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labwork [276]
2 years ago
7

Two containers hold the same radioactive isotope. Container A contains 1000 atoms, and container B contains 500 atoms. which con

tainer has greater radioactive decay?The rate of decay of atoms in container A is greater than the rate of decay of atoms in container B.
Chemistry
1 answer:
enyata [817]2 years ago
8 0

Answer:

The rate of decay of atoms in container A is greater than the rate of decay of atoms in container B.

Explanation:

From the question,

Container A contains 1000 atoms

Container B contains 500 atoms

<u>The rate of decay of atoms in container A is greater than the rate of decay of atoms in container B.</u>

The reason for such is due to the difference in the concentration of the isotopes. Container A which contains higher number of atoms will have the more changes of the release of the neutron as the changes of the hitting and splitting increases as the density of the atoms increases.

<u>Thus, the atoms in the container A will therefore decay faster than the atoms in the container B. </u>

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The terms motif (fold) and domain describe levels of protein organization more complicated than primary or secondary structure.
posledela

Answer:

Each specific property of motif and domain is explained.

Explanation:

Domain;

  • May retain a 3D structure when separated from rest of the protein.          
  • Unit of tertiary structure because alpha helix and beta sheets are units of secondary structure.
  • Stable globular units like pyruvate kinase
  • May be distinct functional units in a protein

Motif;

  • Repetetive supersecondary structure because they contain cluster of secondary structure.
  • Beta Alpha Beta unit is an example of motif
  • Clusters of secondary structure

Both Motif and Domain;

  • Stabilized by hydrophobic interactions like hydrogen bonding stabilize the both.
  • Depends on primary structure like the arrangement of amino acid in polypeptide chain determine the secondary and tertiary structure of proteins.

8 0
2 years ago
Determine the empirical formula for compounds that have the following analyses: a. 66.0% barium and 34.0% chlorine b. 80.38% bis
almond37 [142]

Answer: a) BaCl_2

b)  BiO_3H_3

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

a) Mass of Ba= 66.06 g

Mass of Cl = 34.0 g

Step 1 : convert given masses into moles.

Moles of Ba =\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{66.06g}{137g/mole}=0.48moles

Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Ba = \frac{0.48}{0.48}=1

For O =\frac{0.96}{0.48}=2

The ratio of Ba: Cl= 1:2

Hence the empirical formula is BaCl_2

b) Mass of Bi= 80.38 g

Mass of O= 18.46 g

Mass of H = 1.16 g

Step 1 : convert given masses into moles.

Moles of Bi =\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{80.38g}{209g/mole}=0.38moles

Moles of O= \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{18.46g}{16g/mole}=1.15moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.16g}{1g/mole}=1.16moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Bi= \frac{0.38}{0.38}=1

For O =\frac{1.15}{0.38}=3

For H=\frac{1.16}{0.38}=3

The ratio of Bi: O: H= 1:3: 3

Hence the empirical formula is BiO_3H_3

6 0
2 years ago
Give two areas where the compressible nature of gas is applied​
galina1969 [7]

Answer:

1. Gases can be easily liquefied into very small volumes and stored in liquid form Eg in LPGA cylinders and used in homes.

2. Balloons can be easily filled with air.

6 0
1 year ago
How many pints of a 30% sugar solution must be added to a 5 pint of a 5% sugar solution to obtain a 20% sugar solution?
ser-zykov [4K]

You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.

You can use a modified dilution formula to calculate the volume of 30 % sugar.

<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3

Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt

(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar

30<em>x</em> + 25 = 20x + 100

10<em>x</em> = 75

<em>x</em> = 75/10 = 7.5

5 0
2 years ago
When balanced, which equation would have the coefficient 3 in front of any of the reactants? Zn + HCl ZnCl2 + H2 H2SO4 + B(OH)3
Gwar [14]

The correct answer is B. H2SO4 + B(OH)3  B2(SO4)3 + H2O

Hope this helps!

6 0
2 years ago
Read 2 more answers
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