All categories

2 years ago

Sodium undergoes beta decay and becomes magnesium, write an equation. please help...

Physics

1 answer:

Tatiana [17]2 years ago

6 0

Remark

There are 2 kinds of Beta Decay. This one is an example of a neutron becomes a proton and the emitted particle is an electron.

$_{11}^{23}\text{Na}\longrightarrow_{12}^{23}\text{Mg}+_{-1}^{ 0}\text{1e-}$

You might be interested in

Theories have both an explanatory and a predictive function. <br> a. True <br> b. False

Simora [160]

Theories have both an explanatory an a predictive function. True

3 0

2 years ago

Read 2 more answers

Part K Find the y-component of the stone's velocity when it is 8.00 m below your hand

Romashka-Z-Leto [24]

The y-component of the stone's velocity when it is 8 m below the hand is 14.86 m / s

v² = u² + 2 a s

s = Displacement

u = Initial velocity

a = Acceleration

u = 8 m / s

s = 8 m

v² = 8² + 2 * 9.8 * 8

v² = 64 + 156.8

v = √ 220.8

v = 14.86 m / s

The equation used to solve the problem is an equation of motion. These equations are designed to locate an object in motion using components such as velocity, displacement, acceleration and time.

Therefore, the y-component of the stone's velocity is 14.86 m / s

To know more about Equations of motion

brainly.com/question/5955789

#SPJ1

4 0

8 months ago

Can anyone tell me the ans of this question with steps of the solution

gavmur [86]

Answer:

Explanation:

100s:160 paces

50s: 80 paces

80*0.80= 64

3 0

2 years ago

Read 2 more answers

You use 35 J of energy to move an object 5 m. What is the weight of the object

Vlad [161]

Explanation:

35 ÷ 5 = 7kg.

=======================

7 0

2 years ago

250 mL 0.1 M HCl solution is mixed with 250 mL

pishuonlain [190]

Answer:

The concentration of OH⁻ in the mixture is 0.05 M

Explanation:

The reaction of neutralization between HCl and NaOH is the following:

H⁺(aq) + OH⁻(aq) ⇄ H₂O(l)

The number of moles of HCl is:

$n_{HCl} = C*V = 0.1 mol/L*0.250L = 0.025 moles$

Similarly, the number of moles of NaOH is:

$n_{NaOH} = C*V = 0.2 mol/L*0.250L = 0.05 moles$

Now, from the reaction of HCl and NaOH we have the following number of moles of NaOH remaining:

$n_{NaOH} = 0.05 moles - 0.025 moles = 0.025 moles$

Finally, the concentration of OH⁻ in the mixture is:

$C =\frac{n_{NaOH}}{V_{T}}=\frac{0.025 moles}{0.250*2 L} = 0.05 moles/L$

Therefore, the concentration of OH⁻ in the mixture is 0.05 M.

I hope it helps you!

8 0

2 years ago