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professor190 [17]
3 years ago
9

223Ra decays by alpha emission with a half-life of 11.43 days. For a 1.0g sample of 223Ra, after one half-life, what mass of 223

Ra remains? What has happened to the remaining mass?
Chemistry
1 answer:
Ray Of Light [21]3 years ago
6 0

Explanation:

The reaction that shows the alpha decay of ^{223}_{88}Ra is:

^{223}_{88}Ra\rightarrow ^{219}_{86}Rn+^{4}_{2}He

Half life is the time at which the concentration of the reactant reduced to half. So,

Mass\ of\ ^{223}_{88}Ra\ remained=\frac {Initial\ concentration}{2}=\frac {1.0\ g}{2}=0.5\ g

The remaining mass is converted to ^{219}_{86}Rn.

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ki77a [65]

Answer:

number three is the answer moderate

4 0
3 years ago
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Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:
rusak2 [61]

Answer:

60.42% is the percent yield of the reaction.

Explanation:

Moles of methane gas at 734 Torr and a temperature of 25 °C.

Volume of methane gas = V = 26.0 L

Pressure of the methane gas = P = 734 Torr = 0.9542 atm

Temperature of the methane gas = T = 25 °C = 298.15 K

Moles of methane gas = n

PV=nRT

n=\frac{PV}{RT}=\frac{0.9542 atm\times 26.0L}{0.0821 atm L/mol K\times 298.15 K}=1.0135 mol

Moles of water vapors at 700 Torr and a temperature of 125 °C.

Volume of water vapor = V' = 23.0 L

Pressure of water vapor = P' = 700 Torr = 0.9100 atm

Temperature of  water vapor = T' = 125 °C = 398.15 K

Moles of water vapor gas = n'

P'V'=n'RT'

n'=\frac{PV}{RT}=\frac{0.9100 atm\times 23.0L}{0.0821 atm L/mol K\times 398.15 K}=0.6402 mol

CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)

According to reaction , 1 mol of methane reacts with 1 mol of water vapor. As we can see that moles of water vapors are in lessor amount which means it is a limiting reagent and formation of hydrogen gas will depend upon moles of water vapors.

According to reaction 1 mol of water vapor gives 3 moles of hydrogen gas.

Then 0.6402 moles of water vapor will give:

\frac{3}{1}\times 0.6402 mol=1.9208 mol of hydrogen gas

Moles of hydrogen gas obtained theoretically = 1.9208 mol

The reaction produces 26.0 L of hydrogen gas measured at STP.

At STP, 1 mole of gas occupies 22.4 L of volume.

Then 26 L of volume of gas will be occupied by:

\frac{1}{22.4 L}\times 26 L= 1.1607 mol

Moles of hydrogen gas obtained experimentally = 1.1607 mol

Percentage yield of hydrogen gas of the reaction:

\frac{Experimental}{Theoretical}\times 100

\%=\frac{ 1.1607 mol}{1.9208 mol}\times 100=60.42\%

60.42% is the percent yield of the reaction.

8 0
3 years ago
The following equilibrium is formed when copper and bromide ions are placed in a solution:
JulsSmile [24]

Answer:

A)

1. Reaction will shift rightwards towards the products.

2. It will turn green.

3. The solution will be cooler..

B) It will turn green.

Explanation:

Hello,

In this case, for the stated equilibrium:

heat + Cu(H_2O)_6 ^{+2} (blue) + 4Br^- \rightleftharpoons 6H_2O + CuBr_4^{-2} (green)

In such a way, by thinking out the Le Chatelier's principle, we can answer to each question:

A)

1. If potassium bromide, which adds bromide ions, is added more reactant is being added to the solution, therefore, the reaction will shift rightwards towards the products.

2. The formation of the green complex is favored, therefore, it will turn green.

3. The solution will be cooler as heat is converted into "cold" in order to reestablish equilibrium.

B) In this case, as the heat is a reactant, if more heat is added, more products will be formed, which implies that it will turn green.

Regards.

3 0
3 years ago
An elastic cord can be stretched to its elastic limit by a load of 2N.If a 35cm length of the cord is extended 0.6cm by a force
Levart [38]

Explanation:

from \: hookes \: law \\ F = k.e \\ but \: e = 0.6 \: cm \\ 2 = k \times 0.6 \\ k = 3.33 \\ when \:F \: is \: 2.5 \\ 2.5 = 3.33 \times e {}^{.}   \\  {e}^{. }  = 0.75 \: cm \\ new \: length = 35 -  {e}^{.}  \\  = 35 - 0.75 \\  = 34.25 \: cm

3 0
3 years ago
A 0.708 g sample of a metal, M, reacts completely with sulfuric acid according to the reaction M ( s ) + H 2 SO 4 ( aq ) ⟶ MSO 4
ollegr [7]

Answer:

The metal has a molar mass of 65.37 g/mol

Explanation:

Step 1: Data given

Mass of the metal = 0.708 grams

Volume of hydrogen = 275 mL = 0.275 L

Atmospheric pressure = 1.0079 bar = 0.9947 atm

Temperature = 25°C

Vapor pressure of water at 25 °C = 0.03167 bar = 0.03126 atm

Step 2: The balanced equation

M(s) + H2SO4(aq) ⟶ MSO4 (aq) + H2(g)

Step 3: Calculate pH2

Atmospheric pressure = vapor pressure of water + pressure of H2

0.9947 atm = 0.03126 atm + pressure of H2

Pressure of H2 = 0.9947 - 0.03126

Pressure of H2 = 0.96344 atm

Step 4: Calculate moles of H2

p*V=n*R*T

⇒ with p = The pressure of H2 = 0.96344 atm

⇒ with V = the volume of H2 = 0.275 L

⇒ with n = the number of moles H2 = TO BE DETERMINED

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

⇒ with T = the temperature = 25°C = 298 Kelvin

n = (p*V)/(R*T)

n = (0.96344 * 0.275)/(0.08206*298)

n = 0.01083 moles

Step 5: Calculate moles of M

For 1 mole of H2 produced, we need 1 mole M

For 0.0108 moles of H2 we need 0.01083 moles of M

Step 6: Calculate molar mass of M

Molar mass M = Mass M / moles M

Molar mass M = 0.708 grams / 0.01083 moles

Molar mass M = 65.37 g/mol

The metal has a molar mass of 65.37 g/mol

5 0
4 years ago
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