Answer:
I make not know because im the 7th grade but Im give my worthy guess I now this its endothermic
Explanation:
The answer should be <span>C. nuclear reaction.</span>
<h3>
Answer:</h3>
7.226 × 10^23 molecules.
<h3>
Explanation:</h3>
- A compound is a substance that is made by two or more atoms from different elements.
- A mole of a compound contains a number of molecules equivalent to Avogadro's number, 6.022 × 10^23.
- That is, one mole of a compound contains 6.022 × 10^23 molecules.
In this case we are given;
Number of moles of H₂O as 1.2 moles
But, 1 mole of H₂O contains 6.022 × 10^23 molecules.
We are required to calculate the number of molecules present;
- To calculate the number of molecules we are going to multiply the number of molecules in one mole by the number of moles.
Number of molecules = 1.2 moles × 6.022 × 10^23 molecules/mole
= 7.226 × 10^23 molecules.
Thus, 1.2 moles of water contains 7.226 × 10^23 molecules.
Answer:
Δ[NH₃]/Δt = 2/3 ( Δ[H₂]/Δt )
Explanation:
For determining rates as a function of reaction coefficients one should realize that these type problems are <u>always in pairs</u> of reaction components. For the reaction N₂ + 3H₂ => 2NH₃ one can compare ...
Δ[N₂]/Δt ∝ Δ[H₂]/Δt, or
Δ[N₂]/Δt ∝ Δ[NH₃]/Δt, or
Δ[H₂]/Δt ∝ Δ[NH₃]/Δt, but never 3 at a time.
So, set up the relationship of interest ( ammonia rate vs. hydrogen rate)... nitrogen rate is ignored.
Δ[H₂]/Δt ∝ Δ[NH₃]/Δt
Now, 'swap' coefficients of balanced equation and apply to terms given then set term multiples equal ...
N₂ + 3H₂ => 2NH₃ => 2(Δ[H₂]/Δt) = 3(Δ[NH₃]/Δt) => 2/3(Δ[H₂]/Δt) = (Δ[NH₃]/Δt)
NOTE => Comparing rates individually of the component rates in reaction process, the rate of H₂(g) consumption is 3/2 times <u>faster</u> than NH₃(g) production (larger coefficient). So, in order to compose an equivalent mathematical relationship between the two, one must reduce the rate of the H₂(g) by 2/3 in order to equal the rate of NH₃(g) production. Now, given the rate of one of the components as a given, substitute and solve for the unknown.
CAUTION => When Interpreting rate of reaction one should note that the rate expression for an individual reaction component defines 'instantaneous' rate or speed. <u>This means velocity (or, speed) does not have 'signage'</u>. Yes, one may say the rate is higher or lower as time changes but that change is an acceleration or deceleration for one instantaneous velocity to another. Acceleration and Deceleration do have signage but the positional instantaneous velocity (defined by a point in time) does not. Thus is reason for the 'e-choice' answer selection without the signage associated with the expression terms.
