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3 years ago

Select all the correct answers.

Chemistry

1 answer:

8_murik_8 [283]3 years ago

7 0

There should be a sufficient amount of the selected isotope in the rock.

The half-life of the isotope must be long enough to capture the age of the rock.

Explanation:

Sully must consider two main aspect before selecting her choice isotope for dating.

There must be sufficient amount of the selected isotope in the rock.

The half - life of the isotope must be long enough to capture the age of the rock.

Radiometric dating gives a rock an absolute numerical age.
The half-life of an isotope is time take for half of a radioactive element to decay.
If the half-life of an isotope is very short, all the parent nuclide would have turned to daughter nuclides.
Also, we must have sufficient amount of both the daughter and parent isotope in the selected rock.

learn more:

Radiometric dating brainly.com/question/7022607

#learnwithBrainly

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A helium filled ballon had a volume of 8.50 L on the ground at 20.0 C and a pressure of 750.0 Torr. After the ballon was release

Marrrta [24]

Answer:

$V_2=12.1L$

Explanation:

Hello!

In this case, according to the given data of volume, pressure and temperature, it is possible to infer this problem can be solved via the combined gas law:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

Thus, regarding the question, we evidence we need V2, but first we make sure the temperatures are in Kelvins:

$T_1=20+273=293K\\\\T_2=-40+273=233K$

Then, we obtain:

$V_2=\frac{P_1V_1T_2}{T_1P_2}\\\\V_2=\frac{0.987atm*8.50L*233K}{293K*0.550atm}\\\\V_2=12.1L$

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5 0

3 years ago

On a distance-time graph, at 2 hours the graph is at a height of 20 meters, and at 3 hours it is at a height of 100 meters. What

Gnoma [55]

Answer:

80 m/hr

Explanation:

Changes from 20 to 100 meters in ONE Hour

changes 80 meters in one hour = 80 m/hr

8 0

2 years ago

For the following reaction, 22.6 grams of nitrogen monoxide are allowed to react with 4.64 grams of hydrogen gas . nitrogen mono

antiseptic1488 [7]

Answer:

- 10.5 g of N₂

- Limiting reagent: NO

- 3.13 g of H₂ remains

Explanation:

First of all we state the reaction: 2NO(g) + 2H₂(g) → 2H₂O(l) + N₂(g)

We need to find out the limiting reactant and the excess reagent

Ratio in the reactants is 2:2. Let's convert the mass to moles:

22.6 g / 30 g/mol = 0.753 moles of NO

4.64 g / 2 g/mol = 2.32 moles of H₂

Certainly the limiting reagent is the NO and the excess reactant is the hydrogen:

- For 0.753 moles of NO, we need 0.753 moles of H₂ (we have 2.32 moles)

- For 2.32 moles of H₂, we need 2.32 moles of NO (and we don't have enough NO, because we only have 0.753 moles)

As the H₂ is the excess reagent, some moles still remains after the reaction is complete → 2.32 mol - 0.753 mol = 1.567 moles

We convert the moles to mass: 1.567 mol . 2g /1mol = 3.13 g of H₂ remains

As the NO is the limiting reagent, we can work with the equation:

We propose this rule of three: 2 moles of NO can produce 1 mol of N₂

Then, 0.753 moles of NO must produce (0.753 . 1) /2 = 0.376 moles of N₂

We convert the moles to mass 0.376 mol . 28 g / 1 mol = 10.5 g

3 0

3 years ago

A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C･g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('

polet [3.4K]

Answer:

$a=5.65cm$

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

$Q_{Pt}=-Q_{Deu}$

We can represent the heats in terms of mass, heat capacities and temperatures:

$m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})$

Thus, we solve for the mass of platinum:

$m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g$