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3 years ago

Determine the number of neutrons in an atom U-233.

2 answers:

5 0

We Know, Atomic number of Uranium = 92
Molar mass in given species = 233

Now, Molar mass = Number of Protons + Number of Neutrons
233 = 92 + x
x = 233 - 92
x = 141

So, your final answer is 141

Hope this helps!

babunello [35]3 years ago

4 0

Answer : The number of neutrons in an atom U-233 is, 141 neutrons.

Explanation :

The given atom is, U-233

The atomic mass number of U = 233

The atomic number of U = 92 (from the periodic table)

Now we have to determine the number of neutrons in U-233.

Number of neutrons = Atomic mass number - Atomic number

Number of neutrons = 233 - 92 = 141

Therefore, the number of neutrons in an atom U-233 is, 141 neutrons

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Explain how you can use boyle's law to determine the new volume of gas when its pressure is increased from 270 kPa to 540 kPa? T

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There are several information's already given in the question. Based on the information's the answer can be easily deduced.

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4 years ago

If an equal number of moles of reactants are used, do the following equilibrium mixtures contain primarily reactants or products

puteri [66]

Answer:

For a: The equilibrium mixture contains primarily reactants.

For b: The equilibrium mixture contains primarily products.

Explanation:

There are 3 conditions:

When $K_{eq}>1$ ; the reaction is product favored.
When $K_{eq}$ ; the reaction is reactant favored.
When $K_{e}=1$ ; the reaction is in equilibrium.

For the given chemical reactions:

For a:

The chemical equation follows:

$HCN(aq.)+H_2O(l)\rightleftharpoons CN^-(aq.)+H_3O^+(aq.);K_{eq}=6.2\times 10^{-10}$

The expression of $K_{eq}$ for above reaction follows:

$K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}$

As, $K_{eq}$ , the reaction will be favored on the reactant side.

Hence, the equilibrium mixture contains primarily reactants.

For b:

The chemical equation follows:

$H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g);K_{eq}=2.51\times 10^{4}$

The expression of $K_{eq}$ for above reaction follows:

$K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}$

As, $K_{eq}>1$ , the reaction will be favored on the product side.

Hence, the equilibrium mixture contains primarily products.

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3 years ago

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Answer:

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