Answer:
1. C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O
2. V = 596L
Explanation:
Butane (C₄H₁₀) reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O) thus:
C₄H₁₀ + O₂ → CO₂ + H₂O
1. The balanced chemical equation is:
C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O
2. 0,360kg of butane are:
360g×
=<em>6,19moles of butane</em>
These moles of butane are:
6,19moles of butane×
= <em>24,8 moles CO₂</em>
Using V=nRT/P
Where:
n are moles (24,8 moles CO₂); R is gas constant (0,082atmL/molK); T is temperature, 20°C (293,15K); and P is pressure (1atm).
Volume (V) is:
<em>V = 596L</em>
I hope it helps!
To determine the moles in 40 grams of magnesium, we need the atomic weight. This can easily be found on a periodic table. For this problem, let's use 24.305 grams/mole.
We are going to set up an equation to determine this problem. In this equation, we want all our units to cancel out except for 'moles.'

In this, we can see that the unit 'grams' will cancel out to leave us with moles.
In solving the equation, we determine that there are approximately 1.65 moles of Magnesium.
I cant help u sorry .......
<span>This means that has great capacity to react with other chemical elements in nature, reacting mainly with sodium, therefore, can only be findings in chemical compounds in most cases.
hope this helps!.</span>
Answer:
280 g Al₂O₃
Explanation:
To find the mass, you need to multiply the given value by the molar mass. This will cause the conversion because the molar mass exists as a ratio; technically, the ratio states that there are 101.96 grams per every 1 mole Al₂O₃. It is important to arrange the ratio in a way that allows for the cancellation of units. In this case, the desired unit (grams) should be in the numerator. The final answer should have 2 sig figs to reflect the given value (2.7 mol).
Molar Mass (Al₂O₃): 101.96 g/mol
2.7 moles Al₂O₃ 101.96 g
------------------------ x ------------------- = 275 g Al₂O₃ = 280 g Al₂O₃
1 mole