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4 years ago

The parent element 247 95 am decays by beta decay which daughter element will formn

1 answer:

6 0

Atomic mass of the parent element =247,
atomic number of the parent element = 95

In the process of β-decay electron leaves the nucleus, so instead of one neutron we get one proton.

Mass of proton≈mass of neutron,
so atomic mass will not change.

Charge of proton =+1, and charge of neutron = 0.
So, we will get atomic number increased by one.

New element (daughter) will have
atomic mass = 247,
and atomic number= 95+1=96

Number 95 - Am (parent),
number 96 - Cm(daughter),

So, from Am-247 we will get Cm-247.

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How, exactly, would altering the pH in a solution affect a living organism and how do organisms prevent large pH fluctuations?

oksano4ka [1.4K]

Answer:

pH solutions cause the water to become more basic or more acidic therefore would force the organisms to adapt to their new situations

4 0

3 years ago

Aluminum chloride can be formed from its elements:

saul85 [17]

<u>Answer:</u> The $\Deltas H^o_{formation}$ for the reaction is -1406.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical reaction for the formation reaction of $AlCl_3$ is:

$2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3 (s)$ $\Delta H^o_{formation}=?$

The intermediate balanced chemical reaction are:

(1) $HCl(g)\rightarrow HCl(aq.)$ $\Delta H_1=-74.8kJ$ ( × 6)

(2) $H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$ $\Delta H_2=-185kJ$ ( × 3)

(3) $AlCl_3(aq.)\rightarrow AlCl_3(s)$ $\Delta H_3=+323kJ$ ( × 2)

(4) $2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)$ $\Delta H_4=-1049kJ$

The expression for enthalpy of formation of $AlCl_3$ is,

$\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4]$

Putting values in above equation, we get:

$\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ$

Hence, the $\Deltas H^o_{formation}$ for the reaction is -1406.8 kJ.

6 0

3 years ago

graduated cylinder is filled with water to a volume of 6.2 ML. an irregular shaped plastic object weighing 1.2 g is placed in th

Aleks04 [339]

Answer:

A. Density of object = 0.86 g/mL

B. The object will float in the water.

Explanation:

The following data were obtained from the question:

Volume of water = 6.2 mL

Mass (m) of object = 1.2 g

Volume of water + Object = 7.59 mL

Density of object =?

Density of water = 1 g/mL

Next, we shall determine the volume of the object. This can be obtained as follow:

Volume of water = 6.2 mL

Volume of water + Object = 7.59 mL

Volume of object =?

Volume of object = (Volume of water + Object) – (Volume of water)

Volume of object = 7.59 – 6.2

Volume of object = 1.39 mL

Therefore, the volume of the object is 1.39 mL

A. Determination of the density of the object.

Mass (m) of object = 1.2 g

Volume (V) of object = 1.39 mL

Density (D) of object =?

Density = mass /volume

Density = 1.2/1.39

Density of object = 0.86 g/mL

B. Determination of whether the object will float or sink.

Density of object = 0.86 g/mL

Density of water = 1 g/mL

From the above, we can see that the density of water is greater than that of the object. This implies that the object is lighter than water. Therefore, the object will float in the water.

8 0

4 years ago

How much percent of earth is water

horsena [70]

71% of the earth is water and the 29% is continents and islands.

6 0

4 years ago

What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the i

UNO [17]

Answer:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

$Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-$

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

$Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)$

The net equation is then:

$Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)$

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

$Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-$

Actual cathode half-equation:

$Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)$

Actual net reaction:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

6 0

3 years ago