The equation that shows the decomposition of silver carbonate is as follows:
<span>2Ag2CO3 .........> 4Ag + 2CO2 + O<span>2
From the periodic table:
mass of silver = 107.8682 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
molar mass of Ag2CO3 = 2(107.8682) + 12 + 3(16) = 275.7364 grams
From the balanced equation above:
2(275.7362) = 551.4728 grams of Ag2CO3 produces 4(107.8682) = 431.4728 grams of Ag
Therefore, to know the mass of Ag produced from 2.76 grams of Ag2CO3, we will simply use cross multiplication as follows:
mass of Ag produced = (2.76*431.4728) / (551.4728) = 2.16 grams
Based on the above calculations, the law of conservation of mass is applied.</span></span>
<span>HofO3 - [H of O(g) + H of O2] = deltaH
HofO3 = -107.2kJ/mol + 249.8 + 0 (Heat of formn.of O2 is 0
= +142.6kJ/mol </span>
Answer:
stationary stationary means not in motion also known as not moving
Answer:
Ligands
Explanation:
Ligands are small molecules that transmit signals in between or within cells. Ligands exert their effects by binding to cellular proteins called receptors.
<span>perpendicular to a uniform 1.50- magnetic field, as shown in the figure.
Part A
Calculate the net force which the magnetic field exerts on the coil.
ANSWER:
=0Correct
Part B
Calculate the torque which the magnetic field exerts on the coil.
ANSWER:
=0Correct
Part C
The coil is rotated through a 30.0 angle about the axis shown, the left side coming out of the plane of the figureand the right side going into the plane. Calculate the net force which the magnetic field now exerts on the coil.(
Hint:
In order to help visualize this 3-dimensional problem, make a careful drawing of the coil when viewedalong the rotation axis.)
ANSWER:
=0Correct
Part D
Calculate the torque which the magnetic field now exerts on the coil.
ANSWER:
=8.09Ă—10
â’2
Correct
Torque on a Current Loop in a Magnetic Field
Learning Goal:
To understand the origin of the torque on a current loop due to the magnetic forces on the current-carrying wires.This problem will show you how to calculate the torque on a magnetic dipole in a uniform magnetic field. We startwith a rectangular current loop, the shape of which allows us to calculate the Lorentz forces explicitly. Then wegeneralize our result. Even if you already know the general formula to solve this problem, you might find itinstructive to discover where it comes from
Part A
Find , the electric field inside the cube.
Hint A.1 Net force on charges in a conductor
Hint not displayed
Hint A.2 Find the magnetic force magnitude
Hint not displayed
Hint A.3 Find the magnetic force direction
Hint not displayed
Hint A.4 Determine the force due to the electric field
Hint not displayed
Express the electric field in terms of , , and unit vectors ( , , and/or ).
ANSWER:
=CorrectNow, instead of electrons, suppose that the free charges have
positive
charge . Examples include "holes" insemiconductors and positive ions in liquids, each of which act as "conductors" for their free charges.
Part B
If one replaces the conducting cube with one that has positive charge carriers, in what direction does the inducedelectric field point?
ANSWER:
CorrectThe direction of the electric field stays the same regardless of the sign of the charges that are free to move in theconductor.Mathematically, you can see that this must be true since the expression you derived for the electric field isindependent of .Physically, this is because the force due to the magnetic field changes sign as well and causes positive charges tomove in the direction (as opposed to pushing negative charges in the direction). Therefore the result isalways the same: positive charges on the side and negative charges on the side. Because the electric fieldgoes from positive to negative charges will always point in the direction (given the original directions of</span>