Why is mendeleevâs periodic table considered a viable scientific model?

Please help me on this I’ll give 30 points

Rufina [12.5K]

Answer:

1. 10.66 moles of Al

2. 2.5 moles of Al₂O₃

3. 143.87 g of SO₂

Explanation:

The balance chemical equation is as follow,

4 Al + 3 O₂ → 2 Al₂O₃

According to balance chemical equation,

3 moles of O₂ required = 4 moles of Al

So,

8 moles of O₂ will require = X moles of Al

Solving for X,

X = 8 mol × 4 mol / 3 mol

X = 10.66 moles of Al

According to balance chemical equation,

4 moles of Al produced = 2 moles of Al₂O₃

So,

5 moles of Al will produce = X moles of Al₂O₃

Solving for X,

X = 5 mol × 2 mol / 4 mol

X = 2.5 moles of Al₂O₃

The balance chemical equation for the oxidation of carbon disulfide is as follow;

CS₂ + 3 O₂ → CO₂ + 2 SO₂

Step 1: Calculate Moles of CS₂ as;

Moles = Mass / M.Mass

Moles = 85.5 g / 76.14 g/mol

Moles = 1.12 moles of CS₂

Step 2: Find out moles of SO₂ as;

According to balance chemical equation,

1 mole of CS₂ produced = 2 moles of SO₂

So,

1.12 moles of CS₂ will produce = X moles of SO₂

Solving for X,

X = 1.12 mol × 2 mol / 1 mol

X = 2.24 moles of SO₂

Step 3: Calculate Mass of SO₂ as;

Mass = Moles × M.Mass

Mass = 2.24 mol × 64.06 g/mol

Mass = 143.87 g of SO₂

4 0

3 years ago

On the Periodic Table, which elements are predominantly (mostly) gases with low densities?

aliya0001 [1]

I’m sorry if I wasted your time but I think it’s alkali metals but I’m
Not sure

5 0

2 years ago

Determine the empirical formulas for compounds with the following percent compositions:

Elis [28]

Answer:

For a: The empirical formula of the compound is $P_2O_5$

For b: The empirical formula of the compound is $KH_2PO_4$

Explanation:

For a:

We are given:

Percentage of P = 43.6 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Phosphorus = $\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{43.6g}{31g/mole}=1.406moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{56.4g}{16g/mole}=3.525moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.406 moles.

For Phosphorus = $\frac{1.406}{1.406}=1$

For Oxygen = $\frac{3.525}{1.406}=2.5$

Converting the moles in whole number ratio by multiplying it by '2', we get:

For Phosphorus = $1\times 2=2$

For Oxygen = $2.5\times 2=5$

Step 3: Taking the mole ratio as their subscripts.

The ratio of P : O = 2 : 5

Hence, the empirical formula for the given compound is $P_2O_5$

For b:

We are given:

Percentage of K = 28.7 %

Percentage of H = 1.5 %

Percentage of P = 22.8 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of K = 28.7 g

Mass of H = 1.5 g

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Potassium = $\frac{\text{Given mass of Potassium}}{\text{Molar mass of Potassium}}=\frac{28.7g}{39g/mole}=0.736moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of hydrogen}}=\frac{1.5g}{1g/mole}=1.5moles$

Moles of Phosphorus = $\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{22.8g}{31g/mole}=0.735moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{47g}{16g/mole}=2.9375moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.735 moles.

For Potassium = $\frac{0.736}{0.735}=1$

For Hydrogen = $\frac{1.5}{0.735}=2.04\approx 2$

For Phosphorus = $\frac{0.735}{0.735}=1$

For Oxygen = $\frac{2.9375}{0.735}=3.99\approx 4$

Step 3: Taking the mole ratio as their subscripts.

The ratio of K : H : P : O = 1 : 2 : 1 : 4

Hence, the empirical formula for the given compound is $KH_2PO_4$

3 0

3 years ago

The effort needed when a simple machine is used is best described by which of these statements?

defon

C.The effort force is most often smaller than the force needed without the machine.

7 0

3 years ago

Photon energy is proportional to which of the following quantities? speed of light in a vacuum frequency of the light emitting t

charle [14.2K]

Answer:

The energy of the photon is directly proportional to the frequency of the light emitting the photon.

Explanation:

Energy of the photon is given by expression:

$E=h\times \nu=\frac{hc}{\lambda }$

E = Energy associated with the photon

$\nu$ = frequency of the light

h = Planck's constant = $6.626\times 10^{-34} J s$

c = speed of the light in vacuum = constant

$\lambda$ =Wavelength of the photon

$E\propto \nu$

The energy of the photon is directly proportional to the frequency of the light emitting the photon.

5 0

3 years ago