When CH₄ is burnt in excess O₂ following products are formed,
CH₄ + 2 O₂ → CO₂ + 2 H₂O
According to equation 1 mole of CH₄ (16 g) reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O. Hence the products are,
1 mole of CO₂ and 2 moles of H₂O
Converting 1 mole CO₂ to grams;
As,
Mass = Moles × M.mass
Mass = 1 mol × 44 g.mol⁻¹
Mass = 40 g of CO₂
Converting 2 moles of H₂O to grams,
Mass = 2 mol × 18 g.mol⁻¹
Mass = 36 g of H₂O
Total grams of products;
Mass of CO₂ = 44 g
+ Mass of H₂O = 36 g
-------------
Total = 80 g of Product
Result:
80 grams of product is formed when 16 grams of CH₄ is burnt in excess of Oxygen.
Answer: noble gasses
Explanation: zoom on on the red section and look at the key for what red means.
It’s 20500000, here are some extras if u need
Answer:
<h2>D. Krypton (Kr)</h2>
<u>Explanation:</u>
because
Neon and Krypton, both are belongs to Inert or Noble gas Group
Answer:
87.5 mi/hr
Explanation:
Because a = Δv / Δt (a = vf - vi/ Δt), we need to find the acceleration first to know the change in velocity so we can determine the final velocity.
vf = 60 mi/hr
vi = 0 mi/hr
Δt = 8 secs
a = vf - vi/ Δt
= 60 mi/hr - 0 mi/hr/ 8 secs
= 60 mi/hr / 8 secs
= 7.5 mi/hr^2
Now that we know the acceleration of the car is 7. 5 mi/hr^2, we can substitute it in the acceleration formula to find the final velocity when the initial velocity is 50 mi/hr after 5 secs.
vi = 50 mi/ hr
Δt = 5 secs
a = 7.5 mi/ hr^2
a = vf - vi/ Δt
7.5 = vf - 50 mi/hr / 5 secs
37.5 = vf - 50
87.5 mi/ hr = vf