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Shalnov [3]
3 years ago
13

What is the difference between an empirical and a molecular formula? Give the empirical formulas of (a) C8H18, (b) P4O10.

Chemistry
1 answer:
PilotLPTM [1.2K]3 years ago
4 0
I have a big house it had 5 floors and it has a pool with a slide and I have a silicone baby doll and y’all know that those things are extremely expensive like about I would say $1,499 dollars
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Use Table 1: Analysis of Decay in the laboratory guide to answer this question: Scientists find a piece of wood that is thought
Sladkaya [172]
<span>We are given the initial amount of 1 million carbon-14 atoms and the final amount which is 1/16 of the current atmospheric 14C levels. Also, the half life of carbon is </span>5,750 years. WE can use the decay formula

Aₓ = A₀e^-(ln2/t1/2)t
1,000,000(1/16) = (1,000,000)e^-(ln2/5750)t
t = 23,000 years
6 0
3 years ago
What is the structure of earth's system?
Ber [7]
Earths structure is a sphere
8 0
3 years ago
A 50.0 mL sample of 12.0 M HCl is diluted to 200 mL. What is true about the diluted solution?
kobusy [5.1K]

The concentration of the solution reduces and the number of moles of solute isn't affected.

Data;

  • V1 = 50mL
  • C1 = 12.0M
  • V2 = 200mL
  • C2 = ?
<h3>Facts about the diluted solution</h3>

1. When the solution is diluted, the concentration changes and this time, the concentration reduces.

Using dilution formula

c_1 v_1 = c_2 v_2\\12 * 50 = c_2 * 200\\c_2 = \frac{600}{200} \\c_2 = 3M

The concentration of the solution reduces.

2. The number of moles remains the same.

When a solution is diluted, the number of moles remains the same because there's no change in the mass of the solute.

Learn more on concentration of a solution here;

brainly.com/question/2201903

7 0
3 years ago
Can someone help me with this question?
Softa [21]

Answer:

5 , I think. sorry if its wrong

8 0
3 years ago
Use the periodic table to determine the electron configuration for iodine (i). express your answer in condensed form.
Dovator [93]
Iodine electron configuration is:

1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 5S^2 4d^10  5P^5
when Krypton is the noble gas in the row above iodine in the periodic table,
we can change 1S^2  2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 by the symbol
[Kr] of Krypton.

So we can write the electron configuration of Iodine:
[Kr] 5S^2 4d^10 5P^5

8 0
3 years ago
Read 2 more answers
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