The reaction between C2H2 and O2 is as follows:
2C2H2 + 5O2 = 4CO2 + 2H2O
After balancing the equation, the reaction ratio between C2H2 and O2 is 2:5.
The moles of O2 in this reaction is 84.0 mol. According to the above ratio, the moles of C2H2 needed to react completely with the O2 is 84.0mole *2/5 = 33.6 mole.
Answer:
The freshwater sources that are generally in continuous motion and follow a defined path are called streams and rivers.
If I were to improve the lab then I will make the following changes:
- The experiment aimed to observe and model the effects of rivers on erosion. So, I can make a virtual model of the river and can compare the velocity, gradients and volume of rivers.
- Comparison between the low and high factors listed can help in computing the effect of the powerful river on erosion.
- The high velocity. gradient and volume of the river will cause more erosion as it exerts more force.
- The low volume, gradient and velocity river will affect in a less manner on erosion.
Explanation:
thats all i know ( correct me if im wrong please)
Explanation:
Below is an attachment containing the solution.
Heat radiates from the fire and cooks the marshmallow because heat transfer.
Answer:
91.16% has decayed & 8.84% remains
Explanation:
A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt
Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹
Time (t) = 1000yrs
A = fraction of nuclide remaining after 1000yrs
A₀ = original amount of nuclide = 1.00 (= 100%)
lnA = lnA₀ - kt
lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426
A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years
Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.
