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coldgirl [10]
3 years ago
11

Use the drop-down menus to name these structures

Chemistry
1 answer:
Masja [62]3 years ago
5 0
You need to add an attachment or something this doesnt give the needed information
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Hydrogen 3 has a half life of 12.32 years a sample of h-3 weighing 3.02 grams is left for 15.0 years what will the final weight
yawa3891 [41]

Answer:

The final mass of sample is 1.3 g.

Explanation:

Given data:

Half life of H-3 = 12.32 years

Amount left for 15.0 years = 3.02 g

Final amount = ?

Solution:

First all we will calculate the decay constant.

t₁/₂ = ln² /k

t₁/₂ =12.32 years

12.32 y =  ln² /k

k = ln²/12.32 y

k = 0.05626 y⁻¹

Now we will find the original amount:

ln (A°/A) = Kt

ln (3.02 g/ A) = 0.05626 y⁻¹ × 15.0 y

ln (3.02 g/ A) = 0.8439

3.02 g/ A = e⁰°⁸⁴³⁹

3.02 g/ A = 2.33

A = 3.02 g/ 2.33

A = 1.3 g

The final mass of sample is 1.3 g.

8 0
3 years ago
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When you add salt to water it makes the solution boil
alukav5142 [94]

Answer:

A i believe

Explanation:

4 0
3 years ago
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What is the average Volume of a single penny from the total 25 penny data?
mojhsa [17]

Answer:

1.16 mL  

Explanation:

Assume the data show that 25 pennies have a total volume of 29.00 mL.

\text{Average volume} = \dfrac{\text{29.00 mL}}{25} = \textbf{1.16 mL}

7 0
3 years ago
The first stage of aerobic respiration is a series of biochemical reactions called the Calvin cycle.
Marina CMI [18]

Answer:

The answer is indeed true

Explanation:

8 0
3 years ago
What is the enthalpy of combustion (per mole) of C4H10 (g)? 
Artemon [7]
The balanced chemical reaction for the complete combustion of C4H10 is shown below:

                    C4H10 + (3/2)O2 --> 4CO2 + 5H2O

The enthalpy of formation are listed below:
          C4H10: -2876.9 kJ/mol
              O2:   none (because it is pure substance)
             CO2: -393.5 kJ/mol
             H2O: -285.8 kJ/mol

The enthalpy of combustion is computed by subtracting the total enthalpy formation of the reactants from that of the products.

               ΔHc = (4)(-393.5 kJ/mol) + (5)(-285.8 kJ/mol) - (-2876.9 kJ/mol)
                       = -<em>126.1 kJ</em>

Thus, the enthalpy of combustion of the carbon is -126.1 kJ. 
5 0
3 years ago
Read 2 more answers
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