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3 years ago

9

Describe and explain the trends in reactivity in different groups of the periodic table, using your knowledge of electronic stru

cture. [6 mark]

1 answer:

Alex777 [14]3 years ago

6 0

Metal period- Reactivity Gets smaller as you move from left to right.
Group- Reactivity gets bigger as you move down a group.

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Is MgS ionic or covalent

Keith_Richards [23]

Answer:

Ionic

Explanation:

Mg has 2 electron in it's outermost orbit, by donating this two electron to Sulfur it get stable mg 2+ electronic configuration while sulfur has 6 electron in it's outermost shell ,so sulfur accept this 2 electron and complete it's octet and become s2–.

6 0

3 years ago

WILL MARK BRANILEST

ahrayia [7]

Answer:

FeCl3 is the limiting reactant

O2 is in excess

Theoretical yield Cl2 = 9.84 grams

The % yield is 96.5 %

Explanation:

Step 1: Data given

Mass of FeCl3 = 15.0 grams

Moles O2 = 4.0 moles

Mass of Cl2 produced = 9.5 grams

Step 2: The balanced equation

4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

Step 3: Calculate moles FeCl3

Moles FeCl3 = mass FeCl3 / molar mass FeCl3

Moles FeCl3 = 15.0 grams / 162.2 g/mol

Moles FeCl3 = 0.0925 moles

Step 4: Calculate limiting reactant

FeCl3 is the limiting reactant. Because we have way more (more than ratio 3:4) moles O2 than FeCl3. It will completely be consumed (0.0925 moles). O2 is in excess. There will react = 0.069375 moles O2

There will remain 4.0 - 0.069375 = 3.930625 moles

Step 5: Calculate moles Cl2

For 4 moles FeCl3 we need 3 moles O2 to produce 2 moles Fe2O3 and 6 moles Cl2

For 0.0925 moles FeCl3 moles we'll have 6/4 * 0.0925 = 0.13875 moles Cl2.

Step 6: Calculate mass Cl2

Mass Cl2 = moles * molar mass

Mass Cl2 = 0.13875 moles * 70.9 g/mol

Mass Cl2 = 9.84 grams

Step 7: Calculate % yield

% yield = (actual yield / theoretical yield) * 100%

% yield = (9.5 grams / 9.84 grams ) * 100%

% yield = 96.5 %

The % yield is 96.5 %

4 0

3 years ago

Is gasoline evaporated a chemical change

Len [333]

No. The only thing that changed was the looks of the gasoline, not the chemical components.

5 0

3 years ago

Read 2 more answers

A stock phenolphthalein solution was prepared by dissolving 0.050 g in a total volume of 100.0 ml. You place one drop (0.050 ml)

devlian [24]

Answer:

the value of molar absorptivity is 229000 $M^{-1} cm^{-1}$

Explanation:

given data

phenol phthalein solution = 0.050 g

total volume = 100.0 ml

dilute = 100.0 ml

diluted sample = 0.18

solution

we get here concentration that is express as

concentration = ( mass of solute × 1000 ) ÷ ( molar mass of solute × volume of solution ) .............1

put here value

concentration = $\frac{0.050\times 1000}{318.32\times 100}$

concentration = 0.00157 M

and here dillution equation is express as

c1 × v1 = c2 × v2 .................2

here c1 and c2 is initial and final concentration

and v1 and v2 is initial and final volume

put here value

0.001571 × 0.050 = c2 × 100

c2 = 7.855 × $10^{-7}$ M

and

now we get molar by absorbance equation that is

A = E × C × l ................3

here A is absorbance and E is molar and c is absorptivity and l is path length

put here value

0.18 = E × 7.855 × $10^{-7}$ × 1

E = 229000 $M^{-1} cm^{-1}$

3 0

3 years ago

Hydrogen cyanide is used to prepare sodium cyanide, which is used in part to obtain gold from gold-containing rock. If a reac- t

Sliva [168]

Answer:

Mass of HCN produced = 6.75 g

Explanation:

Reaction is as follows:

$2NH_3+3O_2+2CH_4 \rightarrow 2HCN + 6H_2O$

First calculate the no. of moles of each chemical species.

molecular mass of $NH_3$ is 17 g/mol

No. of mol of $NH_3$ = 11.5/17 = 0.676 mol

Molecular mass of $O_2$ = 32 g/mol

No. of mol of $O_2$ = 12/32 = 0.375

Molecular mass of $CH_4$ = 16 g/mol

No. of mol of $CH_4$ = 10.5/16 = 0.656 mol

from the balanced chemical reaction, it is clear that 2-moles ammonia reacts with 3 moles oxygen and 2 moles methane to form 2 moles of HCN.

or, 1-mol ammonia reacts with 1.5 mol oxygen and 1 mol methane to form 1 mol of HCN.

Thus, no. of oxygen present is less than required and so it will act as limiting reagent.

From the chemical equation,

3 moles oxygen produces 2 moles HCN

or one mole oxygen produces (2/3) moles HCN

0.375 moles oxygen produces (2/3) × 0.375 HCN = 0.25 moles of HCN

molar mass of HCN = 27 g/mole

Mass = mol × molar mass

mass of HCN = 27 × 0.25 = 6.75 g

8 0

3 years ago