All categories

3 years ago

4. Explain how radioactive isotopes are used to detect health problems.

1 answer:

6 0

Radioisotopes are widely used to diagnose disease and as effective treatment tools. For diagnosis, the isotope is administered and then located in the body using a scanner of some sort.

You might be interested in

What’s the difference between heterogeneous and homogeneous mixtures?

Ronch [10]

Answer:

Mixtures can be either homogeneous or heterogeneous. A mixture in which its constituents are distributed uniformly is called homogeneous mixture, such as salt in water. A mixture in which its constituents are not distributed uniformly is called heterogeneous mixture, such as sand in water.

5 0

3 years ago

Which reaction will occur? 2NaBr+I2 —> 2NaI + Br2

Veronika [31]

Answer:

This is a single replacement reaction because I replaces Br.

7 0

3 years ago

Read 2 more answers

Which of the substances shown below most likely exhibit hydrogen bonding with other identical

navik [9.2K]

Huhl. Iidudididuudufurururuururururu

6 0

2 years ago

Read 2 more answers

"To determine the amount of heroin in the mixture, you dissolve 1.00 g of the white powdery mixture in water in a 100.0-mL volum

UkoKoshka [18]

Explanation:

Formula to calculate osmotic pressure is as follows.

Osmotic pressure = concentration × gas constant × temperature( in K)

Temperature = $25^{o} C$

= (25 + 273) K

= 298.15 K

Osmotic pressure = 531 mm Hg or 0.698 atm (as 1 mm Hg = 0.00131)

Putting the given values into the above formula as follows.

0.698 = $C \times 0.082 \times 298.15 K
$

C = 0.0285

This also means that,

$\frac{\text{moles}}{\text{volume (in L)}}$ = 0.0285

So, moles = 0.0285 × volume (in L)

= 0.0285 × 0.100

= $2.85 \times 10^{-3
}$

Now, let us assume that mass of $C_{12}H_{23}O_{5}N$ = x grams

And, mass of $C_{12}H{22}O_{11}$ = (1.00 - x)

So, moles of $C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}$

= $\frac{x}{369}$

Now, moles of $C_{12}H_{22}O_{11} = \frac{(1.00 - x)}{342}$

= $\frac{x}{369} + \frac{(1.00 - x)}{342}$

= $2.85 \times 10^{-3}$

= x = 0.346

Therefore, we can conclude that amount of $C_{12}H_{23}O_{5}N$ present is 0.346 g and amount of $C_{12}H_{22}O_{11}$ present is (1 - 0.346) g = 0.654 g.

4 0

3 years ago

Is chemistry hard in grade 11<br>explain to me everything plsss thanks

damaskus [11]

Answer: yes it’s hard

Explanation:

8 0

3 years ago

Read 2 more answers