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2 years ago

6

Do all forms of radioactive decay change the identity of the original element?

1 answer:

hoa [83]2 years ago

5 0

Explanation:

<h3>yes, Radioactive decay involves the emission of a particle and/or energy as one atom changes into another. In most instances, the atom changes its identity to become a new element.</h3>

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Describe how Gregor mendel's methods helped endure the accuracy of his results

aliina [53]

Not sure good luck on finding someone too help you

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3 years ago

What is the difference between Dalton’s and modern atomic theory

stira [4]

Answer:

Dalton says atoms of a given element are identical in mass and the modern one says atoms of a given element are identical in average mass. ... Modern theory says they atoms can be subdivided, created or destroyed by ordinay means.

hope this helps!

4 0

2 years ago

Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta

BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, $C_{Ag}$ = 78 wt%

concentration of Pd, $C_P_d$ = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, $A_A_g$ = 107.87 g/mol

atomic weight of iron, $A_P_d$ = 106.4 g/mol

Step 1: determine the average density of the alloy

$\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }$

$\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3$

Step 2: determine the average atomic weight of the alloy

$A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }$

$A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole$

Step 3: determine unit cell volume

$V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}$

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

$V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell$

Step 4: determine the unit cell edge length

Vc = a³

$a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm$ = 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0

3 years ago

Mercury has a density of 13.6 g/mL. How many milliliters of mercury have a mass of 350 g? *

Colt1911 [192]

solution:

density x volume=mass volume x volume

13.6gml x 3.55ml

thus,the object has a mass of 98.3g

7 0

2 years ago

Which type of power can be used to reduce the production of carbon dioxide?

klasskru [66]

A. solar. Hope it helped!

8 0

2 years ago

Read 2 more answers