HELP NEEDED. 30 POINTS Which is the SI symbol for volume? A. m B. g C. L D. к

PLZ HELP ...The half-life of polonium-218 is 3.0 minutes. If you start with 40.0 g, how long will it be before only 4.5 g remain

Elanso [62]

Answer:- 9.4 minutes.

Solution:- Radioactive decay obeys first order reaction kinetics and the equation used to solve this type of problems is:

$lnN=-kt+lnN_0$

where, k is decay constant and t is the time. $N_0$ is the initial amount of the radioactive substance and N is the remaining amount.

Since the value of decay constant is not given, so we need to calculate it first from given half life by using the formula:

$k=\frac{0.693}{t_1_/_2}$

where $t_1_/_2$ stands for half life.

Given half life is 3.0 minutes.

So, $k=\frac{0.693}{3.0min}$

$k=0.231min^-^1$

Let's plug in the values in the first order reaction equation and solve it for t.

$ln4.5g=-0.231min^-^1(t)+ln40.0g$

It could also be written as:

$ln(\frac{4.5g}{40.0g})=-0.231min^-^1}$

$-2.18=-0.231min^-^1}$

$t=\frac{-2.18}{-0.231min^-^1}$

k = 9.4 min

So, the radioactive substance would take 9.4 minutes to decay from 40.0 grams to 4.5 grams.

4 0

3 years ago

If you needed to make 100 mL of a 0.2 M fruit drink solution from the 1.0 M fruit drink

telo118 [61]

Answer:

We take 20.0 mL of the 1.0 M fruit drink solution and then add 80.0 mL of water to make 100 mL of a 0.2 M fruit drink solution.

Explanation:

Using the rule that: the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution. (MV) before dilution = (MV) after dilution. M before dilution = 1.0 M, V before dilution = ??? mL. M after dilution = 0.2 M, V after dilution = 100 mL. ∴ V before dilution = (MV) after dilution / M before dilution = (0.2 M)(100 mL) / (1.0 M) = 20.0 mL. So, we take 20.0 mL of the 1.0 M fruit drink solution and then add 80.0 mL of water to make 100 mL of a 0.2 M fruit drink solution

8 0

3 years ago

How many liters of 0.615 M NaOH will be needed to raise the pH of 0.385 L of 5.13 M sulfurous acid (H2SO3) to a pH of 6.247?

givi [52]

Answer:

Volume of NaOH required = 3.61 L

Explanation:

H2SO3 is a diprotic acid i.e. it will have two dissociation constants given as follows:

$H2SO3\leftrightarrow H^{+}+HSO3^{-}$ --------(1)

where, Ka1 = 1.5 x 10–2 or pKa1 = 1.824

$HSO3^{-}\leftrightarrow H^{+}+SO3^{2-}$ --------(2)

where, Ka2 = 1.0 x 10–7 or pKa2 = 7.000

The required pH = 6.247 which is beyond the first equivalence point but within the second equivalence point.

Step 1:

Based on equation(1), at the first eq point:

moles of H2SO3 = moles of NaOH

$i.e. \ 5.13\ moles/L*0.385L = moles\ NaOH\\therefore, \ moles\ NaOH = 1.98\ moles\\V(NaOH)\ required = \frac{1.98\ moles}{0.615\ moles/L} =3.22L$

Step 2:

For the second equivalence point setup an ICE table:

$HSO3^{-}+OH^{-}\leftrightarrow H2O+SO3^{2-}$

Initial 1.98 ? 0

Change -x -x x

Equil 1.98-x ?-x x

Here, ?-x =0 i.e. amount of OH- = x

Based on the Henderson buffer equation:

$pH = pKa2 + log\frac{[SO3]^{2-} }{[HSO3]^{-} } \\6.247 = 7.00 + log\frac{x}{(1.98-x)} \\x=0.634 moles$

Volume of NaOH required is:

$\frac{0.634\ moles}{0.615 moles/L}=0.389L$

Step 3:

Total volume of NaOH required = 3.22+0.389 =3.61 L

3 0

3 years ago

10.

Amanda [17]

The best answer is:

a. Extrusive igneous rocks

In general, rocks that are formed by the cooling of magma are classified as igneous rocks. The key word however in this case is "lava," which is magma that reached the surface. This is what makes it an extrusive igneous rock, because it formed in the surface.

6 0

3 years ago

Read 2 more answers

A tank can hold up to 2500 liters of liquid. The tank initially contains 1000 liters of a brine solution with 50kg of salt. A br

Bogdan [553]

Answer:

67.93 kg.

Explanation:

See the attached pictures for detailed explanation.

8 0

3 years ago