Answer:
four outer planetsThe Outer Planet the four outer planets and the Sun, with sizes to scale. From left to right, the outer planets are Jupiter, Saturn, Uranus, and Neptune. The gas giants are mostly made of hydrogen and helium. These are the same elements that make up most of the Sun.
Explanation:
I hope this helps you
Ari
mwah
Just look at the number in front also called coefficient (you have to balance the equations first, but all the questions here are balanced, so no worries). for q1.
in the balanced equation, the number in front of aluminum oxide is 2 (2 - this number Al2O3) and for aluminium is 4 as in (4 Al). so the ratio is 2:4. simplified it is 1:2. or write it out fully
2 Al2O3: 4 Al
ignore everything after the number.
2:4
same as 1:2
Aluminium oxide to oxygen
2 Al2O3: 3 O2
2:3
aluminum to oxygen
4 Al: 3 O2
4:3
question 2
Mercury oxide to Mercury
2 HgO : 2 Hg
2:2
same as 1:1
Mercury oxide to oxygen
2 HgO : O2
since oxygen in this case does not have a number written in front of it, the default is 1.
2: 1.
you should be able to do the rest
Answer is: mass of unused sulfur is 5.87 grams.
Balanced chemical reaction: C + 2S → CS₂.
m(C) = 12.0 g; mass of carbon.
m(S) = 70.0 g; mass of sulfur.
n(C) = m(C) ÷ M(C).
n(C) = 12 g ÷ 12 g/mol.
n(C) = 1 mol; amount of substance.
n(S) = m(S) ÷ M(S).
n(S) = 70 g ÷ 32.065 g/mol.
n(S) = 2.183 mol.
From chemical reaction: n(C) : n₁(S) = 1 : 2.
n₁(S) = 1 mol · 2 = 2 mol.
Δn(S) = n(S) - n₁(S).
Δn(S) = 2.183 mol - 2 mol.
Δn(S) = 0.183 mol; amount of unused sulfur.
Δm(S) = 0.183 mol · 32.065 g/mol.
Δm(S) = 5.87 g.
We are given the molar concentration of an aqueous solution of weak acid and the pH ofthe solution, and we are asked to determine the value of Ka for the acid.
The first step in solving any equilibrium problem is to write the equation for the equilibriumreaction. The ionization of benzoic acid can be written as seen in the attached image (1).
The equilibrium-constant expression is the equation number (2)
From the measured pH, we can calculate pH as seen in equation (3)
To determine the concentrations of the species involved in the equilibrium, we imagine that thesolution is initially 0.10 M in HCOOH molecules. We then consider the ionization of the acidinto H+ and HCOO-. For each HCOOH molecule that ionizes, one H+ ion and one HCOO- ionare produced in solution. Because the pH measurement indicates that [H+] = 1x 10^-4 M atequilibrium, we can construct the following table as seen in the equation number (4)
To find the value of Ka, please see equation (5):
We can now insert the equilibrium concentrations into the expression for Ka as seen in equation (6)
Therefore, 2.58x10^-4 M is the concentration of benzoic acid to have a pH of 4.0