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3 years ago

Which equation represents a transuranium

Chemistry

1 answer:

Maksim231197 [3]3 years ago

3 0

Answer:

1) Since you have not provided the equations to select the right one, I am going to explain you the relevant facts that are used to solve this question.

2) The transuranium elements are the chemiical elements with atomic number greater than that of the uranium.

The atomic number of uranium is 92. So, the transuranium elements are the elements with atomic number 93 or greater.

This are some of the transuranium elements:

Neptunio - 93

Plutonium - 94

Americium - 95

Curium - 96

Berkelium - 97

Californium - 98

Einstenium - 99

And so all the known elements (the last one is the 118).

3) In a nuclear reaction the total mass number ( shown as superscript to the left of the symbol) and total atomic number (shown as subscript to the left of the symbol) are conserved.

4) Beta decay is the release of a beta particle, which is an electron (considered massles and with charge - 1). So, the beta decay is represented with the symbol:

β, which means 0 mass and charge - 1.

-1

5) This is, then, an example of a β decay equation for one transuranium element:

239 239 0

Np → Pu + β

93 94 -1

As you see 239 = 239 + 0 and 93 = 94 - 1, showing that the total mass number ( shown as superscript to the left of the symbol) and the total atomic number (shown as subscript to the left of the symbol) are conserved.

Explanation:

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Answer:

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Balance this equation. Pb(NO3)2(aq)+NaCl(aq) -> NaNO3(aq)+PbCl2(s)

const2013 [10]

Answer:

Pb(NO3)2(aq) + 2NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)

Explanation:

Pb(NO3)2(aq)+NaCl(aq) -> NaNO3(aq)+PbCl2(s)

This is how it starts out.

Left:

2 NO3s
1 Pb
1 Na
1 Cl

Right

1 Na
1 NO3
1 Pb
2 Cl

So the place to start with this equation is to bring the Cls up to 2

Pb(NO3)2(aq)+2NaCl(aq) -> NaNO3(aq)+PbCl2(s)

But the Nas are now out of kilter.

Pb(NO3)2(aq)+ 2NaCl(aq) -> NaNO3(aq)+PbCl2(s)

Now the right has a problem. There's only 1 Na

Pb(NO3)2(aq) + 2 NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)

Check it out. It looks like we are done.

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2 years ago

If 42.7 of 0.208 M hydrochloric acid are needed to completely neutralize a solution of calcium hydroxide, how many grams of calc

Montano1993 [528]

Answer:

0.329 g

Explanation:

In the context of this problem, we have a chemical reaction between hydrochloric acid and calcium hydroxide. HCl is the acid here and calcium hydroxide is the base. Hence, we have an acid-base reaction, also known as neutralization reaction.

In a neutralization reaction, water is produced as a product, as well as a salt that we obtain after we exchange the cations: calcium bonds to chloride and hydrogen bonds to hydroxide (the latter is the formation of water). This means we also produce calcium chloride as a product. The overall reaction represents this as:

$Ca(OH)_2(aq)+2 HCl (aq)\rightarrow CaCl_2 (aq)+2 H_2O (l)$

Firslt of all, we wish to find the number of moles of HCl present. Having molarity and volume, this is done by applying the molarity formula. It states that molarity is equal to the rate between moles and volume:

$c_{HCl}=\frac{n_{HCl}}{V_{HCl}}$

Rearranging for moles of HCl, n:

$n_{HCl}=c_{HCl}V_{HCl}$

Based on stoichiometry of the balanced chemical equation, notice that 1 mole of calcium hydroxide reacts with 2 moles of HCl, meaning:

$n_{Ca(OH)_2}=\frac{1}{2} n_{HCl}=\frac{1}{2}c_{HCl}V_{HCl}$

Now that we have the expression for moles, we may also express moles of calcium hydroxide as the ratio between its mass and molar mass:

$n_{Ca(OH)_2}=\frac{m_{Ca(OH)_2}}{M_{Ca(OH)_2}}$

Using the last two equations, we see that:

$\frac{1}{2}c_{HCl}V_{HCl}=\frac{m_{Ca(OH)_2}}{M_{Ca(OH)_2}}\\\therefore m_{Ca(OH)_2}=\frac{1}{2}c_{HCl}V_{HCl}M_{Ca(OH)_2}$

Substitute the given data, as well as the molar mass of calcium hydroxide:

$m_{Ca(OH)_2}=\frac{1}{2}\cdot0.208 M\cdot0.0427 L\cdot74.093 g/mol=0.329 g$

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3 years ago

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3 years ago

Which equation represents a transuranium​

Which equation represents a transuranium