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3 years ago

A 50-gram sample has a half-life of 12 days. How much material will remain after 12 days?

2 answers:

8 0

A 50-gram sample with a half-life of 12 days will have a remaining mass of 25 grams after its 12-day half-life. Every cycle of a half-life, the sample will lose half of its mass, so if the half-life, itself, is 12 days and the time period passing is 12 days, one half-life has passed and the material will be halved.

Rom4ik [11]3 years ago

7 0

Answer:25 grams

Explanation:

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There is no connection between the study of chemistry and physics. A. true B. false

9966 [12]

B. False, have a connection between chemistry and physis

4 0

3 years ago

The vapor pressure of liquid chloroform, CHCl3, is 100. mm Hg at 283 K. A 0.380 g sample of liquid CHCl3 is placed in a closed,

Katyanochek1 [597]

Answer:

100 mm Hg

Explanation:

From the question we are told that

The vapor pressure of CHCl3, is $P = 100 \ mmHg = \frac{100}{760}= 0.13156 \ atm$

The temperature of CHCl3 is $T = 283 \ K$

The volume of the container is $V_c = 380mL = 380 *10^{-3}\ L$

The temperature of the container is $T_c = 283 \ K$

The mass of CHCl3 is m = 0.380 g

Generally the number of moles of CHCl3 present before evaporation started is mathematically represented as

$n = \frac{m }{M }$

Here M is the molar mass of CHCl3 with the value $M = 119.38 \ g/mol$

=> $n = \frac{ 0.380 }{119.38 }$

=> $n = 0.00318 \ mols$

Generally the number of moles of CHCl3 gas that evaporated is mathematically represented as

$n_g = \frac{PV}{RT}$

Here R is the gas constant with value $R = 0.08206 L \ atm /mol\cdot K$

$n_g = \frac{0.13156* 380 *10^{-3} }{0.08206 * 283}$

$n_g = 0.00215 \ mols$

Given that the number of moles of CHCl3 evaporated is less than the number of moles of CHCl3 initially present , then it mean s that not all the liquid evaporated

At equilibrium the temperature of CHCl3 will be equal to the pressure of air so the pressure at equilibrium is 100 mmHg

4 0

3 years ago

Use this equation for the following problems: 2NaN3 --> 2Na+3N2

olchik [2.2K]

Answer:

1) 65.0

2) 16.434 L = 16434 mL.

Explanation:

2NaN₃ → 2Na + 3N₂,

It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂.

23.6 L of N₂ contains → ??? g of N₂.

∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.

We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.

We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

using cross multiplication:

2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

??? mol of NaN₃ produce → 0.775 moles of N₂.

∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.

Finally, we can get the grams of NaN₃ needed:

mass = no. of moles x molar mass = (0.517 mol)(65.0 g/mol) = 33.6 g.

Q2: How many mL of N₂ result if 8.3 g Na are also produced?

We need to get the no. of moles of 8.3 g Na using the relation:

n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.

We can get the no. of moles of N₂ produced with 0.36 mol of Na:

using cross multiplication:

2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

0.36 moles of Na produced with → ??? moles of N₂.

∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.

We can get the mass of 0.54 mol of N₂:

mass = no. of moles x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.

Now, we can get the mL of 15.12 g of N₂:

using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

??? L of N₂ contains → 15.12 g of N₂.

∴ The volume of N₂ result = (1.0 L)(15.12 g)/(0.92 g) = 16.434 L = 16434 mL.

4 0

3 years ago

What is the molarity of a solution that contains 0.082 mol KI in a 2.03 L solution?

astraxan [27]

Answer:

molarity of the KI solution = 0.04 mol/L

Explanation:

Molarity (M) is the concentration of a solution expressed as the number of moles of solute per liter of solution.

The law that we can applied to calculate the M is:

M = n / V

n - number of moles

V- volume of the solution (liters)

Then insert in the equation the values from the question;

M = 0.082 mol / 2.03 L = 0.04 mol/L

4 0

3 years ago

Which statement describes how NO2- reacts in this equilibrium:

WITCHER [35]

The statement which describes how NO2- reacts in this equilibrium:
H2SO3(aq) + NO2-(aq) HSO3-(aq) + HNO2(aq
is the second option - B. as a Brønsted-Lowry base by accepting a proton.
This is because bases take proton H+ in order to become HNO2.

6 0

3 years ago

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