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3 years ago

Which atomic model shows the atom as a positively charged sphere with negatively charged electrons spread throughout

Chemistry

1 answer:

tankabanditka [31]3 years ago

5 0

Answer:

B) Thomson's Model

Explanation:

Dalton's Model simply consisted of small spheres that represented atoms, and Bohr's model had a nucleus in the center with electrons orbiting in strict orbits, not randomly spread throughout. Thomson's plum pudding model, however, had an overall positively charged atom with many electrons spread throughout. If you search up "Plum Pudding Model" you can visually understand this.

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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

Bess [88]

Answer:

4Al + 3O₂ → 2Al₂O₃

Explanation:

4Al + 3O₂ → 2Al₂O₃

Only this reaction above obeys the law of conservation of mass. The others flout the rule.

The law of conservation of mass states that "matter is neither created nor destroyed in a chemical reaction but are simply transformed from one form to another".

By this law, the number of atoms on both sides of expression must be the same;

Number of atoms

Elements Left hand side Right hand side

Al 4 4

O 6 6

8 0

2 years ago