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vladimir2022 [97]
3 years ago
10

You have a radioactive isotope with a mass of 400 grams. The sample has a half life of 5 days. How much is left after 25 days ha

ve passed?
Chemistry
1 answer:
Alexeev081 [22]3 years ago
4 0

Answer:

Amount left after 25 days = 12.5 g

Explanation:

Given data:

Mass of sample = 400 g

Half life of sample = 5 days

Mass left after 25 days = ?

Solution:

First of all we will calculate the number of half lives passes in given time period.

Number of half lives = Time elapsed / Half life

Number of half lives = 25 days/ 5 days

Number of half lives = 5

At time zero = 400 g

At 1st half life = 400 g/2 = 200 g

At 2nd half life = 200 g/2 = 100 g

At 3rd half life = 100 g/2 = 50 g

At 4th half life = 50 g/2 = 25 g

At 5th half life = 25 g/2 = 12.5 g

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galina1969 [7]

Answer:

Robert Boyle

Explanation:

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6 0
2 years ago
What is the percent composition of nitrogen in N 2 O
mash [69]

Answer:

63.6%

Explanation:

The given compound is:

     N₂O;

The problem here is to find the percent composition of nitrogen in the compound.

First find the molar mass of the compound:

 Molar mass of N₂O = 2(14) + 16  = 44g/mol

So;

 Percentage composition of Nitrogen  = \frac{2 x 14}{44}  x 100  = 63.6%

5 0
3 years ago
How many grams of oxygen are produced when 7.65 moles of water is decomposed
maksim [4K]

Answer:

The answer to your question is 122.4 g of O₂

Explanation:

Data

mass of O₂ = ?

moles of H₂O = 7.65

Process

1.- Write the balanced chemical reaction

                   2H₂O  ⇒  2H₂  +  O₂

2.- Convert the moles of H₂O to grams

molar mass of H₂O = 2 + 16 = 18 g

                    18 g of H₂O ---------------- 1 mol

                      x                ----------------- 7.65 moles

                      x = (7.65 x 18) / 1

                      x = 137.7 g H₂O

3.- Calculate the grams of O₂

                 36 g of H₂O -------------------- 32 g of O₂

              137.7 g of H₂O -------------------  x

                        x = (32 x 137.7) / 36

                       x = 122.4 g of O₂

 

6 0
3 years ago
What is a vocab word for Friction?
Leona [35]

Answer: Friction is the resistance to motion of one object moving relative to another. It is not a fundamental force, like gravity or electromagnetism. Instead, scientists believe it is the result of the electromagnetic attraction between charged particles in two touching surfaces.

Explanation:

7 0
2 years ago
A student is preparing to perform a series of calorimetry experiments. She first wishes to determine the calorimeter constant (C
laila [671]

Answer:

  • First choice: 99 J/K

Explanation:

<u>1) First law of thermodynamic (energy balance)</u>

  • Heat released by the the hot water (345K ) = Heat absorbedby the cold water (298 K) + Heat absorbed by the calorimeter

<u>2) Energy change of each substance:</u>

  • General formula:

     

Heat released or absorbed = mass × Specific heat × change in temperature

  • density of water: you may take 0.997 g/ ml as an average density for the water.

  • mass of water: mass = density × volume = 50.0 ml × 0.997 g/ml = 49.9 g

  • Specif heat of water: 1 cal / g°C

  • Heat released by the hot water:

       Heat₁ = 49.9 g × 1 cal / g°C × (345 K - 317 K) = 49.9 g × 1 cal / g°C × (28K)

  • Heat absorbed by the cold water:

       Heat₂ = 49.9 g × 1 cal / g°C × (317 K - 298 K) = 49.9 g × 1 cal / g°C × (19K)

  • Heat absorbed by the calorimeter

       Heat₃ = Ccal × (317 K - 298 K) = Ccal × (19K)

<u>4) Balance</u>

  • Heat₁ = Heat₂ + Heat₃

49.9 g × 1 cal / g°C × (28 K) = 49.9 g × 1 cal / g°C × (19 K) + Ccal × (19 K)

  • Solve for Ccal

Ccal = [49.9 g × 1 cal / g°C × (28 K) - 49.9 g × 1 cal / g°C × (19 K) ] / 19K

Ccal = 23.6 cal/ K

  • Convert to cal / K to Joule / K

  • 1 cal = 4.18 Joule

       23.6 cal / K × 4.18 J / cal = 98.6 J/K

Which rounded to 2 signficant figures leads to 99 J/k, which is the first choice.

4 0
3 years ago
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