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3 years ago

How one atom undergoing fission can cause up to three other atoms to undergo fission.

Chemistry

1 answer:

mamaluj [8]3 years ago

7 0

Answer:

In order to initiate most fission reactions, an atom is bombarded by a neutron to produce an unstable isotope, which undergoes fission. When neutrons are released during the fission process, they can initiate a chain reaction of continuous fission which sustains itself.

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In a constant-pressure calorimeter, 55.0 mL of 0.340 M Ba(OH), was added to 55.0 mL of 0.680 M HCI. The reaction caused the temp

Brut [27]

Answer:

Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O

The reactants are present in equimolar amounts, so there is no excess or limiting reactants.

(0.0500 L) x (0.600 mol/L HCl) x (2 mol H2O / 2 mol HCl) = 0.0300 mol H2O

(4.184 J/g·°C) x (50.0 g + 50.0 g) x (25.82 - 21.73)°C = 1711.256 J

(1711.256 J) / (0.0300 mol H2O) = 57042 J/mol = 57.0 kJ/mol H2O

Explanation:

4 0

3 years ago

Which of the following changes occurs in the nucleus when a position particle is given off

sineoko [7]

Explanation:

When the position particle is given off, the nucleus' atom's turn to a cluster and the wavelength becomes shorter. Hope this helps!

8 0

3 years ago

What mass of ag2co3 could you produce from 12.7 g agno3 assuming that it is the limiting reagent?

zhenek [66]

10.3 g of $Ag_{2} CO_{3}$ will produce from 12.7 g $AgNO_{3}$ .

In simple terms, a limiting reagent is a reactant that is completely used up in the reaction.

It is also referred to as a limiting reactant or limiting agent.

Now, according to the question,

Since it is a limiting reagent, it will react fully.

The mass of $AgCO_{3}$ that will be produced will be:-

Equivalent mole of $AgNO_{3}$ = Equivalent weight of $Ag_{2} CO_{3}$ = 12.7/169.97 = x*2/274x = 10.3 g.

Hence, 10.3 g of $Ag_{2} CO_{3}$ will be produced.

It is used to restrict the reaction.

It tells you the estimated amount of compound to be used.

It brings quantitative understanding to chemical reactions.

More information on limiting reagents can be found here :

brainly.com/question/11848702

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7 0

2 years ago

Use the de Broglie's Wave Equation to find the wavelength of an electron moving at 7.3 × 106 m/s. Please show your work. Note: h

ad-work [718]

Answer:

$\lambda=9.96\times 10^{-11}\ m$

Explanation:

The expression for the deBroglie wavelength is:

$\lambda=\frac {h}{m\times v}$

Where,

$\lambda$ is the deBroglie wavelength

h is Planck's constant having value $6.62607\times 10^{-34}\ Js$

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

v is the speed of electron.

Given that v = $7.3\times 10^6\ m/s$

Applying in the equation as:

$\lambda=\frac {h}{m\times v}$

$\lambda=\frac{6.62607\times 10^{-34}}{9.11\times 10^{-31}\times 7.3\times 10^6}\ m$

$\lambda=\frac{10^{-34}\times \:6.626}{10^{-25}\times \:66.503}\ m$

$\lambda=\frac{6.626}{10^9\times \:66.503}\ m$

$\lambda=9.96\times 10^{-11}\ m$

6 0

3 years ago

Why is co2 collected by downward delivery in preparation

ozzi

Answer:

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5 0

4 years ago