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bixtya [17]
3 years ago
9

If a uranium atom undergoes both alpha and gamma decay, what happens to it?

Chemistry
2 answers:
WARRIOR [948]3 years ago
5 0

If a uranium atom undergoes both alpha and gamma decay, then it means that there will be formation of one helium particle which is also known as alpha particle and gamma decay is the radiation or release of energy. Whereas in a radioactive reaction, Uranium-235 absorbs a neutron and splits into two new atoms.

hope this helps

Alex_Xolod [135]3 years ago
5 0

Answer: _{92}^{235}\textrm{U}+_0^1\textrm{n}\rightarrow _{56}^{143}Ba+_{36}^{90}Kr+3_0^1\textrm{n}

Explanation: Nuclear fission : It is a process which involves the conversion of a heavier nuclei into two or more small and stable nuclei along with the release of energy.

General representation of an element is given as:   _Z^A\textrm{X}

where,

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

Alpha decay : When a larger nuclei decays into smaller nuclei by releasing alpha particle. In this process, the mass number and atomic number is reduced by 4 and 2 units respectively.

Gamma radiations does not carry any charge and are electrically neutral.

The equation for the decay will be written as:_{92}^{235}\textrm{U}+_0^1\textrm{n}\rightarrow _{56}^{143}Ba+_{36}^{90}Kr+3_0^1\textrm{n}

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What is/are defined in elements of the same group?
rusak2 [61]

Answer:

C

Explanation:

Elements in the same groups have similar chemical/physical properties. The reason the answer isn't electron configuration is that if this was the case, there would be no reason for transition metals to exist as they can have a varying amount of electrons in their outer shell.

For example, all alkali metals (Group 1) turn into positive ions when reacting and Halogens (Group 7) all turn into negative ions after reacting.

4 0
3 years ago
Science question pleace answer!!
mylen [45]

Answer:

the humidity 80% at the temperature (20° C ) will have more water vapour because relative humidity is more at lesser temperature (20° C) and lower at higher temperature (40°C )

<em>i</em><em> </em><em>hope</em><em> </em><em>it</em><em> </em><em>helped</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

6 0
3 years ago
How many grams of glucose(C6H12O6) are produced by 10.0 grams of water(H2O)?
GaryK [48]

Answer:

16.7 g of glucose

Explanation:

Convert grams of H₂O to moles.  <em>0.555 mol</em>

We can use the chemical equation to determine the theoretical yield.  Based on the equation, for every 6 moles of H₂O, 1 mole of glucose is produced.  The ratio of glucose moles to H₂O moles is 1/6.  

Multiply the moles of H₂O by the ratio to find moles of glucose.  <em>0.0925 mol</em>

Convert moles of glucose to grams.  <em>16.7 g</em>

3 0
3 years ago
In the reaction of aluminum with iron III oxide, if you start with 54.2 grams of aluminum, how many grams of iron III oxide are
Sliva [168]

Answer:

m_{Fe_2O_3}=160.41gFe_2O_3

Explanation:

Hello!

In this case, since the reaction between aluminum and iron (III) oxide is:

2Al+Fe_2O_3\rightarrow Al_2O_3+2Fe

In such way, since there is 2:1 mole ratio between aluminum (atomic mass = 26.98 g/mol) and iron (III) oxide (molar mass = 159.70 g/mol), we'll be able to compute the mass of the required reactant as shown below:

m_{Fe_2O_3}=54.2gAl*\frac{1molAl}{26.98gAl}* \frac{1molFe_2O_3}{2molAl}*\frac{159.70gFe_2O_3}{1molFe_2O_3}\\\\m_{Fe_2O_3}=160.41gFe_2O_3

Best regards!

5 0
3 years ago
What volume of 0.500 M HNO3(aq) must completely react to neutralize 100.0 milliliters of 0.100 M KOH(aq)?
asambeis [7]
KOH+ HNO3--> KNO3+ H2O<span>
From this balanced equation, we know that 1 mol HNO3= 1 mol KOH (keep in mind this because it will be used later).

We also know that 0.100 M KOH aqueous solution (soln)= 0.100 mol KOH/ 1 L of KOH soln (this one is based on the definition of molarity).

First, we should find the mole of KOH:
100.0 mL KOH soln* (1 L KOH soln/ 1,000 mL KOH soln)* (0.100 mol KOH/ 1L KOH soln)= 1.00*10^(-2) mol KOH.

Now, let's find the volume of HNO3 soln:
1.00*10^(-2) mol KOH* (1 mol HNO3/ 1 mol KOH)* (1 L HNO3 soln/ 0.500 mol HNO3)* (1,000 mL HNO3 soln/ 1 L HNO3 soln)= 20.0 mL HNO3 soln.

The final answer is </span>(2) 20.0 mL.<span>

Also, this problem can also be done by using dimensional analysis. 

Hope this would help~ </span>
6 0
4 years ago
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