Select all that reasons that the reaction contents does not conduct at this low point.

An atom of beryllium (m = 8.00 u) splits into two atoms of helium (m = 4.00 u) with the release of 92.2 kev of energy. if the or

galben [10]

The kinetic energy of the products is equal to the energy liberated which is 92.2 keV. But let's convert the unit keV to Joules. keV is kiloelectro volt. The conversion that we need is: 1.602×10⁻¹⁹ <span>joule = 1 eV

Kinetic energy = 92.2 keV*(1,000 eV/1 keV)*(</span>1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules

From kinetic energy, we can calculate the velocity of each He atom:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
v = 5.367×10¹¹ m/s

4 0

2 years ago

Read 2 more answers

NO2 can react with the NO in smog, forming a bond between the N atoms. Draw the structure of the resulting compound, including f

ELEN [110]

First, let's write down the balanced chemical reaction between the given reactants:

NO₂ + NO → N₂O + O₂

The Lewis structure of the main product is shown in the attached picture. To determine the formal charge of each element, the formula is as follows:

Formal Charge = Valence electrons - Non-bonding valence electrons - (Bonding electrons/2)

For the leftmost N:
Formal charge = 5 - 2 - 6/2 = 0
For the middle N:
Formal charge = 5 - 0 - 8/2 = 1
For O:
Formal charge = 6 - 6 - 2/2 = -1

6 0

2 years ago

A 2400.-gram sample of an aqueous solution contains 0.012 gram of nh3. What is the concentration of nh3 in the solution

Kryger [21]

Answer : The concentration of $NH_3$ in the solution is, $2.94\times 10^{-4}M$

Explanation :

First we have to calculate the volume of aqueous solution that is water.

Density of water = 1.00 g/mL

Mass of water = 2400 g

$\text{Volume of water}=\frac{Mass}{Density}=\frac{2400g}{1.00g/mL}=2400mL$

Now we have to calculate the concentration of ammonia solution.

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Mass of }NH_3\times 1000}{\text{Molar mass of }NH_3\times \text{Volume of solution (in mL)}}$

Molar mass of $NH_3$ = 17 g/mole

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{0.012g\times 1000}{17g/mole\times 2400mL}=3.12mole/L=2.94\times 10^{-4}M$

Therefore, the concentration of $NH_3$ in the solution is, $2.94\times 10^{-4}M$

7 0

2 years ago

Recall that your hypothesis is that these values are the fraction of atoms that are still radioactive after n half-life cycles.

jolli1 [7]

Answer : A= 0.5, B = 0.25 , C = 0.125, D = 0.015625 and E = 0.00390625

Explanation :

Half life of a substance is defined as the amount of time taken by the substance to reduce to half of its original amount.

Here n represents the number of half lives.

The amount of substance that remains after n half lives can be calculated using the given formula, $0.5^{n}$

So when we have n =1,

Fraction of substance that remains = 0.5¹ = 0.5.

That means after first half life over, the amount of substance that remains is 0.5 times that of original.

Therefore we have A = 0.5

When n = 2, we have 0.5² = 0.25

So when 2 half lives are over, the amount of substance that remains is 0.25 times that of original

Therefore B = 0.25

When n = 3, we have 0.5³ = 0.125

So when 3 half lives are over, the amount of substance that remains is 0.125 times that of original.

Therefore we have C = 0.125

When n = 6 , we have 0.5⁶ = 0.015625

So D = 0.015625

When n = 8, we have 0.5⁸ = 0.00390625

Therefore E = 0.00390625

The values for A, B, C, D and E are 0.5, 0.25, 0.125, 0.015625 and 0.00390625 respectively.

8 0

2 years ago

Read 2 more answers

Who helps recycle nutrients through an ecosystem? A. producers and decomposers B. producers and consumers C. consumers and decom

s2008m [1.1K]

D. Producers im sure of it hope im right

8 0

2 years ago

Read 2 more answers