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alukav5142 [94]
1 year ago
9

The isotope \left.\begin{array}{r}212 \\ 83\end{array}\right? Bi has a half-life of 1.01 yr. What mass (in mg) of a 2.00-mg samp

le will remain after 3.75X10³ h?
Chemistry
1 answer:
anyanavicka [17]1 year ago
5 0

Half-life is the length of time it takes for half of the radioactive atoms of a specific radionuclide to decay. A good rule of thumb is that, after seven half-lives, you will have less than one percent of the original amount of radiation.

<h3>What do you mean by half-life?</h3>

half-life, in radioactivity, the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay (change spontaneously into other nuclear species by emitting particles and energy), or, equivalently, the time interval required for the number of disintegrations per second of a radioactive.

<h3>What affects the half-life of an isotope?</h3>

Since the chemical bonding between atoms involves the deformation of atomic electron wavefunctions, the radioactive half-life of an atom can depend on how it is bonded to other atoms. Simply by changing the neighboring atoms that are bonded to a radioactive isotope, we can change its half-life.

Learn more about half life of an isotope here:

<h3>brainly.com/question/13979590</h3><h3 /><h3>#SPJ4</h3>
You might be interested in
A sample of Ne gas has a pressure of 654 mmHg with an unknown volume. The gas has a pressure of 345 mmHg when the volume is 495m
Lemur [1.5K]

Answer:

The initial volume of Ne gas is 261mL

Explanation:

This question can be answered using Ideal Gas Equation;

However, the following are the given parameters

Initial Pressure = 654mmHg

Finial Pressure = 345mmHg

Final Volume = 495mL

Required

Initial Volume?

The question says that Temperature is constant;

This implies that, we'll make use of Boyle's law ideal gas equation which states;

P_1V_1 = P_2V_2

Where P_1 represent the initial pressure

P_2 represent the final pressure

T_1 represent the initial temperature

T_2 represent the final temperature

P_1 = 654mmHg\\P_2 = 345mmHg\\V_2 = 495mL

Substitute these values in the formula above;

654 * V_1 = 345 * 495

654V_1 = 170775

Divide both sides by 654

\frac{654V_1}{654} = \frac{170775}{654}

V_1 = \frac{170775}{654}

V_1 = 261.123853211

V_1 = 261mL (Approximated)

<em>The initial volume of Ne gas is 261mL</em>

7 0
3 years ago
Given 450.98 g of Cu(NO3)2, how many moles of Ag can be made? Provide your final answer rounded to two decimal places.
vivado [14]

Answer:

4.82 moles of Ag.

Explanation:

We'll begin by calculating the number of mole in 450.98 g of Cu(NO₃)₂. This can be obtained as follow:

Molar mass of Cu(NO₃)₂ = 63.5 + 2[14 + (16×3)]

= 63.5 + 2[14 + 48]

= 63.5 + 2[62]

= 63.5 + 124

= 187.5 g/mol

Mass of Cu(NO₃)₂ = 450.98 g

Mole of Cu(NO₃)₂ =?

Mole = mass /Molar mass

Mole of Cu(NO₃)₂ = 450.98 / 187.5

Mole of Cu(NO₃)₂ = 2.41 moles

Next, we shall determine the number of mole of Cu needed to produce 450.98 g (i.e 2.41 moles) of Cu(NO₃)₂. This can be obtained as follow:

Cu + 2AgNO₃ —> Cu(NO₃)₂ + 2Ag

From the balanced equation above,

1 mole of Cu reacted to produce 1 mole of Cu(NO₃)₂.

Therefore, 2.41 moles of Cu will also react to produce 2.41 moles of Cu(NO₃)₂.

Thus, 2.41 moles of Cu is needed for the reaction.

Finally, we shall determine the number of mole of Ag produced from the reaction. This can be obtained as follow:

From the balanced equation above,

1 mole of Cu reacted to produce 2 moles of Ag.

Therefore, 2.41 moles of Cu will react to produce = 2× 2.41 = 4.82 moles of Ag.

Thus, 4.82 moles of Ag were obtained from the reaction.

6 0
3 years ago
14. What is the pH of a 0.24 M solution of sodium propionate, NaC3H502, at 25°C? (For
Murrr4er [49]

Answer:

9.1

Explanation:

Step 1: Calculate the basic dissociation constant of propionate ion (Kb)

Sodium propionate is a strong electrolyte that dissociates according to the following equation.

NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻

Propionate is the conjugate base of propionic acid according to the following equation.

C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻

We can calculate Kb for propionate using the following expression.

Ka × Kb = Kw

Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰

Step 2: Calculate the concentration of OH⁻

The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.

[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M

Step 3: Calculate the concentration of H⁺

We will use the following expression.

Kw = [H⁺] × [OH⁻]

[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/1.4 × 10⁻⁵ = 7.1 × 10⁻¹⁰ M

Step 4: Calculate the pH of the solution

We will use the definition of pH.

pH = -log [H⁺] = -log 7.1 × 10⁻¹⁰ = 9.1

5 0
3 years ago
The molar mass of h2o is 18.01 g/mom . The molar mass of o2 is 32.00 g/moo .what mass of h2o on grams must react to produce 50.0
Julli [10]
Oxygen can be obtained from water using electrolysis process as follows:
2 H2O .............> 2H2 + O2

It is given that: 
molar mass of water = 10.01 grams and molar mass of O2 = 32 grams

From the balanced chemical equation, we can conclude that:
2 moles of water produce 1 mole of oxygen
2 x 18.01 = 36.02 grams of water produce 32 grams of oxygen.

To calculate how many grams of water must react to produce 50 grams of oxygen, we can use cross multiplication as follows:
mass of required water = 56.28125 grams
4 0
3 years ago
When 3.51 g of phosphorus was burned in chlorine, the product was a phosphorus chloride. Its vapor took 1.77 times as long to ef
kumpel [21]

Answer:

<em>molar mass of the phosphorus chloride = 138.06 g/mol</em>

<em></em>

Explanation:

<em>mass of phosphorus will be the same as mass of CO2, since it is stated that they are of equal amount.</em>

mass = 3.51 g

<em>lets assume that it took the CO2 1 sec to effuse, then the time taken by the phosphorus chloride will be 1.77 sec</em>

From this we can say that

rate of effusion of CO2 = 3.51/1 = 3.51 g/s

rate of effusion of the phosphorus chloride = 3.51/1.77 = 1.98 g/s

<em>From graham's equation of effusion</em>,

\frac{Rc}{Rp} = \sqrt{\frac{Mp\\}{Mc} }

Rc = rate of effusion of CO2 = 3.51 g/s

Rp = rate of effusion of phosphorus chloride = 1.98 g/s

Mc = molar mass of CO2 = 44.01 g/mol

Mp = molar mass of the phosphorus chloride = ?

Imputing values into the equation, we have

\frac{3.51}{1.98} = \sqrt{\frac{Mp\\}{44.01} }

1.77 = \frac{\sqrt{Mp} }{6.64}

11.75 = \sqrt{Mp}

Mp = 11.75^{2}

Mp = <em>molar mass of the phosphorus chloride = 138.06 g/mol</em>

5 0
3 years ago
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