F = MA A = change in velocity Constant velocity = no change in velocity = Zero acceleration Ignoring friction, 5 kg * 0 m/s^2 = 0 N Therefore, there is no force (0 N) required to keep the object in motion if the surface friction is ignored.
I hope my answer has come to your help.
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The answer is the fourth choice, or Have membrane-bound organelles. Hope this helps you!
Please mark as Brainliest if this was correct!
-Belle
Answer:
False.
Explanation:
Rate of reaction also depends on the number of collision that is being happening in reacting molecules. So concentration of the reactants is also important to decide the rate of reaction.
The rate of reaction is dependent on the concentration as well.
As the concentration of reactants decreases with time number of collisions also decreases .
An example of a reaction that occurs within the core of a nuclear reactor is the nuclear fission reaction given:
- ²³⁵₉₂U + ¹₀n ---> ⁹⁰₃₈Sr + ¹⁴³₅₄ + 3 ¹₀n
<h3>What is a nuclear reactor?</h3>
A nuclear reactor is a device which produces electrical energy as a result of the nuclear reactions that take place within it.
In a nuclear reactor, the reaction that takes place within the core is a nuclear fission chain reaction.
In a nuclear fission reaction, the nucleus of larger atoms are split into the nucleus of smaller atoms when fast moving neutrons are used to bombard the nucleus of the large atom. The fission of the nucleus of the large atom results in the formation of atoms of lighter nucleus as well as more protons which then bombard more nucleus of the large atoms resulting in a chain reaction.
The chain reaction occurring within the nuclear reactor core is controlled by the insertion of boron rods which absorbs the excess neutrons produced.
An example of a reaction that occurs within the core of a nuclear reactor is given below:
²³⁵₉₂U + ¹₀n ---> ⁹⁰₃₈Sr + ¹⁴³₅₄ + 3 ¹₀n
Learn more about nuclear fission at: brainly.com/question/913303
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Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
