Answer:
Heating a substance increases the speed at which molecules move.
Explanation:
A chemical element that has an atomic number less than 58 and an atomic mass greater than 135.6m is barium (atomic no. 56 and atomic mass137.13 ) and lanthanum (atomic no. 57 and atomic mass 135.6).
<h3>Give a brief introduction about Barium and Lanthanum.</h3>
Barium is an element with the symbol Ba and atomic number 56. It is an alkaline earth metal that is soft and silvery, and it is the fifth element in group 2. Barium is never found in nature as a free element due to its extreme chemical reactivity. Oil well drilling fluid uses barium sulfate as an insoluble ingredient. It is employed as an X-ray radiocontrast agent in a purer form to image the human gastrointestinal tract. Barium compounds that dissolve in water have been employed as rodenticides despite being hazardous.
Chemical element lanthanum has the atomic number 57 and the symbol La. It is a silvery-white, ductile, soft metal that slowly tarnishes when exposed to air. It serves as the eponym for the group of 15 related elements in the periodic table between lanthanum and lutetium, of which lanthanum is the first and prototype. The rare earth elements traditionally include lanthanum.
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Mass of PH3= 6.086 g
<h3>Further explanation</h3>
Given
6.0 L of H2
Required
mass of PH3
Solution
Reaction
P4 + 6H2 → 4PH3
Assumed at STP ( 1 mol gas=22.4 L)
Mol of H2 for 6 L :
= 6 : 22.4 L
= 0.268
From the equation, mol PH3 :
= 4/6 x moles H2
= 4/6 x 0.268
= 0.179
Mass PH3 :
= 0.179 x 33,99758 g/mol
= 6.086 g
Answer:
<h3>601.93 g/mol</h3>
<h3>explanation:</h3>
Problem: The Ba3(PO4)2 (molar mass = 601.93 g/mol) precipitate that formed from a salt mixture has a mass of 0.667 g.
<u>Answer:</u> The amount remained after 151 seconds are 0.041 moles
<u>Explanation:</u>
All the radioactive reactions follows first order kinetics.
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process = 151 sec
= initial amount of the reactant = 0.085 moles
[A] = amount left after decay process = ?
Putting values in above equation, we get:
![4.82\times 10^{-3}=\frac{2.303}{151}\log\frac{0.085}{[A]}](https://tex.z-dn.net/?f=4.82%5Ctimes%2010%5E%7B-3%7D%3D%5Cfrac%7B2.303%7D%7B151%7D%5Clog%5Cfrac%7B0.085%7D%7B%5BA%5D%7D)
![[A]=0.041moles](https://tex.z-dn.net/?f=%5BA%5D%3D0.041moles)
Hence, the amount remained after 151 seconds are 0.041 moles