Answer:
<em>C</em><em> </em><em>.</em><em> </em><em>the</em><em> </em><em>dramatic</em><em> </em><em>scenery</em><em> </em><em>created</em><em> </em><em>by</em><em> </em><em>volcanic</em><em> </em><em>eruptions</em><em> </em><em>attracts</em><em> </em><em>tourists</em><em>. </em>
The homologous structures and the analogous structures of different species.
Answer:
Acceleration = (change in speed) / (time for the change)
Change in speed= (0 - 26 km/hr) = -26 km/hr
(-26 km/hr) x (1,000 m/km) x (1 hr / 3,600 sec) = -7.222 m/sec
Average acceleration = (-7.222 m/s) / (22 min x 60sec/min) = -0.00547 m/sec²
Average speed during the stopping maneuver =
(1/2) (start speed + end speed) = 13 km/hr = 3.6111 m/sec
Explanation:
Answer:
0.4 moles of water produced by 6.25 g of oxygen.
Explanation:
Given data:
Mass of oxygen = 6.25 g
Moles of water produced = ?
Solution:
Chemical equation;
2H₂ + O₂ → 2H₂O
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 6.35 g/ 32 g/mol
Number of moles = 0.2 mol
Now we will compare the moles of oxygen with water:
O₂ : H₂O
1 : 2
0.2 : 2×0.2 = 0.4 mol
0.4 moles of water produced by 6.25 g of oxygen.
