Answer:
10 atm.
Explanation:
Using the combined gas law equation as follows;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (atm)
P2 = final pressure (atm)
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question,
P1 = 5 atm
P2 = ?
V1 = 4L
V2 = 2L
T1 = 25°C = 25 + 273 = 298K
T2 = 25°C = 298K
Using P1V1/T1 = P2V2/T2
5 × 4/298 = P2 × 2/298
20/298 = 2P2/298
Cross multiply
298 × 20 = 298 × 2P2
5960 = 596P2
P2 = 5960 ÷ 596
P2 = 10 atm.
Different fabrics rub together, and electrons may rub off
Answer:
D is the correct answer.
Explanation:
I took the test on A pex. Everyone make sure you look at your answers. They are NOT always in the same order.
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).
Answer: 
atoms of hydrogen are there in
35.0 grams of hydrogen gas.
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 
of particles.
To calculate the moles, we use the equation:
1 mole of hydrogen 
= 
atoms
17.5 mole of hydrogen = 
atoms
There are atoms of hydrogen are there in
35.0 grams of hydrogen gas.
