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Katarina [22]
4 years ago
9

The time required for a radioactive mass to reduce by a factor of 2 is called:

Chemistry
2 answers:
umka2103 [35]4 years ago
7 0

Answer:

half life would be the correct choice

Explanation:

Degger [83]4 years ago
4 0

The time required for a radioactive mass to reduce by a factor of 2 is called half-life.

You might be interested in
What temperature kelvin would .97mol of gas be to occupy 26L at 751.2mmHg?
crimeas [40]

Answer: 322.56 Kelvin

Explanation:

Use the Ideal Gas Law

PV=nRT

R is the gas constant

T is the temperature in Kelvins

P is the pressure in atmospheres

V is the volume in liters

n is the number of moles of gas

First, the mm of mercury need to be converted to atmospheres using the conversion factor 1atm = 760 torr.

751.2mmHg (torr) =0.988atm

Now plug everything in

(0.988)(26)=(0.97)(0.0821)T\\T=322.56K

5 0
3 years ago
What is the polyatomic ion
yanalaym [24]
A polyatomic ion is a charged combination of individual atoms. The combination is regarded as a single unit when considering reactivity with other elements. Example: OH- which is a polyatomic ion made from one oxygen and hydrogen atom to form the negative hydroxide ion.
7 0
3 years ago
What are the formulas for the compounds formed by the C3+, 02- pairs of ions;
choli [55]
Answer : Cr203

Explanation:
5 0
3 years ago
A sample of argon gas occupies a certain volume at 11°C. At what temperature would the volume of the gas be three times as big?
stiv31 [10]
V₁/T₁ = V₂/T₂ = const

T₁=11+273=284K
V₂=3V₁

V₁/T₁=3V₁/T₂

T₂=3T₁

T₂=3*284=852K

t₂=852-273=579°C
4 0
3 years ago
Read 2 more answers
The percent composition of artificial snow is 38.3 % C, 3.22 %H, 24.4 % Na, and the remainder is oxygen. Determine the empirical
BabaBlast [244]

Answer:

C₃H₃NaO₂

Explanation:

The percentage composition of the artificial snow is given.

The smallest whole number ratio of the atoms in a compound gives the empirical formula.

To find the empirical formula of the artificial snow, we use the method below;

Percentage composition of oxygen  = 100 - (38.3+3.33+ 24.4) = 33.97

Elements                            C                     H              Na             O

Percentage

Composition                   38.3                   3.22         24.4          33.97

Number of

moles                           38.3/12                  3.22/1     24.4/23     33.97/16

                                       3.19                     3.22        1.06             2.12

Divide by

the smallest               3.19/1.06                3.22/1.06  1.06/1.06     2.12/1.06

                                        3                             3               1                2

The empirical formula of the compound is C₃H₃NaO₂

7 0
3 years ago
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