What is the main idea of the article? *

The fossil record provides evidence of past life as well as clues to the past climate, environment, and abiotic factors in an ar

murzikaleks [220]

That is true because fossils can tell many things.

5 0

3 years ago

Read 2 more answers

Given the chemical formulas of the following compounds, name each compound and state the rules you used to determine each name.

Svetlanka [38]

When naming an ionic compound, write the name of the cation, which is the metal first. Then, write the name of the anion, which is the nonmetal. However, you remove the last 2-3 letters and replace suffixes.

1. RbF --> Rubidium Fluoride
Change fluorine to fluoride
2. CuO --> Copper (II) Oxide
Change oxygen to oxide. Oxide has a charge of -2. Since no subscripts are written, it means they have the same opposite charge. So, we use Copper (II).
3. (NH₄)₂C₂O₄ ---> Ammonium Oxalate
NH₄ is ammonia, but we change it to ammonium for polyatomic ions.

5 0

3 years ago

Will the electronegativity of tellurium be larger or smaller than that of antimony?

ddd [48]

Answer:

Tellurium has a higher electronegativity than antimony because it has one more proton which creates a stronger attraction between the protons and electrons.

Explanation:

Hope this helped! Stay safe and goodluck on what you're doing! ♥♥♥

7 0

3 years ago

What is the Ka of a weak acid (HA) if the initial concentration of weak acid is 4.5 x 10-4 M and the pH is 6.87? (pick one)

rewona [7]

Answer:

Ka = $4.04 \times 10^{-11}$

Explanation:

Initial concentration of weak acid = $4.5 \times 10^{-4}\ M$

pH = 6.87

$pH = -log[H^+]$

$[H^+]=10^{-pH}$

$[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M$

HA dissociated as:

$HA \leftrightharpoons H^+ + A^{-}$

(0.00045 - x) x x

[HA] at equilibrium = (0.00045 - x) M

x = $1.35 \times 10^{-7}\ M$

$Ka = \frac{[H^+][A^{-}]}{[HA]}$

$Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}$

0.000000135 <<< 0.00045

$Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}$

5 0

3 years ago

Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at

valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100 - x

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,

$121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042$

$120.9038x+122.9042\left(100-x\right)=12176.01$

Solving for x, we get that:

x = 57.2 %

Thus, percentage abundance of 121 Sb is = 57.2 %

percentage abundance of 123 Sb is = 100 - 57.2 % = 42.8 %

6 0

3 years ago