Ask question

Ask question

All categories

konstantin123 [22]

3 years ago

8

The process by which a radioactive isotope loses protons or other materials from its nucleus is called

1 answer:

trasher [3.6K]3 years ago

6 0

The process is called decay.

You might be interested in

Elements within the same group of the periodic table behave similarly because they have the same number of what

Darina [25.2K]

Same number of ions
Yeeeeeeeeee

4 0

3 years ago

Read 2 more answers

32.33 mL of 1.031 M potassium hydroxide were required to reach the endpoint of a titration of 50.00 mL of nitric acid. What was

tester [92]

The molar concentration of the nitric acid solution was 0.6666 mol/L.

<em>Balanced equation</em>: KOH + HNO_3 → KNO_3 + H_2O

<em>Moles of KOH</em>: 32.33 mL KOH × (1.031 mmol KOH /1 mL KOH)

= 33.33 mmol KOH

<em>Moles of HNO_3</em>: 33.33 mmol KOH× (1 mmol HNO_3/1 mmol KOH)

= 33.33 mmol HNO_3

<em>Concentration of KOH</em>: <em>c </em>= "moles"/"litres" = 33.33 mmol/50.00 mL

= 0.6666 mol/L

4 0

3 years ago

8. A sample of potassium chlorate (KCIO,) was heated in a test tube and decomposed 2KC?(s) 302 (g) + 2KCIO, (s) The oxygen was c

dangina [55]

Answer:

Partial pressure of $O_{2}$ in the gas was 733 torr and mass of $KClO_{3}$ in the sample was 2.12 g.

Explanation:

a) Total pressure of gas = (partial pressure of water vapour)+(partial pressure of $O_{2}$ )

Here partial pressure of water vapour is 21 torr and total pressure of gas is 754 torr.

So, partial pressure of $O_{2}$ = (total pressure of gas)-(partial pressure of water vapour) = (754 torr) - (21 torr) = 733 torr

b) Lets assume that $O_{2}$ behaves ideally. Hence-

PV=nRT

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin

here P = 733 torr = $(733\times 0.001316)atm$ = 0.9646 atm

V = 0.65 L, R = 0.082 L.atm/(mol.K), T=(273+22)K = 295 K

So, $n=\frac{PV}{RT}$

= $\frac{(0.9646 atm)\times (0.65 L)}{(0.082 L.atm/(mol.K))\times (295 K)}$

= 0.0259 moles

As 3 moles of $O_{2}$ are produced from 2 moles of $KClO_{3}$ therefore 0.0259 moles of $O_{2}$ are produced from $(\frac{2\times 0.0259}{3})$ moles or 0.0173 moles of $KClO_{3}$ .

Molar mass of $KClO_{3}$ = 122.55 g

So mass of $KClO_{3}$ in sample = $(0.0173\times 122.55)g$

= 2.12 g

7 0

3 years ago

Ifa dozen nails weighs 1 pound, how many nails are in 7.6 pounds?

NikAS [45]

Then there should be mostly 91 nails

4 0

3 years ago

Working on-board a research vessel somewhere at sea, you have (carefully) isolated 12.5 micrograms (12.5 ×10–6 g) of what you ho

german

Answer:

The value is $Z = 311.33 \ g/mol$

Explanation:

From the question we are told that

The mass of saxitoxin is $m = 12.5 mg = 12.5 * 10^{-6} g$

The volume of water is $V = 3.10 mL = 3.10 *10^{-3} L$

The osmotic pressure is $P = 0.236 = \frac{0.236}{760} = 3.105 * 10^{-4} atm$

The temperature is $T = 19^oC = 19 + 273 = 292 \ K$

Generally the osmotic pressure is mathematically represented as

$P = C * T * R$

Here R is the gas constant with value

$R = 0.0821 ( L .atm /mol. K)$

and C is the concentration of saxitoxin

So

$3.105 * 10^{-4} = C * 0.0821 * 292$

$C = 1.295 *10^{-5} mol/L$

Generally the number of moles of saxitoxin is mathematically represented as

$n = C * V$

=> $n = 1.295 *10^{-5} *3.10 *10^{-3}$

=> $n = 4.015 *10^{-8} \ mol$

Generally the molar mass of saxitoxin is mathematically represented as

$Z = \frac{m}{n}$

=> $Z = \frac{12.5 * 10^{-6}}{ 4.015 *10^{-8}}$

=> $Z = 311.33 \ g/mol$

5 0

3 years ago