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fredd [130]
3 years ago
9

Use Table 1: Analysis of Decay in the laboratory guide to answer this question: Scientists find a piece of wood that is thought

to be from an ancient fire circle. They find that the wood contains an amount of carbon-14 (14C) that is approximately 1/16 of the current atmospheric 14C levels. Determine approximately how many years ago the tree was chopped down to be used for the firewood. If you started with 1 million carbon-14 atoms, how many atoms would remain in the wood? 14C has a t1/2 of 5,750 years.
Chemistry
1 answer:
Sladkaya [172]3 years ago
6 0
<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
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0.50 g of hydrogen chloride (HCl) is dissolved in water to make 4.0 L of solution. What is the pH of the resulting hydrochloric
svlad2 [7]

Explanation:

Given the mass of HCl is ---- 0.50 g

The volume of solution is --- 4.0 L

To determine the pH of the resulting solution, follow the below-shown procedure:

1. Calculate the number of moles of HCl given by using the formula:

number of moles of a substance=\frac{given mass of the substance}{its molecular mass}

2. Calculate the molarity of HCl.

3. Calculate pH of the solution using the formula:

pH=-log[H^+]

Since HCl is a strong acid, it undergoes complete ionization when dissolved in water.

HCl(aq)->H^+(aq)+Cl^-(aq)

Thus, [HCl]=[H^+]

Calculation:

1. Number of moles of HCl given:

number of moles of a substance=\frac{given mass of the substance}{its molecular mass}\\=0.50g/36.5g/mol\\=0.0137mol

2. Concentration of HCl:

Molarity of HCl=\frac{number of moles of HCl}{its molar mass}\\=\frac{0.0137 mol}{4.0 L} \\= 0.003425 M

3. pH of the solution:

pH=-log[H^+]\\=-log(0.003425)\\=2.47

Hence, pH of the given solution is 2.47.

3 0
3 years ago
So I am lost and I don't know how what any of those conversion factor things are and I'm wondering how to do #2,3, and 4. (I did
dalvyx [7]

Explanation:

#2.

A centigram is 1/100 of a gram, so that means a gram equals 100 centigrams.

Therefore you multiply 72.4 grams by 100/1 (or just 100), and get 7240 cg.

You did that one right but put the wrong unit in the answer. It is is cg ( centigrams).

#3.

1 liter is equal to 1000 milliliters, and I kiloliter is equal to 1000 liters. So one kiloliter is 1000*1000 milliliters or 1,000,000 milliliters.

The conversion factor would be

1/1000000

#4.

1 gigabyte is equal to 10^9 bytes.

I byte is equal to 10^9 bytes.

So 1 gigabyte is 10^9 * 10^9 nanobytes, or 10^18.

The conversion factor would be (1*10^18)/1.

6 0
3 years ago
Why do the deprotonated organic acid (RCO2-) and protonated organic base (RNH3+) go into the aqueous layer?
rodikova [14]

Answer:

The deprotonated organic acid (RCO2-) and protonated organic base (RNH3+) go into the aqueous layer because Organic compounds that are neither acids or bases do not react with either NaOH or HCl and, therefore remain more soluble in the organic solvent and are not extracted.

Explanation:

4 0
3 years ago
Genetics and reproduction
Zepler [3.9K]

genetics and reproduction is all about dna.

7 0
2 years ago
At constant pressure, which of these systems do work on the surroundings? A ( s ) + B ( s ) ⟶ C ( g ) A(s)+B(s)⟶C(g) 2 A ( g ) +
Tju [1.3M]

Correct question:

At constant pressure, which of these systems do work on the surroundings?

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

(c) A ( g ) + B ( g ) ⟶ C ( g )

(d) 2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

Answer:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

Explanation:

Work done by a system on the surroundings at a constant pressure is given as;

W = -PΔV

Where;

ΔV is gas expansion, that is final volume of the gas minus initial volume of the gas must be greater than zero.

Part (a)

A ( s ) + B ( s ) ⟶ C ( g )

ΔV = 1 - (0) = 1 (expansion)

Part (b)

2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

ΔV = 5 - ( 2+ 2) = 1 (expansion)

Part (c)

A ( g ) + B ( g ) ⟶ C ( g )

ΔV = 1 - ( 1 + 1) = -1 (compression)

Part (d)

2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

ΔV = 3 - ( 4) = -1 (compression)

Thus, systems where there is gas expansion are in part (a) and part (b). The correct answers are:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

4 0
2 years ago
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