Carbon-14 emits beta radiation and decays with a half life of 5730 years. Assume you start with 2.5x10^15 grams of Carbon-14, how many grams remain at
the end of 5 half lives?
A. 1.1x10^-10 g
B. 2.5x10^-10 g
C 1.3x10^-10 g
D. 5.0x10^-10 g

Answers

Answer 1

The amount remaining at the end of 5 half-lives is 7.81×10¹³ g

From the question given above, the following data were obtained:

Half-life (t½) = 5730 yearsOriginal amount (N₀) = 2.5×10¹⁵ gNumber of half-lives (n) = 5Amount remaining (N) =?

The amount remaining can be obtained as follow:

N = 1/2ⁿ × N₀

N = 1/2⁵ × 2.5×10¹⁵

N = 1/32 × 2.5×10¹⁵

N = 0.03125 × 2.5×10¹⁵

N = 7.81×10¹³ g

Therefore, the amount remaining after 5 half-lives is 7.81×10¹³ g

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In the conclusion to the snake mole lab, you have to convert 3.45 moles of (NH4)2S04 to
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Molar mass = 132.14g/mol

Calculate the energy of a proton of light with a frequency of 8.5x1014 Hz.

Answers

Answer:

[tex] \huge{5.632 \times {10}^{ - 19} \: J}[/tex]

Explanation:

The energy of the proton can be found by using the formula

E = hf

where

E is the energy in J

f is the frequency in Hz

h is the Planck's constant which is

6.626 × 10-³⁴ Js

From the question we have

E = 6.626 × 10-³⁴ × 8.5 × 10¹⁴

We have the final answer as

[tex]5.632 \times {10}^{ - 19} \:J[/tex]

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which acid is part of the white coating on sour patch kids?

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Tartaric acid is the white, powdery substance that coats tart candies such as Sour Patch Kids. COmbustion analysis of a 12.01g sample of tartaric aid, which contains carbon, hydrogen and oxygen, produces 14.08g CO2 and 4.32g

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Answer the following questions.
P1) Matching.
Complete the table with the data from the lab. First, calculate and enter the mole ratio in the chart, then enter the volumes of the precipitates.

Answers

The mole ratio can be seen below and the volume of each precipitate is 60 mL.

A mole ratio refers to a conversion factor that compares the quantities of two chemicals in moles in a chemical laboratory experiment.

Mole ratio For 1:

= 10 mL : 50 mL= 1 : 5 mL

Mole ratio For 2:

15 mL : 45 mL= 1 : 3 mL

Mole ratio For 3:

20 mL : 40 mL1 : 2 mL

Mole ratio For 4:

30 mL : 30 mL1 mL : 1 mL

Mole ratio For 5:

40 mL : 20 mL2 : 1 mL

Mole ratio For 6:

45 mL : 15 mL3 : 1 mL

Mole ratio For 7:

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Since the parameters from the left side of the diagram are not shown, we will assume that the volumes for each precipitate are the addition of both volumes in each column.

By doing so, we have:

1.

(10 +50) mL = 60 mL

2.

(15 + 45) mL = 60 mL

3.

(20 + 40)mL = 60 mL

4.

(30 + 30) mL = 60 mL

5.

(40 + 20)mL = 60 mL

6.

(45 + 15) mL = 60 mL

7.

(50 + 10)mL = 60 mL

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Answers

Answer:

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Explanation:

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Knowing this, divide your mass by volume:

10 g / 2 ml = 5 g/ml

Final answer:

Destiny = 5 g/ml

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Answer:

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Hope you could get an idea from here.

Doubt clarification - use comment section.

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Answer:

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Explanation:

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Explanation:

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From the question given above, the following data were obtained:

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Molecular formula = empirical × n = molar mass

[CH₃O]n = 62

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Divide both side by 31

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Answers

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5 Stoichiometry Chemistry questions !



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3. The accompanying diagram represents a cell in water. Formulas of molecules that can move freely across the cell membrane are shown. Some molecules are located inside the cell and others are in the water outside the cell.



(Picture) diagram



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Answers

Answer:

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Learn more about diatomic molecules:

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The diagram in question is attached below.

Butane (C4 H10(g), Delta. Hf = â€"125. 6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Delta. Hf = â€"393. 5 kJ/mol ) and water (H2 O, Delta. Hf = â€"241. 82 kJ/mol) according to the equation below. 2 upper C subscript 4 upper H subscript 10 (g) plus 13 upper o subscript 2 (g) right arrow 8 upper C upper O subscript 2 (g 0 plus 10 upper H subscript 2 upper O (g). What is the enthalpy of combustion (per mole) of C4H10 (g)? Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants. â€"2,657. 5 kJ/mol â€"5315. 0 kJ/mol â€"509. 7 kJ/mol â€"254. 8 kJ/mol.

Answers

The enthalpy of this reaction is -5315 KJ/mol.

The equation of the reaction is;

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

We know that the enthalpy of reaction can be obtained from the enthalpy of formation of the reactants and products as follows;

ΔHrxn = ΔHf(products) - ΔHf(reactants)

We have the following information from the question;

ΔHf C4H10 = - 125. 6 kJ/mol

ΔHf CO2 = - 393. 5 kJ/mol

ΔHf H2O = - 241. 82 kJ/mol

ΔHf O2 = 0 KJ/mol

Hence;

[(8 × (- 393. 5 )) + (10 × (-  241. 82))] - [2( - 125. 6))]

= -5315 KJ/mol

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