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4 years ago

Arrange the following elements in order of decreasing first ionization energy: S, Ca, F, Rb, and Si.

Chemistry

1 answer:

kherson [118]4 years ago

8 0

Answer:

The concentration of energy needed to withdraw an electron from an atom’s mole in the gas phase is known as the ionization energy of an atom. It is more accurately termed as the first ionization energy. The ionization energy upsurges from left to right through a period and from top to bottom in the groups.

Of the given elements S, Ca, F, Rb, and Si, the S, and Si belong to the third period, and the atomic radius of S is less in comparison to Si, F belongs to the second period, Rb belongs to the fifth period, and Ca belongs to the fourth period. Thus, the decreasing order of first ionization energy, that is, from largest to smallest is F > S > S > Ca > Rb.

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g The most common position for an double bond in an unsaturated fatty acid is delta _________(fill in the number).

Stolb23 [73]

Answer:

The most common position for an double bond in an unsaturated fatty acid is delta 9 (Δ⁹)

Explanation:

Unsaturated fatty acids are carboxylic acids which contains one or more double bonds. The chain length as well as the number of double bonds is written separated by a colon. The positions of the double bonds are specified starting from the carboxyl carbon, numbered as 1, by superscript numbers following a delta (Δ). For example, an 18-carbon fatty acid containing a single double bond between carbon number 9 and 10 is written as 18:1(Δ⁹).

In most monounsaturated fatty acids, the double bond is between C-9 and C-10 (Δ⁹), and the other double bonds of polyunsaturated fatty acids are generally Δ¹² and Δ¹⁵. This positioning is due to the nature of the biosynthesis of fatty acids. In the mammalian hepatocytes, double bonds are introduced easily into fatty acids at the Δ⁹ position, but cannot introduce additional double bonds between C-10 and the methyl-terminal end. However, plants are able to introduce these additional double bonds at the Δ¹² and Δ¹⁵ positions.

5 0

4 years ago

HELP PLS!

olya-2409 [2.1K]

I have the same question right now and i think that it is D

7 0

4 years ago

In the reaction below, 22 g of H2S with excess

BartSMP [9]

Answer:

24%

Explanation:

From the question, the limiting reactant is H2S.

The reaction equation is;

2H2S + O2 → 2S + 2H2O

Number of moles of H2S reacted = 22g/34 g/mol = 0.647 moles

According to the reaction equation;

2 moles of H2S yields 2 moles of sulphur

0.647 moles of H2S yields 0.647 moles of sulphur

So;

Theoretical yield of sulphur = 0.647 moles * 32 g/mol = 20.7 g

Actual yield = 5 g

% yield = actual yield/theoretical yield * 100

% yield =5 g/20.7 g * 100

% yield = 24%

7 0

3 years ago

3. Given 20g of Barium Hydroxide, how many grams of

anastassius [24]

The number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g

First, we will write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

Ba(OH)₂ + 2NH₄NO₃ → 2NH₄OH + Ba(NO₃)₂

This means, 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

Now, we will calculate the number of moles of barium hydroxide present.

Mass of barium hydroxide (Ba(OH)₂) = 20 g

Using the formula

$Number\ of\ moles = \frac{Mass}{Molar\ mass}$

Molar mass of Ba(OH)₂ = 171.34 g/mol

∴ Number of moles of Ba(OH)₂ present = $\frac{20}{171.34}$

Number of moles of Ba(OH)₂ present = 0.116727 mole

Now,

Since 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

Then,

0.116727 mole of barium hydroxide will react with 2 × 0.116727 mole of ammonium nitrate

2 × 0.116727 = 0.233454 mole

∴ Number of moles of NH₄NO₃ required is 0.233454 mole

Now, for the mass of ammonium nitrate (NH₄NO₃) required

From the formula

Mass = Number of moles × Molar mass

Molar mass of NH₄NO₃ = 80.043 g/mol

∴ Mass of NH₄NO₃ required = 0.233454 × 80.043

Mass of NH₄NO₃ required = 18.68636 g

Mass of NH₄NO₃ required ≅ 18.7g

Hence, the number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g

Learn more on determining mass of reactant required here: brainly.com/question/11232389

6 0

2 years ago

A 12.630 g milk chocolate bar is found to contain 8.315 g of sugar. part a part complete how many milligrams of sugar does the m

kolezko [41]

Answer:
A 12.630 bar of chocolate contains 8315 milligrams of sugar

Explanation:
From the basics of conversion, we can find that:
1 gram is equal to 1000 mg
Using cross multiplication, we can find how many milligrams are present in 8.315 grams as follows:
1 gram .............> 1000 mg
8.315 grams ....> ?? mg
amount in milligrams = (8.315*1000) / 1 = 8315 milligrams

Hope this helps :)

5 0

4 years ago