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lorasvet [3.4K]
3 years ago
8

Medical diagnosis and ______ are common uses of radioactive elements in health care.

Chemistry
1 answer:
KengaRu [80]3 years ago
6 0
<span>The half-life of Carbon 14 and radionuclides are used to estimate the absolute (versus relative) age of pre-history items </span>
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7. what is the density of an object having a mass of 8.0 g and a volume of 25 cm^3? a) 0.32g/cm^3 b) 2.0 g/cm^3 c) 3.1 g/cm^3 d)
natulia [17]
D = m / V

D = 8.0 g  / 25 cm³

D = 0.32 g/cm³

Answer A
6 0
3 years ago
Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec
Svetach [21]

The overall reaction is given by:

Br_{2}(g) + 2 NO(g) \rightarrow  2 NOBr(g)

The fast step reaction is given as:

NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})

The slow step reaction is given as:

NOBr_{2}(g) + NO(g) \rightarrow  2 NOBr(g) (slow step k_{2})

Now, the expression for the rate of reaction of fast reaction is:

r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]

The expression for the rate of reaction of slow reaction is:

r_{2}=k_{2}[NOBr_{2}] [NO]

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as [NOBr_{2}] takes place in this reaction.

The expression of rate of formation is:

\frac{d(NOBr)}{dt}=r_{2}

= k_{2}[NOBr_{2}][NO]    (1)

Now, consider that the fast step is always is in equilibrium. Therefore, r_{1}=0

k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]

[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]

Substitute the value of [NOBr_{2}] in equation (1), we get:

\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]

=k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]

= \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]

Thus, rate law of formation of NOBr in terms of reactants is given by \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}].









4 0
3 years ago
What is the formula for the compound made from mercury (II) and the nitrate ion.
Kazeer [188]

Answer:

Hg(NO3)2

Explanation:

Hope it helps! :)

5 0
3 years ago
Read 2 more answers
If a sample of CO gas at 1.977 atm has a volume of 517.4 mL and the pressure is changed to
Mekhanik [1.2K]

Answer:

The answer to your question is V2 = 333.9 ml

Explanation:

Data

Pressure 1 = P1 = 1.977 atm

Volume 1 = V1 = 517.4 ml

Pressure 2 = P2 = 3.063 atm

Volume 2 = V2 = ?

Process

To solve this problem use Boyle's law

              P1V1 = P2V2

-Solve for V2

              V2 = P1V1/P2

-Substitution

              V2 = (1.977 x 517.4) / 3.063

-Simplification

              V2 = 1022.9 / 3.063

-Result

              V2 = 333.9 ml

3 0
3 years ago
Please Help with this!!!!!
inessss [21]

Answer:

a) 4.69

Explanation:

v=mp

m=10.55

p=2.25

v=10.55/2.25=4.69

7 0
4 years ago
Read 2 more answers
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