How old is a bone if it presently contains 0.3125g of C-14, but it was estimated to have originally contained 80.000g of C-14?

Read this excerpt from Emerson’s essay “Self-Reliance” and decide whether it is biased. Explain your answer as well as what you

nadezda [96]

Answer:

In this excerpt, Emerson is in favor of, and stresses the importance of, individual intellect and moral development. He urges readers to look within—into their inner selves to seek inspiration and attain self-reliance rather than imitating others or adopting age-old ideas, which society has practiced over time. He uses examples of great thinkers such as Moses, Milton, and Plato, who thought for themselves and didn’t let society and established ideas influence them. Emerson further stresses that people should not dismiss their own ideas, only to discover them in others’ works, but learn to nurture and accept their potential. When people are able to accept and trust their own thoughts, they will be able to come up with ideas as brilliant and original as those of any wise man.

Explanation:

Hope this helps.

3 0

3 years ago

An example of a high explosive is what?

Mashutka [201]

Answer:

High explosives do not need to be contained to make their bang. Nitroglycerine, trinitrotoluene, and RDX are high explosives.

Explanation:

6 0

2 years ago

5. Calculate the decrease in temperature when 6.00 L at 20.0 °C is compressed to 4.00 L.

Kazeer [188]

Answer: 195.3 K

Explanation: Use Charles Law to get T2

Charles Law V1 / T1 = V2 / T2

Derive the equation to find T2

T2 = V2T1 / V1

Convert temperature into K

T1 = 20.0 °C + 273 = 293 K

Substitute the values:

4.00 L x 293 K / 6.00 L = 195. 3 K

5 0

3 years ago

Please help me like now please

777dan777 [17]

Answer:

1-1) NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

1-2) 0.5 mole of CO2

2-1) 2C4H10 + 13O2 --> 8CO2 + 10H2O

2-2) 4 mol CO2

Explanation:

Question 1

NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

To balance the equation, count the number of atoms on both sides of the equation

(1 Na, 1+3+1H, 1+1+1C, 3+2Oxygen) --> (1 Na, 1+1+1C, 3+2H, 2+1+2Oxygen)

Combining the pluses will give you the following

(1 Na, 5H, 3C, 5Oxygen) --> (1 Na, 3C, 5H, 5Oxygen)

Both sides are the same, therefore the chemical equation is balanced (originally).

From the equation, we can see that 1 mole of NaHCO3 produces 1 mole of CO2.

So that means 0.5 mole of NaHCO3 would produce 0.5 mole of CO2.

Question 2

C4H10 + O2 --> CO2 + H2O

Again, count the number of atoms on both sides of the equation

(4C, 10H, 2O) --> (1C, 2H, 3O) This time left does not equal right side

You now need to find factors that can make both sides equal.

C4H10 + O2 --> 4CO2 + H2O Now the C is balanced, let's recount 

(4C, 10H, 2Oxygen) --> (4C, 8+1Oxygen, 2H) H&O is still not balanced

C4H10 + O2 --> 4CO2 + 5H2O Now the H is balanced, let's recount

(4C, 10H, 2Oxygen) --> (4C, 8+5Oxygen, 10H) O is still not balanced

C4H10 + (13/2)O2 --> 4CO2 + 5H2O Now the O is balanced

(4C, 10H, 13Oxygen) --> (4C, 13Oxygen, 10H)

But because 13/2 is a fraction, we want to eliminate that by multiplying every reactant and product by 2 (the denominator).

2C4H10 + 13O2 --> 8CO2 + 10H2O Now it's completely balanced!

(8C, 20H, 28Oxygen) --> (8C, 28Oxygen, 20H) Yayy! It's balanced.

Now, 2 mol C4H10 produces 8 mol CO2.

So 1 mol C4H10 produces 4 mol CO2.

6 0

2 years ago

1. Show that heat flows spontaneously from high temperature to low temperature in any isolated system (hint: use entropy change

Inga [223]

Answer:

1 ) Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

= Q/tc - Q/th

2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

Explanation:

1) To show that heat flows spontaneously from high temperature to low temperature

example :

Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )

Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment

Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )

Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

= Q/tc - Q/th

2) Entropy change for Decomposition of mercuric oxide

2HgO (s) → 2Hg(l) + O₂ (g)

Δs = positive

there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C

hence ΔSdecomposition = S⁻ Hg - S⁻ HgO =

Δh of reaction = 181.6 KJ

Temp = 500 + 273 = 773 k

hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

8 0

3 years ago