Answer:
17.934 kg of water
Explanation:
If balanced equation is not given; this format can come in handy.
For any alkane of the type : CₙH₂ₙ₊₂ , it's combustion reaction will follow:
2CₙH₂ₙ₊₂ + (3n+1) O₂ → (2n)CO₂ + 2(n+1) H₂O
For butane:
2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l)
2 moles of butane gives 10 moles of water.
1 mol of any substance has Avogadro number(N) of molecules in it( 6.022 x 10²³)
Mass of 1 mole of any substance is equal to it's molar mass
So, if 2 x N molecules of butane gives 10 x 18 g of water.
Then 1.2 x 10²⁶ molecules will give:

= 17.934 x 10³ g of water
= 17.934 kg of water
Answer:
C) the study of the composition of the atom.
Explanation:
A research can be defined as a systematic investigation or careful consideration of study with respect to a particular problem using scientific methods such as collection of data, documenting critical information, analysis of data, and the establishment of facts in order to reach new conclusions.
Similarly, a basic research is an approach to research that's typically theoritical and it's aimed at developing a theory, searching for the truth or gain a better understanding about a phenomenon, subject, or basic laws on nature.
In this context, an example of basic research is the study of the composition of the atom.
An atom can be defined as the smallest unit comprising of matter that forms all chemical elements. Thus, atoms are basically the building blocks of matters and as such determines or defines the structure of a chemical element.
Generally, atoms are typically made up of three distinct particles and these are protons, neutrons and electrons.
Because a solubility is calculated for saturated solutions we can write:
36.3 g NaCl ---100g water
x g NaCl ---- 300.0 g water
x=36.3*300/100=108.9 g NaCl
Answer:
CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓
Explanation:
We identify the reactants:
CuBr₂ and Pb(CH₃COO)₂
The products will be: Cu(CH₃COO)₂ and PbBr₂
You may know these information:
Salts from acetate are soluble.
Bromide can make solid salts with these cations: Ag⁺, Pb²⁺, Hg₂²⁺, Cu⁺
PbBr₂ is formed, so this will be our precipitate
The equation is:
CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓