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2 years ago

I-131 undergoes beta-minus decay. The chemical symbol for the new element is.

Chemistry

1 answer:

sesenic [268]2 years ago

3 0

The chemical element that is formed when I-131 undergoes beta-minus decay is Xe.

<h3>What is a beta minus decay?</h3>

A beta minus decay is one in which an electron is changed into a proton. The atomic number of the daughter nucleus will increase by one unit and is one place after the parent in the periodic table.

The chemical element that is formed when I-131 undergoes beta-minus decay is Xe.

Learn more about beta decay:brainly.com/question/25455333

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A sample of flammable liquid is placed into an enclosed cylinder which is then fitted with a movable piston. Initially the cylin

polet [3.4K]

Answer:

12.09 L

Explanation:

Step 1: Convert 826.1 mmHg to atm

We will use the conversion factor 760 mmHg = 1 atm.

826.1 mmHg × 1 atm/760 mmHg = 1.087 atm

Step 2: Convert 427.8 J to L.atm

We will use the conversion factor 101.3 J = 1 L.atm.

427.8 J × 1 L.atm/101.3 J = 4.223 L.atm

Step 3: Calculate the change in the volume

Assuming the work done (w) is 4.223 L.atm against a pressure (P) of 1.087 atm, the change in the volume is:

w = P × ΔV

ΔV = w/P

ΔV = 4.223 L.atm/1.087 atm = 3.885 L

Step 4: Calculate the final volume

V₂ = V₁ + ΔV

V₂ = 8.20 L + 3.885 L = 12.09 L

5 0

2 years ago

Which elements can have expanded octets?

PSYCHO15rus [73]

Answer:

Sulfur, phosphorus, silicon, and chlorine are common examples of elements that form an expanded octet.

Explanation:

Phosphorus pentachloride (PCl5) and sulfur hexafluoride (SF6) are examples of molecules that deviate from the octet rule by having more than 8 electrons around the central atom

7 0

2 years ago

Help please !!!!!!!!

Elena L [17]

Answer:

Option B. 2096.1 K

Explanation:

Data obtained from the question include the following:

Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹

Entropy (S) = +614 JK¯¹mol¯¹

Temperature (T) =.?

Entropy is related to enthalphy and temperature by the following equation:

Change in entropy (ΔS) = change in enthalphy (ΔH) / Temperature (T)

ΔS = ΔH / T

With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:

ΔS = ΔH / T

614 = 1287000/ T

Cross multiply

614 x T = 1287000

Divide both side by 614

T = 1287000/614

T = 2096.1 K

Therefore, the temperature at which the reaction will be feasible is 2096.1 K

5 0

3 years ago

A mixture of helium and methane gases, at a total pressure of 821 mm Hg, contains 0.723 grams of helium and 3.43 grams of methan

Nookie1986 [14]

Answer:

For 1: The partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

For 2: The mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

Explanation:

For 1:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For helium:

Given mass of helium = 0.723 g

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

$\text{Moles of helium}=\frac{0.723g}{4g/mol}=0.181mol$

For methane gas:

Given mass of methane gas = 3.43 g

Molar mass of methane gas = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane gas}=\frac{3.43g}{16g/mol}=0.214mol$

To calculate the mole fraction , we use the equation:

$\chi_A=\frac{n_A}{n_A+n_B}$ .......(2)

To calculate the partial pressure of gas, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ......(3)

For Helium gas:

We are given:

$n_{He}=0.181mol\\n_{CH_4}=0.214mol$

Putting values in equation 2, we get:

$\chi_{He}=\frac{0.181}{0.181+0.214}=0.458$

Calculating the partial pressure by using equation 3, we get:

$p_T=821mmHg\\\\\chi_{He}=0.458$

Putting values in equation 3, we get:

$p_{He}=0.458\times 821mmHg=376mmHg$

For Methane gas:

We are given:

$n_{He}=0.181mol\\n_{CH_4}=0.214mol$

Putting values in equation 2, we get:

$\chi_{CH_4}=\frac{0.214}{0.181+0.214}=0.542$

Calculating the partial pressure by using equation 3, we get:

$p_T=821mmHg\\\\\chi_{CH_4}=0.542$

Putting values in equation 3, we get:

$p_{CH_4}=0.542\times 821mmHg=445mmHg$

Hence, the partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

For 2:

We are given:

Partial pressure of nitrogen gas = 363 mmHg

Partial pressure of carbon dioxide gas = 564 mmHg

Total pressure = (363 + 564) mmHg = 927 mmHg

Calculating the mole fraction of the gases by using equation 3:

For nitrogen gas:

$363=\chi_{N_2}\times 927\\\\\chi_{N_2}=\frac{363}{927}=0.392$

For carbon dioxide gas:

$564=\chi_{CO_2}\times 927\\\\\chi_{CO_2}=\frac{564}{927}=0.608$

Hence, the mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

6 0

3 years ago

How many moles are in 25.2g of KMnO4?

Zolol [24]

When finding the moles in a compound you have to know the grams. In this case, 25.2 grams are given for KMnO4. To find the moles you would divide the amount of grams by the molar mass of KMnO4. The molar mass of KmnO4 is 158.034. You you would now divide 25.2 by 158.034 which is 0.15946 moles. Depending on how many decimal places the questions asks for is dependent on you. I just went with 5 significant figures.

4 0

3 years ago

Read 2 more answers