The atomic mass of rubidium listed in the periodic table to determine the mass of Rb−87 is 86.13 amu.
<h3>What is atomic mass?</h3>
The atomic mass is the weight of the roton neutron and electron present inside the nucleus and shells of an atom and the elements are arranged on the basis of this only.
The mass of Rb−85 is 84.9118 amu for 1 amu it will be
amu = 84.9118 / 85 = 0.99
so, the amu for Rb−87 will be,
AMU = 87 × 0.99 = 86.13 amu.
Therefore, the mass of Rb−87 is 86.13 amu for atomic mass of rubidium listed in the periodic table.
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Answer:
1. They will react.
2. 2HCI + Zn —> ZnCl2 + H2
Explanation:
1. From the question given above,
We can see that Zn is higher than H in the activity series.
NOTE: Elements higher (i.e at the top) in the activity series will displace those lower (i.e at the bottom) in the series. Thus, Zn is higher in the series than H. Therefore, Zn will displace H from solution. Hence, they will react.
2. The product obtained from the reaction can be seen as follow:
HCI + Zn —> ZnCl2 + H2
The above equation can be balance as follow:
There are 2 atoms of H on the right side and 1 atom on the left side. It can be balance by putting 2 in front of HCl as shown below:
2HCI + Zn —> ZnCl2 + H2
Answer: acid react with metals to form H2.
Explanation: metals which are more electropositve than hydrogen will always displaced hydrogen from acids. The equation below show how hydrogen is displaced from acid
Zn + 2HCl -—> ZnCl2 + H2
you have a picture so i can see it
Answer:
i only found the 49, hope it still helps, mate
49. In this question (t½) of C-14 is 5730 years, which means that after 5730 years half of the sample would have decayed and half would be left as it is.
After 5730 years ( first half life) 70 /2 = 35 mg decays and 35 g remains left.
After another 5730 years ( two half lives or 11460 years) 35 /2 = 17.5mg decays and 17.5 g remains left .
After another 5730 years ( three half lives or 17190 years) 17.5 /2 = 8.75mg decays and 8.75g remains left.
after three half lives or 17190 years, 8.75 g of C-14 will be left.