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Solnce55 [7]
3 years ago
13

A researcher claims that an ancient scroll originated from greek scholars in about 500 bce. a measure of its carbon-14 decay rat

e gives a value that is 89% of that found in living organisms. how old is the scroll? (the half-life of carbon-14 is 5730 yr.) express the time in years as an integer.
Chemistry
1 answer:
Naddik [55]3 years ago
8 0
The amount of substance present in a certain object with a given half-life in terms of h can be expressed through the equation,

     A(t) = (A(o))(0.5)^(t/h)

where A(t) is the amount of substance after t years and A(o) is the original amount. In this item we are given that A(t)/A(o) is equal to 0.89. Substituting the known values,

     0.89 = (0.5)(t / 5730 years)

The value of t from the equation is 963.34 years.

<em>Answer: 963 years</em>
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What is the percent composition by mass of oxygen in Ca(NO3)2 (gram-formula= 164 g/mol)?
Neko [114]
To find this, we will use this formula:

Molar mass of element
------------------------------------ x 100
Molar mass of compound

So, first lets calculate the mass of the compound as a whole. We use the atomic masses on the periodic table to determine this.

Ca: 40.078 g/mol
N2 (there is two nitrogens): 28.014 g/mol
O6 (there are six nitrogens: 3 times 2): 95.994 g/mol

When we add all of those numbers up together, we get 164.086. That is the molar mass for the whole compound. However, we are trying to figure out what percent of the compound oxygen makes up. From the molar mass, we know that 95.994 of the 164.086 is oxygen. Lets plug those numbers into our equation!

95.994
-----------
164.086

When we divide those two numbers, we get .585. When we multiply that by 100, we get 58.5.

So, the percent compostition of oxygen in Ca(NO3)2, or, calcium nitrate, is 58.5%.
5 0
3 years ago
A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at
lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

Explanation:

<u>Step 1:</u> Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

<u> Step 2:</u> Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

<u>Step 3:</u> Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25  = 8.2 °C

q = 1200 * 4.184 * 8.2 =  41170.56 J

<u>Step 4</u>: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J  = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

<u>Step 5</u>: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

<u>Step 6:</u> Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

8 0
2 years ago
Where and where did BP Texas City Disaster happened
salantis [7]
The Texas City Refinery explosion occurred on March 23, 2005, when a hydrocarbon vapor cloud was ignited and violently exploded at the ISOM isomerization process unit at BP's Texas City refinery in Texas City, Texas, killing 15 workers, injuring 180 others and severely damaging the refinery.
4 0
2 years ago
A stable community that no longer goes through ecological changes?
Tanya [424]
This is called a climax community. 
6 0
3 years ago
The oxidation of copper(I) oxide, Cu 2 O ( s ) , to copper(II) oxide, CuO ( s ) , is an exothermic process. 2 Cu 2 O ( s ) + O 2
ICE Princess25 [194]

Answer:

The energy released as heat when 9.94 g Cu 2 O ( s ) undergo oxidation at constant pressure is -10.142 kJ

Explanation:

Here we have

2Cu₂O ( s ) + O₂ ( g ) ⟶ 4 CuO ( s ) Δ H ∘ rxn = − 292.0 kJ mol

In the above reaction, 2 Moles of Cu₂O (copper (I) oxide) react with one mole of O₂ to produce 4 moles of CuO, with the release of − 292.0 kJ/mol of energy

Therefore,

1 Moles of Cu₂O (copper (I) oxide) react with 0.5 mole of O₂ to produce 2 moles of CuO, with the release of − 146.0 kJ  of energy

We have 9.94 g of Cu₂O with molar mass given as 143.09 g/mol

Hence the number of moles in 9.94 g of Cu₂O is given as

9.94/143.09 = 6.95 × 10⁻² moles of Cu₂O

6.95 × 10⁻² moles of Cu₂O will therefore produce 6.95 × 10⁻² ×  − 146.0 kJ mol  or -10.142 kJ.

3 0
3 years ago
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