Answer:
NaOBr (or) Na⁺ ⁻OBr
Explanation:
The Oxo-Acids of Bromine are as follow,
Hypobromous Acid = HOBr
Bromous Acid = HOBrO
Bromic Acid = HBrO₃
Perbromic Acid = HBrO₄
When these acids are converted to their conjugate bases their names are as follow,
Hypobromite = ⁻OBr
Bromite = ⁻OBrO
Bromate = ⁻OBrO₂
Perbromate = ⁻OBrO₃
According to rules, the positive part of ionic compound is named first and the negative part is named second. So, Sodium Hypobromite has a chemical formula of Na⁺ ⁻OBr or NaOBr.
The mass of carbon in 1 liter of mixture = 1.108 g
<h3>What is the mass of carbon in 1 liter of the mixture?</h3>
The mass of carbon in 1 liter of the mixture is determined as follows:
First the moles of gas is determined using the ideal gas formula:
n = (1 * 1)/(0.08205L * 298)
n = 0.0409 mole of total gas
mass of gas is then determined using the formula:
mass = 1 * 1.375
mass = 1.375 g
Let x = mass of CH₄ and y = mass of C₄H₁₀
x + y = 1.375 g
nCH₄ + nC₄H₁₀ = ntotat
moles = mass/molar mass
x + y = 1.695 => y = 1.695 - x
(x/molar mass of CH₄) + [(1.375 - x)/ molar mass C₄H₁₀ = 0.0409
x/16 + (1.375 - x)/58 = 0.0409
x = 0.380 g CH₄
y = 1.375 - 0.380
y = 0.995 g of C₄H₁₀
mass of C in CH₄ = 12/16 * 0.380 = 0.285
mass of C in C₄H₁₀ = 48/58 * 0.995 = 0.823
Mass of carbon in 1 liter of mixture = 0.285 + 0.823
Mass of carbon in 1 liter of mixture = 1.108 g
In conclusion, the carbon is the major component in the mixture.
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Answer:
3 molecules of water with 4 hydrogen atoms left over
Explanation:
Formula for water is H2O
2 hydrogen per 1 Oxygen
we have 10 H and 3 O
From the calculation, the pH of the solution is 4.85.
<h3>What is the pH?</h3>
The pH is defined as the hydrogen ion concentration of the solution. We have the ICE table as;
HA + H2O ⇔ H3O^+ + A^-
I 0.335 0 0
C -x +x +x
E 0.335 - x x x
Ka = 1 * 10^-14/Kb
Ka = 1 * 10^-14/1.8 × 10^–5
Ka = 5.56 * 10^-10
Ka = [H3O^+] [A^-]/[HA]
But [H3O^+] = [A^-] = x
5.56 * 10^-10 = x^2/ 0.335 - x
5.56 * 10^-10(0.335 - x ) = x^2
1.86 * 10^-10 - 5.56 * 10^-10x = x^2
x^2 + 5.56 * 10^-10x - 1.86 * 10^-10 = 0
x=0.000014 M
Now;
pH = -log 0.000014 M
pH = 4.85
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