Answer:
56
Explanation:
1 mole of gas at STP occupies 22.4 L of the gas
2.5 mole of the gas at STP occupies 22.4×2.5 L of the gas
so 2.5 mole of the gas at STP occupies 56 L of the gas .
<h3>
Answer:</h3>
2000 atoms
<h3>
Explanation:</h3>
We are given the following;
Initial number of atoms of radium-226 as 8000 atoms
Time taken for the decay 3200 years
We are required to determine the number of atoms that will remain after 3200 years.
We need to know the half life of Radium
- Half life is the time taken by a radio active material to decay by half of its initial amount.
- Half life of Radium-226 is 1600 years
- Therefore, using the formula;
Remaining amount = Original amount × 0.5^n
where n is the number of half lives
n = 3200 years ÷ 1600 years
= 2
Therefore;
Remaining amount = 8000 atoms × 0.5^2
= 8000 × 0.25
= 2000 atoms
Thus, the number of radium-226 that will remain after 3200 years is 2000 atoms.
Answer:
Four times the original amount if only one orange was used
Explanation:
We can assume that the oranges all have equal voltages. Connecting them in series will have an increasing effect on the voltage delivered. In our case, this will produce 4 times the voltage of the circuit when only one orange is used.
Whenever simple cells are connected in series, the voltages of the individual cells are added up to form the voltage of the whole circuit.
Let us assume that the voltage of each of the oranges is approximately 0.9 volts. The Voltage produced when the 4 oranges are joined in series is 0.9 + 0.9 + 0.9 + 0.9 = 3.6 volts
Answer:
Where the products are H2O and Ba(NO3)2
Explanation:
A base, as, barium hydroxide (Ba(OH)2) reacts with an acid (HNO3), producing water (H2O), and the related salt (Ba(NO3)2) in a reaction called <em>neutralization reaction.</em>
The balanced reaction is:
Ba(OH)2 + 2 HNO3 → 2 H2O + Ba(NO3)2
<em>Where the products are H2O and Ba(NO3)2</em>
Answer:
9 : 8
Explanation:
Aluminum oxide has the following formula Al₂O₃.
Next, we shall determine the mass of Al and O₂ in Al₂O₃. This can be obtained as follow:
Mass of Al in Al₂O₃ = 2 × 27 = 54 g
Mass of O₂ in Al₂O₃ = 3 × 16 = 48 g
Finally, we shall determine the mass ratio of Al and O₂. This can be obtained as follow:
Mass of Al = 54 g
Mass of O₂ = 48 g
Mass of Al : Mass of O₂ = 54 : 48
Mass of Al : Mass of O₂ = 54 / 48
Mass of Al : Mass of O₂ = 9 / 8
Mass of Al : Mass of O₂ = 9 : 8
Therefore, the mass ratio of Al and O₂ in Al₂O₃ is 9 : 8
