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Arturiano [62]
2 years ago
13

An element with a mass number of 11 and an atomic number of 5 has how many neutrons?

Chemistry
1 answer:
Goryan [66]2 years ago
7 0

Answer:

6 neutrons

Explanation:

6 neutrons

Boron having an atomic number of 5 means that it will have 5 protons. 11 atomic mass units in total. Neutrons also have a atomic mass unit of 1. So there are 6 neutrons

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Which agents cause both chemical and physical weathering
shutvik [7]
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8 0
3 years ago
Can someone please help me
Lilit [14]
It is A. Barium

Explanation: I did that already
3 0
3 years ago
A solution is prepared by dissolving 27.7 g of cacl2 in 375 g of water. the density of the resulting solution is 1.05 g/ml. the
bagirrra123 [75]
The answer is 6.88.
Solution:
We can calculate for the percent composition of CaCl2 by mass by dividing the mass of the CaCl2 solute by the mass of the solution and then multiply by 100. The total mass of the resulting solution is the sum of the mass of CaCl2 solute and the mass of water solvent. Therefore, the percent composition of CaCl2 by mass is 
     % by mass = (mass of the solute / mass of the solution)*100 
                        = mass of solute / (mass of the solute + mass of the solvent)*100
                        = (27.7 g CaCl2 / 27.7g + 375g) * 100 
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5 0
3 years ago
Read 2 more answers
What is the symbol for the element whose atoms have 40 electrons each?
inysia [295]
<span>The symbol for the element whose atoms have 40 electrons each is Zr. This is the element zirconium. In the atoms of a pure element, the number of positively charged protons is normally equal to the number of negatively charged electrons. Hence, the number of electrons in the atom can be inferred from the atomic number, which corresponds to the number of protons in an atom. The atomic number of zirconium is 40.</span>
8 0
3 years ago
Balance this equation. If a coefficient of "1" is required, choose "blank" for that box. CO + O2 → CO2
liberstina [14]

Answer:

2co + o2 > 2co2

Explanation:

so that is tye answer

3 0
3 years ago
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