<h3>You will pay $ 30876800</h3>
We'll begin by calculating the mass in ounce (oz) of a cube foot (ft³) of gold. This can be obtained as follow:
<h3 />
Density of gold = 19298 oz/ft³
Volume of gold = 1 ft³
<h3>Mass of gold =?</h3>
Density = mass /volume
19298 = mass / 1
<h3>Mass of gold = 19298 oz</h3>
Finally, we shall determine the cost of 19298 oz of gold. This can be obtained as follow:
1 oz = $ 1600
Therefore,
19298 oz = 19298 × 1600
19298 oz = $ 30876800
Therefore, a solid cube foot of gold (i.e 19298 oz) will cost $ 30876800
Learn more: brainly.com/question/15407624
Answer:
What mass (g) of barium iodide is contained in 188 mL of a barium iodide solution that has an iodide ion concentration of 0.532 M?
A) 19.6
B) 39.1
C) 19,600
D) 39,100
E) 276
The correct answer to the question is
B) 39.1 grams
Explanation:
To solve the question
The molarity ratio is given by
188 ml of 0.532 M solution of iodide.
Therefore we have number of moles = 0.188 × 0.532 M = 0.100016 Moles
To find the mass, we note that the Number of moles = 
from which we have
Mass = Number of moles × molar mass
Where the molar mass of Barium Iodide = 391.136 g/mol
= 0.100016 moles ×391.136 g/mol = 39.12 g
For the answer to the question above asking w<span>hen an atom of n-14 is bombarded by an alpha particle, the single product is?
</span> <span>You're starting with 14/7 N, correct?
An alpha particle is two protons, two neutrons, which is 4/2, correct?
</span><span>So I</span> think the answer to your question is the third one which is <span>c. 18/9 f </span>
<span>You have to use a Newman projection to make sure that the H on C#2 is anti-coplanar with the Br on C#1. (Those are the two things that are going to be eliminated to make the alkene.)
My Newman projection looks like this when it's in the right configuration:
Front carbon (C#2) has ethyl group straight up, H down/left, and CH3 down/right
Back carbon (C#1) has H straight down, Ph up/left, and Br up/right.
Then when you eliminate the H from C#2 and the Br from C#1, you will have Ph and the ethyl group on the same side of the molecule, and you'll have the remaining H and CH3 on the same side of the molecule.
This is going to give you (Z)-2-methyl-1-phenyl-1-butene.</span>
