Question:
a. a direct linear relationship
b. an inverse linear relationship
c. a direct nonlinear relationship
d. an inverse nonlinear relationship
Answer:
The correct option is;
d. An inverse nonlinear relationship
Explanation:
From the universal gas equation, we have;
P·V = n·R·T
Where we have the temperature, T and the number of moles, n constant, therefore, we have
P×V = Constant, because, R, the universal gas constant is also constant, hence;
P×V = C
Since P varies with V then the graphical relationship will be an inverse nonlinear as we have
V P C
1 5 5
2 2.5 5
3 1.67 5
4 1.25 5
5 1 5
6 0.83 5
7 0.7 5
8 0.63 5
9 0.56 5
10 0.5 5
Where:
V = Volume
P = Pressure
C = Constant = 5
P = C/V
The graph is attached.
Jot down the formula of the ionic compound. Let's say the ionic compound you're working with is NaCl.
Write the name of the metal.
Add the name of the non-metal with an –ide ending.
Combine the cation and anion names.
Practice naming more simple ionic compounds.
Answer: The concentration of 
will be 
after 416 seconds have passed.
Explanation:
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = 
t = age of sample = ?
a = let initial amount of the reactant = 
a - x = amount left after decay process =
The concentration of will be after 416 seconds have passed.
QUICK ANSWER
J.J. Thomson's cathode ray experiment was a set of three experiments that assisted in discovering electrons. He did this using a cathode ray tube or CRT. It is a vacuum sealed tube with a cathode and anode on one side.
<h3>
Answer:</h3>
19.3 g/cm³
<h3>
Explanation:</h3>
Density of a substance refers to the mass of the substance per unit volume.
Therefore, Density = Mass ÷ Volume
In this case, we are given;
Mass of the gold bar = 193.0 g
Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm
We are required to get the density of the gold bar
Step 1: Volume of the gold bar
Volume is given by, Length × width × height
Volume = 0.50 cm × 10.0 cm × 2.0 cm
= 10 cm³
Step 2: Density of the gold bar
Density = Mass ÷ volume
Density of the gold bar = 193.0 g ÷ 10 cm³
= 19.3 g/cm³
Thus, the density of the gold bar is 19.3 g/cm³
