The half-life of element X is 500 years. If there are initially 8 g of X, how much will remain after 1500 years

Answers

Answer 1

Answer:

1 g

Explanation:

From the formula;

N/No = (1/2)^t/t1/2

Where;

N= mass of radioactive element left after a time t = the unknown

No= mass of radioactive element originally present in the sample = 8g

t= time taken for N mass of the sample to remain = 1500

t1/2= half-life of the radioactive element = 500 years

Substituting values, we have;

N/8 = (1/2)^1500/500

N/8 = (1/2)^3

N/8 = 1/8

N= 1/8 ×8

N= 1 g

Therefore; mass of radioactive element left after 1500 years is 1 g


Related Questions

A sample of gas occupies a volume of 7.50 L at 0.988 atm and 301 K. At what temperature is the volume of the gas 4.00 L if the pressure is kept constant.

Answers

Answer:

160.53L

Explanation:

Since Pressure is kept constant we can use charles law

V1/T1 =V2/T2

What is the mass number of an atom with 24 protons and 30 neutrons?

Answers

Answer:

54

Explanation:

Mass number = protones + neutrons

Mass number = 24 + 30

Mass number = 54

What is the maximum number of electrons in the second principal energy level?
02 32 8 18​

Answers

Answer:

8 electrons

Explanation:

The second principal energy level has two sublevels: 2s and 2p

2s : 2 electrons

2p : 6 electrons (3 sublevels × 2 electrons each = 6 electrons)

It can hold a maximum of 8 electrons.

Hope this helps. :)

Five mol of calcium carbide are combined with 10 mol of water in a closed, rigid, high-pressure vessel of 1800 cm3 internal empty volume. Acetylene gas is produced by the reaction:

Answers

Answer:

CaC₂  +  2H₂O  →  C₂H₂  + Ca(OH)₂

Explanation:

In order to find out the reaction, we must know the reactants.

For this situation, we make acetylene gas from carbide calcium CaC₂ and H₂O (water); therefore the reactants are:

- CaC₂  and  H₂O

Acetylene is one of the products made  →  C₂H₂

So the reaction can be formed as this:  CaC₂  +  H₂O  →  C₂H₂

We missed the calcium, and this reaction also makes, Calcium Hydroxide, so the complete equation must be:

CaC₂  +  H₂O  →  C₂H₂  + Ca(OH)₂

This is unbalanced, because we have 1 O in left side and 2 in right side so we add 2 in water so now, we get the complete reaction:

1 mol of calcium carbide reacts to 2 mol of water in order to produce 1 mol of acetylene and 1 mol of calcium hydroxide.

How much MnO2(s) should be added to excess HCl(aq) to obtain 195 mL Cl2(g) at 25 °C and 715 Torr g

Answers

THIS IS THE COMPLETE QUESTION

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as described by the chemical equation.

How much MnO2(s) should be added to excess HCl(aq) to obtain 185 mL of Cl2(g) at 25 °C and 715 Torr?

Answer:

0.62901mol of MnO2(s) should be added

Explanation:

Given:

P = 715/760 = 0.94078atm

v=195ml=0.195l

n = ? moles have to find

R = 0.0821 L atm/K/mole

T = 25 + 273 = 298 K

Then we will make use of below formula

PV = nRT

Insert the values

0.94078*0.195=n 0.0821*298

24.466n=0.1740443

n=0.174/24.466

n=0.007235 nb of moles of cl2

as 1 mole of Cl2 were obtained from 1 mole of MnO2

so 0.007235 of chlorine must have come from

0.007235 moles of MnO2

1 mole of MnO2 = 86.94 g/mole

so 0.007235 moles of MnO2== 86.94* 0.007235

=0.62901

4Ga + 3S2 ⇒ 2Ga2S3


How many grams of Gallium Sulfide would form if 20.5 moles of Gallium burned?

Answers

Answer:

2415.9g (corrected to 1 d.p.)

Explanation:

(Take the atomic mass of Ga=69.7 and S=32.1)

Assuming Ga is the limiting reagent (because the question did not mention the amount of sulphur burnt),

From the balanced equation, the mole ratio of Ga:Ga2S3 = 4: 2 = 2: 1, which means, every 2 moles of Ga burnt, 1 mole of Ga2S3 is produced.

Using this ratio, let y be the no. of moles of Ga2S3 produced,

[tex]\frac{2}{1} =\frac{20.5}{y}[/tex]

y = 20.5 / 2

= 10.25 mol

Since mass = no. of moles x molar mass,

the mass of Ga2S3 produced = 10.25 x (69.7x2 + 32.1x3)

= 2415.9g (corrected to 1 d.p.)

g what would happen to the solubility of a gas in a solution if the pressure above the solution is increased

Answers

Answer: The solubility of gas increases in a solution if the pressure above the solution is increased

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

[tex]C=K_H\times p[/tex]

where,

C = solubility

[tex]K_H[/tex] = Henry's constant

p = partial pressure

As the solubility is directly proportional to the pressure, thus increasing the pressure increases the solubility.

Suppose you need to prepare 141.9 mL of a 0.223 M aqueous solution of NaCl. What mass of NaCl do you need to use to make the solution?

Answers

Answer:

1.811 g

Explanation:

The computation of the mass need to use to make the solution is shown below:

We know that molarity is

[tex]Molarity = \frac{Number\ of\ moles}{Volume\ in\ L}[/tex]

So,

[tex]Number\ of\ moles = Molarity\ \times Volume\ in\ L[/tex]

[tex]= 0.223\times 0.141[/tex]

= 0.031 moles

Now

[tex]Mass = moles \times Molecualr\ weight[/tex]

where,

The Molecular weight of NaCl is 58.44 g/mole

And, the moles are  0.031 moles

So, the mass of NaCL is

[tex]= 0.031 \times 58.44[/tex]

= 1.811 g

We simply applied the above formulas

what is the reduction half equation of Fe(s)+ 2 HC2H3O2(aq) → Fe(C2H3O2)2(aq) + H2(g)

Answers

Answer:

2 H⁺ + 2e = H₂    ( reduction )

Explanation:

Fe( s ) + 2 CH₃COOH = Fe ( OOCCH₃ ) ₂ + H₂

Fe( s ) = Fe⁺² + 2e     ( oxidation )

2 H⁺ + 2e = H₂    ( reduction )

Lead can be prepared from galena [lead(II) sulfide] by first heating with oxygen to form lead(II) oxide and sulfur dioxide. Heating the metal oxide with more galena forms the metal and more sulfur dioxide. Write a balanced equation for the overall reaction by adding the balanced equations for the two steps.

Answers

Answer:

2 PbS(s) + 1.5 O₂(g) + PbO(s) ⇒ 2 SO₂(g) + 3 Pb(s)

Explanation:

Lead can be prepared from galena [lead(II) sulfide] by first heating with oxygen to form lead(II) oxide and sulfur dioxide. The corresponding chemical equation is:

PbS(s) + 1.5 O₂(g) ⇒ PbO(s) + SO₂(g)

Heating the metal oxide with more galena forms the metal and more sulfur dioxide. The corresponding chemical equation is:

2 PbO(s) + PbS(s) ⇒ 3 Pb(s) + SO₂(g)

We can get the overall reaction by adding both steps and canceling what is repeated on both sides.

2 PbS(s) + 1.5 O₂(g) + 2 PbO(s) ⇒ PbO(s) + 2 SO₂(g) + 3 Pb(s)

2 PbS(s) + 1.5 O₂(g) + PbO(s) ⇒ 2 SO₂(g) + 3 Pb(s)

For each of the following ground state electron configurations, determine what is incorrect. N: 1s²2s¹2p³ A) The configuration is not in the ground state. B) The configuration is missing an electron in the 2s orbital. C) The configuration has too many electrons in the 2p orbital.

Answers

Answer:

B

Explanation:

It should be 2s² which means the answer is the configuration is missing an electron in the 2s orbital.

Answer:

B.

Explanation:

It should be 2s² which means the answer

is the configuration is missing an electron

in the 2s orbital.

2 Points
Which element has the fewest valence electrons?
A. Magnesium (Mg)
B. Neon (Ne)
C. Chlorine (CI)
O D. Silicon (Si)
SUSNE

Answers

Answer:

A. Magnesium (Mg)

Explanation:

As the elements go left in the periodic table, they have fewer valence electrons.

Magnesium has 2 valence electrons.

Neon has 8 valence electrons.

Chorine has 7 valence electrons.

Silicon has 4 valence electrons.

Magnesium has the fewest valence electrons.

Answer:

A

Explanation:

Magnesium

Calculate the mass percent by volume of 330.1 g of glucose (C₆H₁₂O₆, MM = 180.2 g/mol) in 325 mL of solution.

Answers

Answer: The mass percent by volume is 101.6%

Explanation:

The solution concentration expressed in percent by volume means that the amount of solute present in 100 parts volume of solution.

It is represented in formula as :

mass percent by volume =[tex]\frac{\text {mass of solute}\times 100}{\text {Volume of solution in ml}}\%[/tex]

Given : mass of glucose = 330.1 g

volume of solution = 325 ml

Thus mass percent by volume =[tex]\frac{330.1g\times 100}{325ml}=101.6\%[/tex]

Thus the mass percent by volume is 101.6%

The process by which the movement of internal bonds converts one type of what organic compound into another is

Answers

Answer: It's is called Reaarrangement

Explanation:

Rearrangement is a chemic reaction in which an atom or group of atoms bond migrate from one carbon atom to another. The movement involves two adjacent atoms and move over a large distance. It is important in functional group transformation.

Reaarrangement can be classified based on the atom or group of atoms it migrate.

Nucleophilic rearrangement which involves the migration of atoms with electron pairs.

Electrophilic migration involved the migration of atom or groups without electron pair.

Free radical involves the atom or group of atoms moving with a lone pair of electron.

Using the determined equivalence point from question 2 and the balanced reaction of acetic acid and sodium hydroxide, calculate the molarity of the acetic acid in your hot sauce packet.
Equivalance point is 3.0 mL NaOH, 0.6 g hot sauce, 0.1 M NaoH.

Answers

Answer:

Molarity of the packet is 0.5M

Explanation:

In the reaction of acetic acid with NaOH:

CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na⁺

1 mole of acetic acid reacts with 1 mole of NaOH.

When you are titrating the acid with NaOH, you reach equivalence point when moles of acid = moles of NaOH.

Moles of NaOH are:

3.0mL = 3.0x10⁻³L ₓ (0.1 mol / L) = 3.0x10⁻⁴ moles of NaOH = moles of CH₃COOH.

Now, you find the moles of acetic acid in the hot sauce packet. But molarity is the ratio between moles of the acid and liters of solution.

As you don't know the volume of your packet, you can assume its density as 1g/mL. Thus, volume of 0.6g of hot sauce is 0.6mL = 6x10⁻⁴L.

And molarity of the packet is:

3.0x10⁻⁴ moles acetic acid / 6x10⁻⁴L =

0.5M

A fictional cubed-shaped bacterium, Bacterius cubis, occupies a volume of 2.0 femtoliters. This particular type of bacteria is known to communicate with its own species by secreting a small molecule called bactoX ( MW=126.9 g/mol ). A. Each bacterium contains 7140 bactoX molecules that can be secreted. How many moles of bactoX are present in a 3.0 μL sample volume that contains 7.512×106 bacterial cells?

Answers

Answer:

There are  [tex]\mathbf{8.90172 \times 10^{-14}}[/tex] moles of bactoX present in a 3.0 μL sample volume that contains 7.512×106 bacterial cells

Explanation:

Given that:

The number of molecules present in one bacterial cell is [tex]7.140 \times 10^3[/tex] molecules

and the sample  contains [tex]7.512 \times 10^6[/tex] molecules.

Number of moles = number of molecules /Avogadro's number

where;

Avogadro's number = 6.023 × 10²³

Number of moles = [tex]\dfrac{7.140 \times 10^3}{6.023 \times 10^{23}}[/tex]

Number of moles = [tex]1.185 \times 10^{-20}[/tex] moles

So;  [tex]1.185 \times 10^{-20}[/tex] moles  is present in one bacteria cell

Similarly;  the sample  contains [tex]7.512 \times 10^6[/tex] molecules.

Therefore; the number of moles present in the bactoX is = [tex]1.185 \times 10^{-20} \times 7.512 \times 10^6[/tex]

= [tex]\mathbf{8.90172 \times 10^{-14}}[/tex] moles

PLEASE ANSWER AS SOON AS POSSIBLE REALLY WOULD APPRECIATE IT

Answers

Answer:

The answer is option D.

Hope this helps you

Identify the ions in the compounds represented in the following formulas. Click in the answer box to open the symbol palette.
Cation Formula Anion Formula
a) NaBr
b) AlCI3
c) Ba3(PO42
d) Mn(NO3)2

Answers

Answer and Explanation:

A ionic compound can be dissociated into its ions: cations (with positive charge) and anions (with negative charge). The net charge of the compound is zero, so the sum of the charges of the ions must be zero.

We can see the ions in the compounds from the dissociation equilibrium, as follows:

a) NaBr  ⇒ Na⁺ + Br⁻

Cation: Na⁺

Anion: Br⁻ (bromide)

b) AlCI₃ ⇒ Al³⁺ + 3 Cl⁻

Cation: Al³⁺

Anion: Cl⁻ (chloride)

c) Ba₃(PO₄)₂⇒ 3 Ba²⁺ + 2 PO₄³⁻

Cation: Ba²⁺

Anion: PO₄³⁻ (phosphate)

d) Mn(NO₃)₂ ⇒ Mn²⁺ + 2 NO₃⁻

Cation: Mn²⁺

Anion: NO₃⁻ (nitrate)

A sample of gas in a cylinder as in the example in Part A has an initial volume of 48.0 L , and you have determined that it contains 1.80 moles of gas. The next day you notice that some of the gas has leaked out. The pressure and temperature remain the same, but the volume has changed to 12.0 L . How many moles of gas (n2) remain in the cylinder

Answers

Answer:

0.45 moles

Explanation:

The computation of the number of moles left in the cylinder is shown below:

As we know that

[tex]\frac{n1}{V1} = \frac{n2}{V2}[/tex]

we can say that

[tex]n2 = n1 \times \frac{V2}{V1}[/tex]

where,

n1 = 1.80 moles of gas

V2 = 12.0 L

And, the V1 = 48.0 L

Now placing these values to the above formula

So, the moles of gas in n2 left is

[tex]= 1.80 \times \frac{12.0\ L}{48.0\ L}[/tex]

= 0.45 moles

We simply applied the above formulas so that the n2 moles of gas could arrive

Convert 120 degrees F to K.
[?]K

Answers

120 Fahrenheit -32 x 5/9 +273.15 = 322

Answer:

322

Explanation:

This is easy

This compound can be made by combining gaseous carbon monoxide with hydrogen gas (with this compound as the only product). What is the maximum mass of this compound that can be prepared if 8.0 kg of hydrogen gas react with 59.0 kg of carbon monoxide gas

Answers

Answer:

Maximum mass of compound produced = 64 Kg

Explanation:

Carbon monoxide react with hydrogen to produce methanol. Equation for the reaction is given as follows:

CO(g) + 2H₂(g) -----> CH₃OH(l)

From the equation of reaction, 1 mole of CO reacts with 2 moles of Hydrogen gas to produce 1 mole of methanol.

Molar mass of CO = 28 g

Molar mass of H₂ = 2 g

Molar mass of methanol = 32 g

Therefore, 28 g of CO reacts with  4 g of H₂ to produce 32 g of methanol

8.0 Kg of hydrogen will react with 8 * (28/4) Kg of CO = 56 Kg of CO.

therefore, hydrogen is the limiting reactant in the reaction under consideration.

8.0 Kg of Hydrogen will react with 56 Kg of CO to produce 8 * (32/4) Kg of methanol =  64 Kg of methanol

Therefore, maximum mass of compound produced = 64 Kg

Which of the following is an inorganic compound?
Na4C
C2H6
C12H22O11
CH3COOH

Answers

Answer:

The answer is option A.

Na4C

Hope this helps you

what is the best course of action if solid material remains in the flask after the heating step of recrystallization

Answers

Answer:

filter the hot mixture.

Explanation:

Solid is stayed undissolved since the arrangement is gotten super saturated. On the off chance that solid molecule is available recrysallization won't happen in this way we need expel the solid molecule by filtarion in hot condition itself . Subsequently, arrangement become totally homogenous and recrysallization item will shaped by moderate cooling

You need to purify 2.0 grams of an impure sample of Acetanilide. The sample is contaminated with aniline. After the purification is complete you isolate 0.8 grams of acetanilide and record a melting point range of 108-110 °C. Complete the following calculations and show your work.

a. Calculate the minimum amount of distilled water you would use to complete the recrystallization.
b. How much acetanilide will still be dissolved in solution even after the sample is cooled to 0 °C?
c. Calculate the % recovery and the % error for the melting point.
d. Why is the percent recovery less than 100%? Describe multiple sources for loss of sample.

Answers

Answer:

Following are the answer to this question:

Explanation:

In the given question an attachment file is missing, that is attached. please find the attached file, and the following are the description of the given points:

a. At 100 degrees in 100 mL 5 g is dissolved.  

For, it required:

[tex]\to 2g = 100 \times \frac{2}{5}[/tex]

         [tex]= 40 \ \ ml \ of \ water.[/tex]  

b. At 0 degrees 100 mL dissolve in 0.3 g.  

So, the dissolve:

[tex]\to 40 \ ml= 0.3\times \frac{40}{100}[/tex]

               [tex]= 0.12g.[/tex]

After refrigeration 0.12 g will still be dissolved.  

c. After dissolving and freezing, precipitation can occur which would still be impure if the cooling is instantaneous. The added solvent was also too hard to recrystallize. The solvent was placed below its place of reservation.  

d. Recovery percentage:

[tex]\to \frac{0.8}{2}\times100[/tex]

[tex]\to 40 \ \%[/tex]

The melting point of acetanilide:

[tex]\to 114.3^{\circ}.[/tex]

Found=109(medium)  

Melting point error percentages:

[tex]= \frac{114.3-109}{114.3}\\\\=4.63 \ \%[/tex]

How much heat is absorbed when 52.39 H2O(l) at 100°C and 101.3 kPa is converted to steam at 100°C? (The molar heat of vaporization of water is
40.7 k/mol.)
2.09 x 1020
O 2.31% 10110
O 1.18 x 102 103
O 1.11% 1021)

Answers

2.09 x 1020 that’s the answer

what are the oxidizing and reducing agents of 2H2S+SO2-->2H2O+3S​

Answers

Answer:

Explanation:

H2S is reducing agent because it is reducing SO2 being oxidized itself.SO2 is oxidizing agent because it is oxidizing H2S being reduced itself.

Calculate the volume of 0.500 M C2H3O2H and 0.500 M C2H3O2Na required to prepare 0.100 L of pH 5.00 buffer with a buffer strength of 0.100 M. The pKa of C2H3O2H is 4.75.

Answers

Answer:

You require 12.8mL of the 0.500M C₂H₃O₂Na and 7.2mL of the 0.500M C₂H₃O₂H

Explanation:

It is possible to obtain pH of a weak acid using H-H equation:

pH = pKa + log₁₀ [A⁻] / [HA]

For the buffer of acetic acid/acetate, the equation is:

pH = pKa + log₁₀ [C₂H₃O₂Na] / [C₂H₃O₂H]

Replacing:

5.00 = 4.75 + log₁₀ [C₂H₃O₂Na] / [C₂H₃O₂H]

1.7783 =  [C₂H₃O₂Na] / [C₂H₃O₂H] (1)

Buffer strength is the concentration of the buffer, that means:

0.1M = [C₂H₃O₂Na] + [C₂H₃O₂H] (2)

Replacing (2) in (1):

1.7783 =  0.1M - [C₂H₃O₂H] / [C₂H₃O₂H]

1.7783 [C₂H₃O₂H] = 0.1M -  [C₂H₃O₂H]

2.7783 [C₂H₃O₂H] = 0.1M

[C₂H₃O₂H] = 0.036M

Also:

[C₂H₃O₂Na] = 0.1M - 0.036M

[C₂H₃O₂Na] = 0.064M

The moles of both compounds you require is:

[C₂H₃O₂Na] = 0.1L × (0.064mol / L) = 0.0064moles

[C₂H₃O₂H] = 0.1L × (0.036mol / L) = 0.0036moles

Your stock solutions are 0.500M, thus, volume of both solutions you require is:

[C₂H₃O₂Na] = 0.0064moles × (1L / 0.500M) = 0.0128L = 12.8mL

[C₂H₃O₂H] = 0.0036moles × (1L / 0.500M) = 0.0072mL = 7.2mL

You require 12.8mL of the 0.500M C₂H₃O₂Na and 7.2mL of the 0.500M C₂H₃O₂H

PLEASE HELP, 5 STARS, 35 POINTS + PRIZE The reaction C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol. At 600.0 K, the rate constant, , is 6.1×10−8 s−1. What is the value of the rate constant at 780.0 K?

Answers

Answer:

346 hope this is helpfull

Calculate the mass in grams for 0.251 moles of Na2CO3

Answers

Answer:

Explanation:

the molar mass for Na2CO3 is 2*23+12+3*16=106 g/mole

106*0.251=26.606 grames

NH4NO2(s)→N2(g)+H2O(l) ---------------- Express your answer as a chemical equation. Identify all of the phases in your answer.

Answers

Answer:

The balanced equation is :

NH4NO2(solid) = N2(gas) + 2 H2O(liquid)

Explanation:

A balanced chemical equation is an equation that has an equal number of atoms and charges on both sides of the equation. The given equation in question is imbalanced as the number of atoms not equal.

In this reaction, solid ammonium nitrite breaks into nitrogen gas and water, reaction known as decomposition.

The correct and balanced equation as follows :

NH4NO2(s) = N2(g) + 2 H2O(l)

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