Someone please help me

A sample of propane(c3h8)has 3.84x10^24 H atoms.

deff fn [24]

Answer:

A) 14. 25 × 10²³ Carbon atoms

B) 34.72 grams

Explanation:

1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.

The sample has 3.84 × 10²⁴ H atoms.

If 8 atoms of Hydrogrn are present in 1 molecule of propane.

3.84 × 10²⁴ H atoms are present in

$\mathfrak{ \frac{3.8 }{8} \times 10 ^{24}}$

= 4.75 × 10²³ molecules of Propane.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

No. of Carbon atoms in 1 molecule of propane = 3

=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³

= 14.25 × 10²³ 

________________________________________

Gram Molecular Mass of Propane(C3H8)

= 3 × 12 + 8 × 1

= 36 + 8

= 44 g

1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.

=> 6.02 × 10²³ molecules of Propane weigh = 44 g

=> 4. 75 × 10²³ molecules of Propane weigh =

$\mathsf{ \frac{44 }{6.02 \times {10}^{23} } \times 4.75 \times {10}^{23} }$

$\mathsf{ = \frac{44 }{6.02 \times \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}} }$

$\mathsf{ = \frac{44 }{6.02 } \times 4.75 }$

= 34.72 g

8 0

3 years ago

Do all metals have more electrons than protons

Furkat [3]

Answer:

no

Explanation:

4 0

4 years ago

How do solve for #13?What is the boiling point of a solution made by dissolving 1.0000 mole of sucrose in 1.0000 kg of water?

exis [7]

What is the boiling point of a solution made by dissolving 1.0000 mole of sucrose in 1.0000 kg of water?

The change in Boiling Point of water can be calculated using this formula:

ΔTb = i * Kb * m

Where i is the van't hoff factor (the number of particles or ions), the kb is a constant (boiling point elevation constant) and m is the molality of the solution.

The kb for water is always 0.515 °C/m. Kb = 0.515 °C/m

The value for i in this case is 1. Since sucrose is a covalent compound and it doesn't dissociate into ions. i = 1

The molal concentration of the solution can be found using this formula:

molality = moles of sucrose/kg of water

molality = 1.000 mol / 1.000 kg of water

molality = 1 m

Now that we know all the values, we can use the formula to find the change in the boiling point of water:

ΔTb = i * Kb * m

ΔTb = 1 * 0.515 °C/m * 1 m

ΔTb = 0.515 °C

Finally, we are asked for the boiling point of the solution, not the change. The boiling point of water at atmospheric pressure is 100.00 °C. If the boiling point rises 0.515 °C when we prepare the solution. The boiling point of the solution is:

Boiling point solution = Boiling point of water + ΔTb

Boiling point solution = 100.000 °C + 0.515 °C

Boiling point solution = 100.515 °C

Answer: The boiling point of the solution is 100.515 °C.

8 0

1 year ago

The mass lost by the positive electron should equal the mass gained by the negative electrode suggest two reasons why the result

vesna_86 [32]

Answer:

The gain in mass by the negative electrode is the same as the loss in mass by the positive electrode. So the copper deposited on the negative electrode must be the same copper ions that are lost from the positive electrode.

3 0

3 years ago

What two parts of an amino acid are involved in a peptide bond?

Lemur [1.5K]

Answer:

The answer to your question is below.

Explanation:

Amino acids are composed by one amino group, one carboxyl group and one chain.

The parts of the amino acid that are involved in a peptide group are the amino group (- NH₂) and the carboxyl group (-COOH).

6 0

3 years ago