What happens if the metal you throw in is MORE REACTIVE than the metal ion in solution?

Suppose you have a spherical balloon filled with air at room temperature and 1.0 atm pressure; its radius is 17 cm. You take the

Sladkaya [172]

Answer: 17.8 cm


Explanation:


1) Since temperature is constant, you use Boyle's law:


PV = constant => P₁V₁ = P₂V₂

=> V₁/V₂ = P₂/P₁

2) Since the ballon is spherical:

V = (4/3)π(r)³

Therefore, V₁/V₂ = (r₁)³ / (r₂)³


3) Replacing in the equation V₁/V₂ = P₂/P₁:

(r₁)³ / (r₂)³ = P₂/P₁

And you can solve for r₂: (r₂)³ = (P₁/P₂) x (r₁)³

(r₂)³ = (1.0 atm / 0.87 atm) x (17 cm)³ = 5,647.13 cm³

r₂ = 17.8 cm

4 0

3 years ago

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Many scientists have been involved in the development of the theory of the structure of

Genrish500 [490]

Answer:

bohr descirbed the atomic structure and found that electrons travel in separate orbits around the nucleus. jj thomson created the plum pudding model.

7 0

2 years ago

Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of

harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g) (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

$n = \frac{m}{M}$

Where:

m: is the mass

M: is the molar mass

For hydrogen we have:

$n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles$

And for nitrogen:

$n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles$

We can see in reaction (1) that 3 moles of hydrogen react with 1 mol of nitrogen, so the number of hydrogen moles needed to react nitrogen is:

$n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles$

Since we have 3.47 moles of hydrogen and we need 7.50 moles to react with all the mass of nitrogen, the limiting reactant is hydrogen.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that 3 moles of hydrogen produces 2 moles of ammonia, so:

$n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles$