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VladimirAG [237]
3 years ago
7

Which represents the balanced nuclear equation for the beta plus decay of C-11?

Chemistry
2 answers:
slavikrds [6]3 years ago
8 0
Nuclear reaction: ¹¹C → ¹¹B + e⁺(positron) + ve(electron neutrino).<span><span><span><span>
</span></span></span></span>Beta decay is radioactive decay<span> in which a beta ray and a neutrino are emitted from an atomic nucleus.
There are two types of beta decay: beta minus and beta plus. In beta minus decay, neutron is converted to a proton and an electron and an electron antineutrino and in beta plus decay, a proton is converted to a neutron and positron and an electron neutrino, so mass number does not change.</span>
Stella [2.4K]3 years ago
5 0

The answer is A, 11/6 C (insert arrow) 11/5 B + 0/+1 e

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A solution 0.20 molar in monomer (styrene) and 4.0 X 10-3 M in benzoyl peroxide initiator is heated at 60°C. kp = 145 liter/mole
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The AE of a system that releases 12.4 J of heat and does 4.2 J of work on its surroundings It is_______ jA. 16.6 B. 12.4 C. 4.2
andre [41]

ANSWER

EXPLANATION

Given that

The energy released by the system is 12.4J

Work done on the surrounding is 4.2J

Follow the steps below to find the change in energy

In the given data, energy is said to be released to the surroundings

Recall, that exothermic reaction is a type of reaction in which heat is released to the surroundings. Hence, change in enthalpy is negative

Step 1; Write the formula for calculating change in energy

\Delta E\text{ }=\text{ q }+\text{ w}

Since heat is released to the surrounding, then q = -12J

Recall, that work done by the system on the surroundings is always negative

Hence, w = -4.2J

Step 2; Substitute the given data into the formula in step 1

\begin{gathered} \text{  }\Delta E\text{ = q + w} \\ \text{  }\Delta E\text{ }=\text{ -12.4 }+\text{ \lparen-4.2\rparen} \\ \text{  }\Delta E\text{ = -12.4 - 4.2} \\ \text{  }\Delta E\text{ }=\text{ -16.6J} \end{gathered}

Therefore, the change i

3 0
1 year ago
If the atmospheric pressure in the laboratory is 1.2 atm, how many moles of gas were in each syringe? (Hint: Choose one volume a
Naily [24]

Answer:

A: 2.525 x 10-4 mol

B: 2.583 x 10-4 mol

Explanation:

Part A:

Data Given:

. Temperature of water (H2O) = 21.3°C

Convert Temperature to Kelvin

T = °C + 273

T = 21.3 + 273 = 294.3 K

volume of (H2O) gaseous state = 5.1 mL

Convert mL to liter

1000 mL = 1L

5.1 ml = 5.1/1000 = 0.0051 L

Pressure = 1.2 atm

. no. of moles = ?

Solution

no. of moles can be calculated by using ideal gas formula

PV = nRT

Rearrange the equation for no. of moles

n=PV/RT......... (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant

where

R = 0.08206 L.atm/ mol. K

Now put the value in formula (1) to calculate no. of moles of

n = 1.2 atm x 0.0051 L / 0.08206 L.atm.mol-1. K-1 x 294.3 K

n = 0.0061 atm.L / 24.162 L.atm.mol-1

n = 2.525 x 10-4 mol

no. of moles of gas (H2O) = 2.525 x 10-4 mol

Part B:

Data Given:

Temperature of water (H2) = 21.3°C

Convert Temperature to Kelvin

T = "C + 273

T= 21.3 + 273 = 294.3 K

volume of (H2) gas = 5.2 mL

Convert mL to liter

1000 mL = 1 L

5.2 ml = 5.2/1000 = 0.0052 L

Pressure = 1.2 atm

. no. of moles = ?

Solution

no. of moles can be calculated by using ideal gas formula

PV = nRT

Rearrange the equation for no. of moles

n= PV / RT......... (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant

where

R = 0.08206 L.atm/mol. K

Now put the value in formula (1) to calculate no. of moles of

n = 1.2 atm x 0.0052 L/0.08206 L.atm.mol-1. K-1 x 294.3 K

n = 0.0062 atm.L/ 24.162 L.atm.mol-1

n = 2.583 x 10-4 mol

I

no. of moles of gas (H2) = 2.583 x 10-4 mol

8 0
4 years ago
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