Answer:
1. 80g
2. 1.188mole
Explanation:
1. We'll begin by obtaining the molar mass of CH4. This is illustrated below:
Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol
Number of mole of CH4 from the question = 5 moles
Mass of CH4 =?
Mass = number of mole x molar Mass
Mass of CH4 = 5 x 16
Mass of CH4 = 80g
2. Mass of O2 from the question = 38g
Molar Mass of O2 = 16x2 = 32g/mol
Number of mole O2 =?
Number of mole = Mass /Molar Mass
Number of mole of O2 = 38/32
Number of mole of O2 = 1.188mole
<u>Answer:</u>
<u>Plasmas of great interest to scientists or manufacturers as</u>
- Plasma is electrically charged gases that contain considerable charged particles that can change the behavior of the substance.
<u>Current uses of plasmas:</u>
- First, it is used to make semiconductors for different types of electronic equipment
- Secondly, they're used in making transmitters for high-temperature films.
<u>Way scientists and engineers hope to use plasmas in the future:</u>
- The scientists are hoping to use plasma in the future to get rid of all hazardous wastes through a process called plasma gasification.
Answer: 1. A, 2.B, 3. D, 4. B, 5. C
Explanation:
i have P.E. too lol have a great day!
Answer:
The balanced equation for this reaction will be
→ 
We can see that 1 mole of methane requires 4 moles of fluorine but we have 0.41 moles of CH4 and 0.56mole of F2
So using the unitary method we will get that
- 1 mole of CH4 → 4 mole of 4 mole of fluorine
- 0.41 mole of methane → 4*0.41 = 1.64 mole of fluorine for complete reaction
but we have only 0.56 mole of fluorine that means fluorine is the limiting reagent and the product will only be formed by only this amount of fluorine.
- 4 moles of fluorine → 1 mole of CF4
- 0.56 mole →
= 0.14mole of CF4
- 4 moles of fluorine → 4 moles of HF
- 0.56 mole of fluorine → 0.56 mole of HF
now to find the heat released we have the formula as
DELTA H = n * Delta H of product - n *delta H of reactant
where n is the moles of the reactant and product.
note: since no information is given about the enthalpies of the species we leave it on general equation also you need to add the product side enthalpy of the species present and similarly on the product side.