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = (0.5)*( $9.3*10^{-5}$ )

rate = $4.65*10^{-5}$ M/s

The value of rate constant can be obtained by using (2).

rate constant = k = $\frac{rate}{[NO]^{2}[O_{2}]}$

k = $\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}$ = 0.83036 $M^{-2}s^{-1}$

k = $\frac{rate}{[NO]^{2}[O_{2}]}$

Substituting the units of rate as M/s and concentrations as M, we get:

$\frac{Ms^{-1} }{M^{3}}$ = $M^{-2}s^{-1}$

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

$rate\alpha [NO]^{2}$

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of $(1.8)^{2}$ = 3.24

5 0

3 years ago

The strength of an atom's attraction for the electrons in a chemical bond is the atom's?

Mazyrski [523]

<span><span>Yes. An element that is highly electronegative pulls more on the electrons in a bond, such as oxygen in H20. This creates a polar bond, where there is a small negative charge on the oxygen, and a small positive charge in between the hydrogens.

</span>Credit goes to "Erin M" answered on yahoo answers a decade ago.
</span>

3 0

3 years ago

Which has a higher boiling point hcl or hi

Likurg_2 [28]

HI.

Hope this helps!

-Payshence

3 0

3 years ago

Read 2 more answers

For the following problems, identify the substance as either an element, a compound, or a mixture. Write E for Element, C for co

andre [41]

Answer:

Explanation:

34. E

35. M

36. C

37. M

38. E

39. C

3 0

3 years ago