All categories

2 years ago

Why do you think the transition metals are 10 columns wide?

2 answers:

4 0

The transition metals are 10 columns wide because they can hold no more than 10 electrons. This is why their width correlates with the number of electrons that are included in these metals. Transition metals include most commonly used metals, such as iron, copper, silver, and gold.

e-lub [12.9K]2 years ago

4 0

Learning and teaching resource for Transition Metals written by PhD ... These d-block elements are the short columns in the center of the periodic ... The group 10 members are nickel (Ni), palladium (Pd), and platinum (Pt). ... elements do have low ionization energies, however, so they exhibit a wide ... Think of a magnet.Transi

You might be interested in

Which of the following best describes where cellular respiration occurs?

snow_lady [41]

I think the answer is C

4 0

2 years ago

Read 2 more answers

[24 points] A sample of soil has a total volume of 205 cm3. The soil mass when saturated is 361 g. A specific yield test was con

telo118 [61]

Answer:

Follows are the solution to the given question:

Explanation:

Dry Soil weight = solid soil weight = $284 \ grams$

solid soil volume = $205 \ cc$

saturated mass soil = $361 \ g$

The weight of the soil after drainage is = $295 \ g$

Water weight for soil saturation = $(361-284) = 77 \ g$

Water volume required for soil saturation = $\frac{77}{1} = 77 \ cc$

Sample volume of water: $= \frac{\text{water density}}{\text{water density input}}$

$= 361- 295 \\\\ = 66 \ cc$

Soil water retained volume = (draining field weight - dry soil weight)

$= 295 - 284 \\\\ = 11 \ cc.$

$\text{POROSITY}= \frac{\text{Vehicle volume}}{\text{total volume Soil}}$

$= \frac{77}{(205 + 77)} \\\\= \frac{77}{(282)} \\\\ = 27.30 \%$

(Its saturated water volume is equal to the volume of voids)

$\text{YIELD SPECIFIC} = \frac{\text{Soil water volume}}{\text{Soil volume total}}$

$= \frac{66}{(205+77)}\\\\= \frac{66}{(282)}\\\\=0.2340\\\\ = 0.23$

$\text{Specific Retention}= \frac{\text{Volume of soil water}}{\text{Total soil volume}}$

$= \frac{11}{282} \\\\= 0.0390 \\\\ = 0.04$

6 0

2 years ago

Enter the coefficients that balance the tarnishing

olga55 [171]

Answer:

2Ag(s) + 1H2S(g) → 1Ag2S(s) + 1H2(g)

Explanation:

Step 1: Data given

Ag(s) + H2S(g) → Ag,S(s) + H2(g)

Ag has and oxidation number of +1

S has an oxidation number of -2

H has an oxidation number of +1

Step 2: The balanced equation

Ag+ + 2H+ + S^2- →

The silver ion and the sulfur ion will bind, but we need 2 silver atoms to bind with 1 sulfur atom.

2 H- atoms will appear as H2.

The balanced equation is:

2Ag+ + 2H+ +S^2- → Ag2S + H2

2Ag(s) + 1H2S(g) → 1Ag2S(s) + 1H2(g)

7 0

2 years ago

How many minutes will it take to plate out 4.50 g of Cu from a solution of Cu(NO3)2 (aq) onto the cathode of an electrolytic cel

ipn [44]

Answer:

It will take 28.5 minutes

Explanation:

Step 1: Data given

Mass of Cu = 4.50 grams

8.00 A of current are used

Molar mass of Cu = 63.5 g/mol

Step 2: Calculate time needed

Cu2+ →Electricity → Cu

we notice a flow of 2 electrons ⇒ This means the Faraday constant = 2F

Since Molar mass of Cu is 63.5 g/mol

63.5 grams of Cu is deposited by 2*96500 C

4.50 grams of Cu ((2*96500)/63.5) * 4.50 = 13677.17 C

Q = It

13677.17 = 8t*60 seconds

t = 28.5 minutes

3 0

2 years ago

Isotope Atomic Mass (amu) Percent Abundance

exis [7]

Answer:

The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.

Explanation:

Given data:

Atomic mass of silicon= ?

Percent abundance of Si-28 = 92.21%

Atomic mass of Si-28 = 27.98 amu

Percent abundance of Si-29 = 4.70%

Atomic mass of Si-29 = 28.98 amu

Percent abundance of Si-30 = 3.09%

Atomic mass of Si-30 = 29.97 amu

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)+(abundance of 2nd isotope × its atomic mass) / 100

Average atomic mass = (92.21×27.98)+(4.70×28.98)+(3.09×29.97) /100

Average atomic mass = 2580.04 +136.21+92.61 / 100

Average atomic mass = 2808.86 / 100

Average atomic mass = 28.08amu.

The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.

7 0

2 years ago

Read 2 more answers