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Lunna [17]
3 years ago
9

Scientists estimate Earth's age to be 4.6 billion years. Samarium has a half-life of approximately 106 billion years. From the p

resent date, approximately how many more years must pass before a sample of samarium that was created at the beginning of Earth's history contains 25 percent of the original parent material? 10 billion 26 billion 100 billion 206 billion
Chemistry
2 answers:
Wewaii [24]3 years ago
8 0

1/4 Ao = Ao * ( 1/2)^(t/106)
t = 212
subtract 4.6 billion years
<span>d)</span>
Alexandra [31]3 years ago
4 0

Answer: Option (D)

Explanation: The following steps are shown how decay takes place-

  1. Samarium (Sm) initially had 100% of its parent material.
  2. Samarium undergoes radioactive decay and becomes half of it i.e 50% of its parent material.
  3. To Decay this 50% of its initial material, it takes 106 billion years.
  4. Again it undergoes radioactive decay and becomes half of it i.e 25%, and during this process it takes another 106 billion years.

So from the above question, the total number of years for Samarium to contain 25% of its parent material will be (106+106)=212 billion years.

it can also be calculated from the radioactive decay formula.

Thus the most approximate option provided here is option (D).

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Tju [1.3M]

Answer:

"Organisms in this tank are attached to the bottom, but use the sun to make food."

Explanation:

<em>"The principal food sources for the benthos are plankton and organic debris from land. In shallow water, larger algae are important, and, where light reaches the bottom, benthic photosynthesizing diatoms are also a significant food source."</em>

6 0
3 years ago
How much carbon dioxide will be formed if 12.5 grams of oxygen reacts with 7.2 grams of propane (C3H8 )? Balanced equation: C3H8
solong [7]

Answer:

10.3125 grams of carbon dioxide will be formed.

Explanation:

The balanced reaction is:

C₃H₈ + 5 O₂→ 3 CO₂ + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles participate in the reaction:

  • C₃H₈: 1 mole
  • O₂: 5 moles
  • CO₂: 3 moles
  • H₂O: 4 moles

Being the molar mass of the compounds:

  • C₃H₈: 44 g/mole
  • O₂: 32 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole

then by stoichiometry of the reaction, the following amounts of mass participate in the reaction:

  • C₃H₈: 1 mole* 44 g/mole = 44 g
  • O₂: 5 moles* 32 g/mole= 160 g
  • CO₂: 3 moles* 44 g/mole= 132 g
  • H₂O: 4 moles* 18 g/mole= 72 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: If by reaction stoichiometry 44 grams of propane react with 160 grams of oxygen, 7.2 grams of propane react with how much mass of oxygen?

mass of oxygen=\frac{7.2 grams of propane*160 grams of oxygen}{44 grams of propane}

mass of oxygen= 26.18 grams

But 26.18 moles of O₂ are not available, 12.5 grams are available. Since you have less mass than you need to react with 7.2 grams of propane, oxygen O₂ will be the limiting reagent.

Then you can apply the following rule of three: if by stoichiometry of the reaction 160 grams of oxygen form 132 grams of carbon dioxide, 12.5 grams of oxygen will form how much mass of carbon dioxide?

mass of carbon dioxide=\frac{12.5 grams of oxygen*132 grams of carbon dioxide}{160 grams of oxygen}

mass of carbon dioxide= 10.3125 grams

<u><em>10.3125 grams of carbon dioxide will be formed.</em></u>

<u><em> </em></u>

<u><em></em></u>

7 0
3 years ago
Find the number of cations present in 7g of sodium prosphate Na=23 g/mol O=16g/mol P=31g/mol​
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, ions. Each magnesium ion is +2 and

each hydroxide ion is -1: two -1 ions are needed for one +2 ion, and the formula for magnesium

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hope that helped!!

3 0
3 years ago
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1 mole of NaCl has a mass of 58.44 grams.

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So, 732.6 kJ of heat will be required for = \frac{58.44g}{30.2kJ}\times 732.6 kJ = 1417.65 grams of NaCl.

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