Answer:
When a front passes over an area, it means a change in the weather. Many fronts cause weather events such as rain, thunderstorms, gusty winds, and tornadoes. At a cold front, there may be dramatic thunderstorms. At a warm front, there may be low stratus clouds. Usually, the skies clear once the front has passed.
Explanation:
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Answer:
Chemistry, the science that deals with the properties, composition, and structure of substances (defined as elements and compounds), the transformations they undergo, and the energy that is released or absorbed during these processes
Answer:We are already given with the mass of the Xe and it is 5.08 g. We can calculate for the mass of the fluorine in the compound by subtracting the mass of xenon from the mass of the compound.
mass of Xenon (Xe) = 5.08 g
mass of Fluorine (F) = 9.49 g - 5.08 g = 4.41 g
Determine the number of moles of each of the element in the compound.
moles of Xenon (Xe) = (5.08 g)(1 mol Xe / 131.29 g of Xe) = 0.0387 mols of Xe
moles of Fluorine (F) = (4.41 g)(1 mol F/ 19 g of F) = 0.232 mols of F
The empirical formula is therefore,
Xe(0.0387)F(0.232)
Dividing the numerical coefficient by the lesser number.
XeF₆
Explanation:
Answer:
The mass of water produced is 4.29 g.
Explanation:
Given;
reacting mass of ethane, C₂H₆ = 2.5 g
reacting mass of oxygen, O₂ = 5.0 g
The balanced combustion reaction is given as follows;
2C₂H₆ + 7O₂ ----------------> 4CO₂ + 6H₂O
Based on the balanced equation above;
requires
7.0 moles of oxygen -----------------------> 2.0 moles of ethane
requires
5.0 g of oxygen --------------------------> x gram of ethane
x = (2 x 5) / 7
x = 1.43 g
Ethane is in excess of 1.07g (2.5 g - 1.43g)
Now, determine the mass of water produced by 1.43 g of ethane;
requires
2.0 moles of ethane -----------------------> 6.0 moles of water
requires
1.43 g of ethane --------------------------> y gram of water
y = (6 x 1.43) / 2
y = 4.29 g
Therefore, the mass of water produced is 4.29 g.