Answer:
2.14 moles of H₂O₂ are required
Explanation:
Given data:
Number of moles of H₂O₂ required = ?
Number of moles of N₂H₄ available = 1.07 mol
Solution:
Chemical equation:
N₂H₄ + 2H₂O₂ → N₂ + 4H₂O
now we will compare the moles of H₂O₂ and N₂H₄
N₂H₄ : H₂O₂
1 : 2
1.07 : 2×1.07 = 2.14 mol
Answer:
The particles begin to vibrate faster and more.
Explanation:
Adding heat to matter increases the energy, thus creating more movement. Eventually, the bucket will melt, turning to a liquid. While it is a sold, it still has particle movement, just not enough to break volume or shape.
Answer:
7,94 minutes
Explanation:
If the descomposition of HBr(gr) into elemental species have a rate constant, then this reaction belongs to a zero-order reaction kinetics, where the r<em>eaction rate does not depend on the concentration of the reactants. </em>
For the zero-order reactions, concentration-time equation can be written as follows:
[A] = - Kt + [Ao]
where:
- [A]: concentration of the reactant A at the <em>t </em>time,
- [A]o: initial concentration of the reactant A,
- K: rate constant,
- t: elapsed time of the reaction
<u>To solve the problem, we just replace our data in the concentration-time equation, and we clear the value of t.</u>
Data:
K = 4.2 ×10−3atm/s,
[A]o=[HBr]o= 2 atm,
[A]=[HBr]=0 atm (all HBr(g) is gone)
<em>We clear the incognita :</em>
[A] = - Kt + [Ao]............. Kt = [Ao] - [A]
t = ([Ao] - [A])/K
<em>We replace the numerical values:</em>
t = (2 atm - 0 atm)/4.2 ×10−3atm/s = 476,19 s = 7,94 minutes
So, we need 7,94 minutes to achieve complete conversion into elements ([HBr]=0).
In Chemistry, to better determine the position of a certain electron, quantum numbers are used. The four quantum numbers are n, l, m and s. In the given above of n= 4, the principal quantum number is 4 and this represents the overall relative energy in the orbital. This means that we are to find the maximum number of electrons in fourth main energy level and the answer is 32.
Answer:
2,780,000mg
Explanation:
Using the Metric Staircase photo provided, you can calculate how many mg there are in 2.78 kg by simply moving the decimal point six places to the right, since the "Kilo" step takes six places to move to the "Milli" step.
2.78 kg
---------------
2 7 8 0 0 0 0 .
we can see that the decimal point is moved to the right six places.
Answer:
2,780,000mg