The correct options are as follows:
1. A.
A synthesis reaction is a type of reaction in which two or more reactants combine together to form only one product. Synthesis reaction always release energy in form of light and heat, therefore, they are usually exothermic reactions. In the option given in A, nitrogen and nitrogen combine together to form ammonia; this is a synthesis reaction.
2. D
A radioactive half life refers to the amount of time it will take for half of an original radioactive isotope to decay.
In the question given above, the half life of the element is 1000. Thus, in 1000 years only half of the original amount will remain. In another 1000 years only 1/4 of the original amount will remain and in another 1000 years only 1/8 of the original amount will remain. Therefore, it will take 3 half lives before 1/8 of the original sample remain.<span />
The average atomic mass of Boron: 10.431 amu
<h3>Further explanation
</h3>
Isotopes are atoms whose no-atom has the same number of protons while still having a different number of neutrons.
So Isotopes are elements that have the same Atomic Number (Proton)
Atomic mass is the average atomic mass of all its isotopes
In determining the mass of an atom, as a standard is the mass of 1 carbon-12 atom whose mass is 12 amu
Mass atom X = mass isotope 1 . % + mass isotope 2.% + ...
The average atomic mass of boron :

Answer:
the pressure would increase
Explanation:
Answer: i would say D or the last one.
Explanation: According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants. The law of conservation of mass is useful for a number of calculations and can be used to solve for unknown masses, such the amount of gas consumed or produced during a reaction.
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Answer:
Explanation:
From the information given:


no of moles of
= 0.01 L × 0.0010 mol/L
no of moles of
= 
no of moles of
= 0.01 L × 0.00010 mol/L
no of moles of
= 
Total volume = 0.02 L
![[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\ \\ \[[Ca^{2+}}] = 0.0005 \ mol/L](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%7D%5D%20%3D%20%5Cdfrac%7B1%5Ctimes10%5E%7B-5%7D%20%5C%20mol%7D%7B0.02%20%5C%20L%7D%20%5C%5C%20%5C%5C%20%20%5C%5C%20%20%5C%5B%5BCa%5E%7B2%2B%7D%7D%5D%20%3D%200.0005%20%5C%20mol%2FL)
![[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%281%5Ctimes%2010%5E%7B-6%7D%20%5C%20mol%29%7D%7B0.02%20%5C%20L%7D)
![[F^{-}] = 5 \times 10^{-5} \ mol/L](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%20%3D%205%20%5Ctimes%2010%5E%7B-5%7D%20%20%5C%20mol%2FL)
![Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}](https://tex.z-dn.net/?f=Q%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5BF%5E-%5D%5E2%20%5C%5C%20%5C%5C%20Q%20%3D%200.0005%20%5Ctimes%20%285%5Ctimes%2010%5E%7B-5%7D%29%5E2%20%5C%5C%20%5C%5C%20Q%20%3D%201.25%20%5Ctimes%2010%5E%7B-12%7D)
Since Q<ksp, then there will no be any precipitation of CaF2