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3 years ago

Why alkali metals easly ionize

1 answer:

7 0

Here’s what I found:

It takes very little energy to remove that outermost electron from an alkali metal. Thus, alkali metals easily lose their outermost electron to become a +1 ion. ... In fact, as you go down the 1A column, the first ionization energies get lower and lower, making cesium the most easily ionized element on the periodic table.

So basically it’s because part of what makes alkali metals so reactive is that they have one electron in their outermost electron layer.

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What was one main point of Dalton’s atomic theory

s344n2d4d5 [400]

Answer:

Everything is composed of atoms! which are the indivisible building blocks of matter and cannot be destroyed. All atoms of an element are identical. The atoms of different elements vary in size and mass.

Explanation:

Hope this helped

8 0

3 years ago

Read 2 more answers

Which isotope has the greatest number of protons?

tino4ka555 [31]

Answer:

The isotope with the greatest number of protons is:

option D: Pu-239, with 94 protons

Explanation:

The number of protons is the atomic number and is a unique number for each type of element.

You can tell the number of protons searching the element in a periodic table and reading its atomic number.

Thus, this is how you tell the number of protons or each isotope

Sample Chemical symbol Element atomic number # of protons

A Pa-238 Pa protactinium 91 91

B U-240 U uranium 92 92

C Np-238 Np neptunium 93 93

D Pu-239 Pu plutonium 94 94

8 0

3 years ago

Convert 65.4 m to mm. Helppp please

Ostrovityanka [42]

Answer:65.4 meters= 65400 millimeters

5 0

3 years ago

Read 2 more answers

The reaction is first order in cyclopropane and has a measured rate constant of k=3.36×10−5 s−1 at 720 k. if the initial cyclopr

sergey [27]

Answer:

0.0277 M.

Explanation:

The integral rate law of a first order reaction:

Kt = ln ([A₀]/[A]),

where, k is the rate constant of the reaction (k = 3.36 × 10⁻⁵ s⁻¹),

t is the time of the reaction (t = 235.0 min = 14100 s),

[A₀] is the initial concentration of cyclopropane ([A₀] = 0.0445 M)

∵ Kt = ln ([A₀]/[A]),

∴ (3.36 × 10⁻⁵ s⁻¹)(14100 s) = ln (0.0445 M)/[A]

Taking the exponential of both sides:

1.6 = (0.0445 M)/[A]

∴ [A] = (0.0445 M)/1.6 = 0.0277 M.

6 0

3 years ago

The half-life of cesium-137 is 30 years. suppose we have a 200-mg sample.

Assoli18 [71]

a) To find the mass after t years:

we will use this formula:

A = Ao / 2^n

when A =the amount remaining

and Ao = the initial amount

and n = t / t(1/2)

by substitution:

∴ A = 200 mg/ 2^(t/30y)

b) Mass after 90 y :

by using the previous formula and substitute t by 90 y

A = 200mg/ 2^(90y/30y)

∴ A = 25 mg

C) Time for 1 mg remaining:

when A= Ao/ 2^(t/t(1/2)

so, by substitution:

1 mg = 200 mg / 2^(t/30y)

∴2^(t/30y) = 200 mg by solving for t

∴ t = 229 y

7 0

4 years ago