Balance the equation

When comparing the successive ionization energies of an element, an unusually big increase in ionization energy is seen when ___

Murljashka [212]

Answer:

The first valence electron is removed.

Explanation:

Ionization energy is the energy that is needed to tear out the last electron of an atom in its ground state or gaseous state.

A(g) + IE ----> A+(g) + 1e-

When we talk about starting an electron we refer to the last valence electron, since it is the one that is furthest from the atomic nucleus, the one that does not feel attracted. As the question asks us the opposite, the ionization energy is going to be extremely high, if the electron to be removed occupies the first valence layer, since it is totally attracted to the nucleus.

8 0

4 years ago

3.0 cm x 4.0 cm x 1.0 cm [?]cm^3

CaHeK987 [17]

Explanation:

Hi there!!

you asked to multiply these all right,

you can simply multiply it ,

=3cm × 4 cm × 1cm

= 12cm^2×1cm (4×3=12)

= 12cm^3 (12×1=12)

Therefore, theanswer is 12 cm^3.

Hope it helps..

4 0

4 years ago

Silver has a density of 10.5g/cm to the 3rd and gold has density of 9.3g/cm to the 3rd. Which would have a greater mass, 5cm to

Akimi4 [234]

The density of gold is 19.3 g/cm³

Ag= silver
Au= gold.
density (Ag)=10.5 g/cm³.
density (Au)=19.3 g/cm³

We have to calculate the mass of 5 cm³ silver:

10.5 g---------------------1 cm³
x--------------------------- 5 cm³

x=(10.5 g * 5 cm³) / 1cm³ ≈52.5 g

And we have to calculate the mass of 5 g gold.

19.3 g------------------------1 cm³
x-------------------------------5 cm³

x=(19.3 g * 5 cm³) / 1cm³≈96.5 g

Then 5 cm³ silver are 52.5 g of mass, and 5 cm³ gold are 96.5 g of mass, therefore, gold would have a greater mass than silver.

answer:5 cm³ gold would have a greater mass than 5 cm³ silver.

3 0

4 years ago

What is the force that accelerates objects towards Earth?

geniusboy [140]

The force that accelerates objects towards Earth is gravity.

6 0

3 years ago

Calculate the pH in titration of a weak acid: What is the pH in titration of formic acid (HCHO2, 0.200 M, 100.0 mL) after the ad

ki77a [65]

Answer:

pH = 12.61

Explanation:

First of all, we determine, the milimoles of base:

0.120 M = mmoles / 300 mL

mmoles = 300 mL . 0120 M = 36 mmoles

Now, we determine the milimoles of acid:

0.200 M = mmoles / 100 mL

mmoles = 100 mL . 0.200M = 20 mmoles

This is the neutralization:

HCOOH + OH⁻ ⇄ HCOO⁻ + H₂O

20 mmol 36 mmol 20 mmol

16 mmol

We have an excess of OH⁻, the ones from the NaOH and the ones that formed the salt NaHCOO, because this salt has this hydrolisis:

NaHCOO → Na⁺ + HCOO⁻

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻ Kb → Kw / Ka = 5.55×10⁻¹¹

These contribution of OH⁻ to the solution is insignificant because the Kb is very small

So: [OH⁻] = 16 mmol / 400 mL → 0.04 M

- log [OH⁻] = pOH → 1.39

pH = 14 - pOH → 12.61

6 0

3 years ago