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3 years ago

Calculate the percent composition of Mg(NO3)2.

Chemistry

1 answer:

dimaraw [331]3 years ago

3 0

Answer:

%M = 16.2%

%N = 18.9%

%O = 64.9%

Explanation:

Mg(NO3)2

Atomic mass of Magnesium M = 24

Atomic mass of Nitrogen N=14

Atomic mass of Oxygen O=16

Molar mass of Mg(NO3)2 = 24 + (14+[16*3})*2

Mg(NO3)2 = 24 + (14+48)*2

Mg(NO3)2 = 24 + 124

Mg(NO3)2 = 148

Percentage composition;

Magnesium M= (24/148) × 100

Magnesium M=0.162 × 100 =16.2%

Oxygen O=(16*6/148) × 100

Oxygen O= 0.649 × 100 = 64.9%

Nitrogen N= (14*2/148) × 100

Nitrogen N=0.189 × 100=18.9%

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Will give medal! What is the reason the group 13 metals have a typical charge of 3+?

Art [367]

My answer:

13 group of the periodic table represented by boron, aluminum and gallium subgroup. It includes gallium, indium, thallium. Typical steper oxidation in the subset gallium 3 is explained by the presence of (n-1)d^10 E-configuration.
Aluminium oxidation degree has +3 an electronic configuration of noble gases S^2P^6

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8 0

3 years ago

If 650. grams of C6H12O6 (FW 180.16) reacts with 650. grams of O2 (FW 32.00) in cellular respiration: C6H12O6 6 O2 --> 6 CO2

cestrela7 [59]

Answer:

O2 is limiting reactant

Explanation:

To find the limiting reactant we need to convert the mass of each reactant to the moles using the formula weight. And, as 1 mole of C6H12O6 reacts with 6 moles of O2, we can know wich reactant will be over first (Limiting reactant) as follows:

<em>Moles C6H12O6:</em>

650g * (1mol/180.16g) = 3.608 moles C6H12O6

<em>Moles O2:</em>

650g * (1mol/32g) = 20.31 moles O2

Now, for a complete reaction of 3.608 moles of C6H12O6 are required:

3.608 moles C6H12O6 * (6mol O2 / 1mol C6H12O6) = 21.65 moles O2

As there are just 20.31 moles of O2,

<h3>O2 is limiting reactant</h3>

4 0

3 years ago

Solid aluminum (AI) and oxygen (O_2) gas react to form solid aluminum oxide (Al_2O_3). Suppose you have 7.0 mol of Al and 9.0 mo

Nimfa-mama [501]

Answer:

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

Explanation:

Step 1: Data given

Numbers of Al = 7.0 mol

Numbers of mol O2 = O2

Molar mass of Al = 26.98 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

4Al(s) + 3O2(g) → 2Al2O3(s)

Step 3: Calculate limiting reactant

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

Al is the limiting reactant, it will be consumed completely (7 moles).

O2 is in excess. There will react 3/4 * 7 = 5.25 moles

There will remain 9-5.25 = 3.75 moles

Step 4: Calculate moles Al2O3

For 4 moles Al we'll have 2moles Al2O3

For 7.0 moles of Al we'll have 3.5 moles of Al2O3 produced

Step 5: Calculate mass of Al2O3

Mass Al2O3 = moles Al2O3 * molar mass Al2O3

Mass Al2O3 = 3.5 moles* 101.96 g/mol

Mass Al2O3 = 356.9 grams

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

3 0

3 years ago

Which statement best describes the compressibility of a gas?

maria [59]

The answer to the question stated above is:
<span> Gas is easily compressible because the molecules of a gas are much further apart than those of a solid.</span>

characteristic properties of gases:

(1) they are easy to compress,

(2) they expand to fill their containers, and

(3) they occupy far more space than the liquids or solids from which they form.

7 0

3 years ago

Please help as soon and best answer gets brainlest

Ann [662]

Answer:

Mass, temperature, and phase.

I think temperature because the higher the temperature of a given quantity of a substance, more is its thermal energy. Similarly, for the same temperature, higher mass of a substance will contain more thermal energy.

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3 years ago