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mestny [16]
3 years ago
11

The radioisotope radon-222 has a half-life of 3.8 days. How much of a 198.6 g sample of radon-222 would be left after approximat

ely 23 days?
Chemistry
1 answer:
bagirrra123 [75]3 years ago
7 0
The radon-222 sample has a half-life of 3.8 days, and we are asked how many times would the mass divide in half after 23 days. First we calculate the amount of times division occurs by taking the number of days and dividing that by the number of days for one half-life to occur: 23/3.8 = 6.05.

We have 198.6 grams of sample, and we are going to divide it in half 6 times to determine how much of it remains after 23 days:
198.6/2 = 99.3 grams
99.3/2 = 49.65 grams
49.65/2 = 24.83 grams
24.83/2 = 12.41 grams
12.41/2 = 6.21 grams
6.21/2 = 3.1 grams

Therefore, we are left with 3.1 grams of radon-222 after 23 days if one half-life equals to 3.8 days.
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In the experiment "Preparation and Standardization of a Sodium Hydroxide Solution", why was the burette rinsed with NaOH before
FromTheMoon [43]

Answer: The correct option is ALL OF THE ABOVE.

Explanation:

TITRATION is a type of volumetric analysis which is used for determining the concentration of solutions. In this process a specific volume of a solution is placed in a conical flask by means of a pipette and small quantities of a second solution is slowly added from a burette until the end point is reached. This is determined by a means of an indicator which shows a characteristic colour change.

During titration, the following precautions should be followed to avoid errors and maintain standardisation in the experiment.

--> Any air bubble in the burette and pipette must be removed during measurement

--> the burette tap should be tightened to avoid leakage.

--> Remove the funnel from the burette before taking any reading to avoid errors in reading the volume.

--> use the base solution such as Sodium Hydroxide Solution to rinse the burette after washing with soap and tap water:

• to remove any air bubble and fill it's tip

• to remove any residual liquid from the water and soap solution which may interfere with the results of the experiment.

• to check if the burette is in good condition.

Therefore all of the above options to the question are correct.

3 0
3 years ago
In the third period of the periodic table sodium is followed by magnesium aluminum silicon and phosphorus which of these element
wariber [46]
Answer:
            Phosphorous has the smallest atomic size.

Explanation:
                   As we know these elements belong to same period means there valence shell is the same. So moving from left to right along the period the shell number remains constant but the number of protons and electrons increases. So, due to increase in number of protons the nuclear charge increases hence attracts the valence electrons more effectively resulting in the decrease of atomic size.

Elements and their atomic radius are as follow,

<span><span>Magnesium          0.160 nm
</span><span>
Aluminium           0.130 nm
</span><span>
Silicon                  0.118 nm
</span><span>
Phosphorus         <span>0.110 nm</span></span></span>
6 0
4 years ago
What happens to elements when they combine together chemically ?
lara [203]

Answer:

When two distinct elements are chemically combined—i.e., chemical bonds form between their atoms—the result is called a chemical compound. Most elements on Earth bond with other elements to form chemical compounds, such as sodium (Na) and Chloride (Cl), which combine to form table salt (NaCl).

7 0
3 years ago
What was the name for the secret code used by many alchemists?
ahrayia [7]

Answer:

A. Zodiac

B. Palingenesis

C. Palabra mysteria

D. Decknamen

The correct answer is D. Decknamen.

Explanation:

5 0
3 years ago
Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6
liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 137ml+137ml=274ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 274 g

T_{final} = final temperature of water = 325.8 K

T_{initial} = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

q=274g\times 4.18J/g^oC\times (325.8-298)K

q=31839.896J=31.84KJ

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0
4 years ago
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