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3 years ago

How many valence electrons do the elements in column 4 have? Are these elements very reactive?

Chemistry

1 answer:

Makovka662 [10]3 years ago

4 0

Answer:

Explanation: Elements in group 14 have 4 valence electrons. Meaning that to achieve a stable octet they need to either gain or lose 4 electrons in total. That would take significantly more energy than other elements, e.g., group 1 (alkaline metals) or group 17 (halogens.)

Explanation:

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Starting from a 1 M stock of NaCl, how will you make 50 ml of 0.15 M NaCl?

WARRIOR [948]

Answer:

We need 7.5 mL of the 1M stock of NaCl

Explanation:

Data given:

Stock = 1M this means 1 mol/ L

A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL

Step 2: Calculate the volume of stock we need

The moles of solute will be constant

and n = M*V

M1*V1 = M2*V2

⇒ with M1 = the initial molair concentration = 1M

⇒ with V1 = the volume we need of the stock

⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L

⇒ with M2 = the concentration of the new solution = 0.15 M

1*V1 = 0.15*(50)

V1 = 7.5 mL

Since 0.0075 L of 1M solution contains 0.0075 moles

50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M

We need 7.5 mL of the 1M stock of NaCl

6 0

3 years ago

ILL GIVE BRAINLY, 5 STARS , FOLLOW AND HEARTS

Jlenok [28]

Answer:

Second one

Explanation:

I think it's the second cause there's no energy w/o heat. So it's the second.

6 0

3 years ago

Read 2 more answers

You wish to measure the iron content of the well water on the new property you are about to buy. You prepare a reference standar

djverab [1.8K]

1.04 ⨯ 10 $^{-4}$ M

<h3>Explanation</h3>

A = ε $\cdot l \cdot c$ by the Beer-Lambert law, where

A the absorbance,
$l$ the path length,
ε the molar absorptivity of the solute, and
$c$ concentration of the solution.

A and ε are the same for both solutions. Therefore, $l \cdot c$ is constant; $l$ is inversely proportional to $c$ . The 100 mL sample would have a concentration 1/4.78 times that of the 45.0 mL reference.

The 13.0 mL standard solution has a concentration of 5.17 ⨯ $10^{-4}$ M. Diluting it to 45.0 mL results in a concentration of $5.17 \times 10^{-4} \times \frac{13.0}{45.0} =$ 1.494 M.

$c$ is inversely related to $l$ for the two solutions. As a result, c₂ = $c_1 \cdot \frac{l_1}{l_2} = 1.494 \times 10^{-4} \times \frac{1}{4.78} =$ 3.126 M.

The 30.0 mL sample has to be diluted by 30.0 / 100.0 times to produce the 100.0 mL solution being tested. The 100.0 mL solution has a concentration of 3.126 M. Therefore, the 30.0 mL solution has a concentration of $3.126 \times \frac{100.0}{30.0} =$ 1.04 ⨯ $10^{-4}$ M.

6 0

3 years ago

What element from the "Periodic Table" is named after an "Austrian Scientist"??

soldier1979 [14.2K]

Answer:

Should be either mendele or fermium

maybe lawrencium not sure but I think it’s lawrencium! Sorry if wrong heh

Explanation: really trying to help! If it’s wrong super sorry!!!

3 0

3 years ago

Insulin is a protein that is used by the body to regulate both carbohydrate and fat metabolism. A bottle contains 125 mL of insu

Gnoma [55]

125x30= 3750mg=3,75g

8 0

4 years ago