Please help me! ASAP

You are performing an experiment that uses 114Ag. 113Ag is radioactive, decays by beta-- emission and has a half-life of 21 minu

maxonik [38]

Answer: The 234.74 grams of sample should be ordered.

Explanation:

Let the gram of 114 Ag to ordered be $N_o$

The amount required for the beginning of experiment = 0.0575 g

Time requires to ship the sample = 4.2hour = 252 min(1 hr = 60 min)

Half life of the sample = ${t_{\frac{1}{2}}$ = 21 min

$\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{21 min}=0.033 min^{-1}$

$\log[N]=\log[N_o]-\frac{\lambda t}{2.303}$

$\log[0.0575 g]=\log[N_o]-\frac{0.033 min^{-1}\times 252 min}{2.303}$

$N_o=234.74 grams$

The 234.74 grams of sample should be ordered.

4 0

3 years ago

What is the half-life of 20 g of a radioactive sample if 5 g remain after 8 minutes?

Anni [7]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 20 g

m (final mass after time T) = 5 g

x (number of periods elapsed) = ?

P (Half-life) = ? (in minutes)

T (Elapsed time for sample reduction) = 8 minutes

Let's find the number of periods elapsed (x), let us see:

$m = \dfrac{m_o}{2^x}$

$5 = \dfrac{20}{2^x}$

$2^x = \dfrac{20}{5}$

$2^x = 4$

$2^x = 2^2$

$\boxed{x = 2}$

Now, let's find the half-life (P) of the radioactive sample, let's see:

$T = x*P$

$8 = 2*P$

$2\:P = 8$

$P = \dfrac{8}{2}$

$\boxed{\boxed{P = 4\:minutes}}\Longleftarrow(Half-Life)\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

5 0

3 years ago

The mass fractions of a mixture of gases are 15 percent nitrogen, 5 percent helium, 60 percent methane, and 20 percent ethane wi

Setler79 [48]

Answer:

Explanation:

mass fraction N₂ : He : CH₄ : C₂H₆ : : 15 : 5 : 60 : 20

mole fraction N₂ : He : CH₄ : C₂H₆ : : 15/28 : 5/4 : 60/16 : 20/30

mole fraction N₂ : He : CH₄ : C₂H₆ : : .5357 : 1.25 : 3.75 : .67

Total mole fractions = .5357 + 1.25 + 3.75 + 0.67 = 6.2057

mole fraction of N₂ = .5357 / 6.2057 = .0877

mole fraction of He = 1.25 / 6.2057 = .20

mole fraction of CH₄ = 3.75 / 6.2057 = .6043

mole fraction of C₂H₆ = .67 / 6.2057 = .108

Partial pressure = total pressure x mole fraction

Partial pressure of N₂ = 1200 kPa x .0877 = 105.24 kPa

Partial pressure of He = 1200 kPa x .20 = 240 kPa

Partial pressure of CH₄ = 1200 kPa x .6043 = 725.16 kPa

Partial pressure of C₂H₆ = 1200 kPa x .108 = 129.6 kPa

6 0

3 years ago

The Chemistry Cafe was out of bread. The cook went next door to the bakery and bought a loaf of bread which has 33 slices. Then,

maria [59]

In the question, we are told that there are;

A loaf containing 33 slices
A loaf containing 33 slices A package of cheese containing 15 slices

We also know that he is making a sandwich that has 2 pieces of both cheese and bread.

Hence;

Total number of bread and cheese = 33 + 15.

Each loaf should have two pieces of each bread and the cheeses make a total of four pieces.

Therefore he can make = 33 + 15/4 = 12 sandwiches.

4 0

2 years ago

Read 2 more answers

What is specific heat capacity

LenaWriter [7]

Answer:

Explanation:

3 0

2 years ago

Read 2 more answers