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3 years ago

10

What is a radioactive isotope

2 answers:

Alla [95]3 years ago

8 0

Answer:

<h2> radionuclide is an atom that has excess nuclear energy, making it unstable. This excess energy can be used in one of three ways: emitted from the nucleus as gamma radiation; transferred to one of its electrons to release it as a conversion electron; or used to create and emit a new particle from the nucleus</h2>

Lelu [443]3 years ago

4 0

Answer:

Is a atom that has to much nuclear energy. Which makes it unstable

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A motorcycle moving initially at a velocity of 20 m/s accelerates

valina [46]

Answer:

40m/s

Explanation:

final velocity is = initial velocity + Acceleration x time

4 0

3 years ago

He pka values of a compound with two ionizable groups are pk1 = 4.10 and pk2 between 7 and 10. a biochemist has 10 ml of a 1.0 m

lyudmila [28]

For the reaction of weak acid:

$HA\rightleftharpoons H^{+}+A^{-}$

The Henderson-hasselbalch equation is:

$pH = pk_a+ log \frac{[A^{-}]}{[HA]}$

The concentration ratio at $pk_1$ is:

$pH = pk_1+ log \frac{[A^{-}]}{[HA]}$

Substituting the values:

$3.2 = 4.1 + log \frac{[A^{-}]}{[HA]}$

$log \frac{[A^{-}]}{[HA]} = - 0.9$

$\frac{[A^{-}]}{[HA]} = 0.126$

Now, for calculating the pk_2 value:

$pH = pk_2+ log \frac{[A^{-}]}{[HA]}$

$8 = pk_2+ log(0.126)$

$pk_2 = 8.9$

Thus, $pk_2 = 8.9$ .

5 0

3 years ago

WILL GIVE BRAINLIEST!!!!!!!!!!!<br><br><br>What are 4 examples of physical properties?<br>

marissa [1.9K]

List of 11 properties.

Hope it helps

5 0

3 years ago

Consider five hypothetical main- group elements, E, G,J,L, and M, that have the outer electron configuration shown below

rewona [7]

The answer is A is iron

6 0

3 years ago

A 6.35 l sample of carbon monoxide is collected at 55.0◦c and 0.892 atm. What volume will the gas occupy at 1.05 atm and 59.0◦c?

Sonja [21]

Answer : The final volume of gas will be, 5.46 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

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

where,

$P_1$ = initial pressure of gas = 0.892 atm

$P_2$ = final pressure of gas = 1.05 atm

$V_1$ = initial volume of gas = 6.35 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $55.0^oC=273+55.0=328K$

$T_2$ = final temperature of gas = $59.0^oC=273+59.0=332K$

Now put all the given values in the above equation, we get:

$\frac{0.892atm\times 6.35L}{328K}=\frac{1.05atm\times V_2}{332K}$

$V_2=5.46L$

Thus, the final volume of gas will be, 5.46 L

7 0

3 years ago