What are the three types of nuclear radiation

why do you think the experimenter conducted this experiment in a lab rather than in the natural world, where this phenomenon was

taurus [48]

Answer:

The experimenter observed this experiment in a lab rather than natural world because it might be dangerous to the atmosphere if he does the experiment in the natural world and it was still an hypothesis so that's why he did it in the lab.

5 0

3 years ago

What volume of O2 collected at 22.0 and 728 mmHg would be produce by the decomposition of 8.15 g KClO3?

adell [148]

Answer:

There is 2.52 L of O2 collected

Explanation:

Step 1: Data given:

Temperature = 22.0 °C

Pressure = 728 mmHg = 728 /760 = 0.958 atm

Mass of KClO3 = 8.15 grams

Molar mass of KClO3 = 122.55 g/mol

Step 2: The balanced equation

2KClO3(s) → 2KCl(s) + 3O2(g)

Step 3: Calculate moles of KClO3

Moles KClO3 = mass KClO3 / molar mass KClO3

Moles KClO3= 8.15 grams / 122.55 g/mol

Moles KClO3 = 0.0665 moles

Step 4: Calculate moles of O2

For 2 moles of KClO3 we'll have 2 moles of KCl and 3 moles of O2 produced

For 0.0665 moles of KClO3 we have 3/2 * 0.0665 = 0.09975 moles

Step 5: Calculate vlume of O2

p*V = n*R*T

V = (n*R*T)/p

⇒ with n = the number of moles O2 = 0.09975 moles

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = 22.0 °C = 273 +22 = 295 Kelvin

⇒ with p = 0.958 atm

V = (0.09975 * 0.08206 * 295) / 0.958

V = 2.52 L

There is 2.52 L of O2 collected

7 0

3 years ago

I need help ASAP

vova2212 [387]

Answer:

1.4 g/cm3

Explanation:

Density = Mass/Volume

Mass = 21g

Volume = 15cm3

Density = 21/15 = 1.4

8 0

3 years ago

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$

Thus it takes $0.0241 \; \text{L}$ of sodium hydroxide to produce this buffer solution.

6 0

3 years ago

Chemistry 10th grade PLEASE really need these answers :(

oksian1 [2.3K]

Answer:

I hope this helped you.

Explanation:

A: 1

B: 3

C: 3

D: 3

E: 5

F: 5

7 0

3 years ago