Answer:
P.E = 25.48 J
Explanation:
Given data:
Mass = 2 Kg
Height = 1.3 m
Potential energy = ?
Solution:
Formula:
P.E = m . g . h
P. E = potential energy
m = mass in kilogram
g = acceleration due to gravity
h = height
Now we will put the values in formula.
P.E = m . g . h
P.E = 2 Kg . 9.8 m /s² . 1.3 m
P.E = 25.48 Kg. m² / s²
Kg. m² / s² = J
P.E = 25.48 J
Answer:
602200000000000000000000
Explanation:
move the decimal place over 23 places to the right
According to Newton's third law, forces come in pairs. There is action-reaction force and an equal (in size - action force) and opposite in direction (reaction force).
<u>Explanation:</u>
On the off chance, when item A applies a force on item B, at that point item B must apply a force of equivalent size and inverse bearing back on object A. This Newton's third law speaks to a specific balance in nature: forces consistently happen two by two.
And, one body can't apply a force on another without encountering a force itself. Now and again, allude to this law freely as activity response, where the force applied is the activity and the force experienced as a result is the response.
Answer:
carbon dioxide and water
Explanation:
Example: Combustion of Methane (CH₄(g))
CH₄(g) + 2O₂(g) => CO₂(g) + 2H₂O(g)**
____________________
Note: The combustion of any hydrocarbon produces CO₂ & H₂O. That is,
Ethane (C₂H₆) + O₂ => CO₂(g) + H₂O(g)
Propane (C₃H₈) + O₂ => CO₂(g) + H₂O(g)
Butane (C₄H₁₀) + O₂ => CO₂(g) + H₂O(g)
The issue remaining is to balance the reaction equation. For these type equation balance Carbon 1st, then Hydrogen and finish with Oxygen. Balancing in this order leaves Oxygen which can be balanced using fractions. If problem requires lowest whole number ratios of elements, simply multiply entire equation by 2 to get standard equation*
______________________
*Standard Equation is defined as the smallest whole number ratios of elements. The 'standard equation' is significant in that it is assumed to be at STP conditions; i.e., 0⁰C (=273K) & 1.0 Atmosphere pressure.
- Ethane (C₂H₆) + 7/2O₂(g) => 2CO₂(g) + 3H₂O(g)
=> 2C₂H₆ + 7O₂(g) => 4CO₂(g) + 6H₂O(g) <= Standard Form of Rxn
- Propane (C₃H₈) + 5O₂(g) => 3CO₂(g) + 4H₂O(g) <= Standard Form of Rxn (no need to balance with the '2' multiple)
- Butane (C₄H₁₀) + 13/2O₂ => 4CO₂(g) + 5H₂O(g)
=> 2C₃H₈ + 13O₂(g) => 4CO₂(g) + 5H₂O(g) <= Standard Form of Rxn
______________________
**Also, note that water, H₂O(g), is listed as a gas. In some cases it will be listed as a liquid, H₂O(l).
22.4L
of any gas contains 1 mol of that gas.
50.75g/10L*22.4L/1 mol= 113.68g/mol- this is the mole weight of your gas
1 mol/113.68g*129.3g=1.137403 mol
Set up a ratio
1.137403mol/x L=1 mol/22.4 L
X=25.477827L, or with sig figs, x=25.5L
