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AleksandrR [38]

3 years ago

13

Suppose that atom x undergoes radioactive decay that results in atom y. If y has a larger atomic number than x, which type of ra

dioactive decay occured? Alpha decay, Beta decay, Gamma decay, or Delta decay??

1 answer:

Zanzabum3 years ago

4 0

Beta decay and I have to write something else to reach the min 20 words so I’m writing this to inform you why a am writing this

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Can you please answer number 8 thanks

MAXImum [283]

A day is added to February every four tears as a corrective measure. The Earth does not exactly revolve around the sun for 365 days. There is a small offset that is corrected by adding a day to February. This happens every four years and it is called Leap year.

4 0

3 years ago

During this reaction, P4 + 5O2 → P4O10, 1.5 moles of product were made in seconds. What is the rate of reaction? 0.011 g/min 210

Ivanshal [37]

The answer is: the rate of reaction is 850 g/min.

n(P₄O₁₀) = 1.5 mol; amount of product.

M(P₄O₁₀) = 283.9 g/mol; molar mass of phosphorus(V) oxide.

m(P₄O₁₀) = n(P₄O₁₀) · M(P₄O₁₀).

m(P₄O₁₀) = 1.5 mol · 283.9 g/mol.

m(P₄O₁₀) = 425.85 g in 30 seconds.

Make proportion: 425 g : 30 s = m(P₄O₁₀) : 60 s.

m(P₄O₁₀) = 850 g in 60 s.

7 0

3 years ago

Read 2 more answers

How many moles of ethylene (C2H4) can react with 12.9 liters of oxygen gas at 1.2 atmospheres and 297 Kelvin? C2H4(g) + 3O2(g) y

7nadin3 [17]

1) Convert 12.9 liters of Oxygen to mol at the given conditions:

PV = nRT ⇒ n = PV/RT

n = [1.2atm*12.9 l] / [0.082 atm l /K mol * 297K]

n = 0.636 mol of O2

2) use the stoichiometry derived from the balanced chemical equation

1mol C2H4 / 3 mol O2 = x mol C2H4 / 0.636 mol O2

x = 0.636 / 3 mol O2 = 0.212 mol O2.

Answer: 0.212 mol O2

5 0

3 years ago

The work function for metallic potassium is 2.3 eV. Calculate the velocity in km/s for electrons ejected from a metallic potassi

mote1985 [20]

Answer:

932.44 km/s

Explanation:

Given that:

The work function of the magnesium = 2.3 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.6022 × 10⁻¹⁹ J

So, work function = $2.3\times 1.6022\times 10^{-19}\ J=3.68506\times 10^{-19}\ J$

Using the equation for photoelectric effect as:

$E=\psi _0+\frac {1}{2}\times m\times v^2$

Also, $E=\frac {h\times c}{\lambda}$

Applying the equation as:

$\frac {h\times c}{\lambda}=\psi _0+\frac {1}{2}\times m\times v^2$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

m is the mass of electron having value $9.11\times 10^{-31}\ kg$

$\lambda$ is the wavelength of the light being bombarded

$\psi _0=Work\ function$

v is the velocity of electron

Given, $\lambda=260\ nm=260\times 10^{-9}\ m$

Thus, applying values as:

$\frac{6.626\times 10^{-34}\times 3\times 10^8}{260\times 10^{-9}}=3.68506\times 10^{-19}+\frac{1}{2}\times 9.11\times 10^{-31}\times v^2$

$3.68506\times \:10^{-19}+\frac{1}{2}\times \:9.11\times \:10^{-31}v^2=\frac{6.626\times \:10^{-34}\times \:3\times \:10^8}{260\times \:10^{-9}}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=7.64538\times 10^{-19}-3.68506\times 10^{-19}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=3.96032462\times 10^{-19}$

$v^2=0.869446\times 10^{12}$

v = 9.3244 × 10⁵ m/s

Also, 1 m = 0.001 km

<u>So, v = 932.44 km/s</u>

5 0

3 years ago

Please help I need a 100

svlad2 [7]

your answer would be C.

6 0

3 years ago

Read 2 more answers