Answer: its 15 its none of those
Explanation:
Answer:
Whether barium chloride solution was pure
Explanation:
We may answer whether barium chloride was pure. The sequence of this experiment might be depicted by the following balanced chemical equations:
Having a total sample of 10.0 grams, we would firstly find the mass percentage of barium in barium chloride:
This means in 10.0 g, we have a total of:
of barium cations.
The precipitate is then formed and we measure its mass. Having its mass determined, we'll firstly find the percentage of barium in barium sulfate using the same approach:
Multiplying the mass we obtained by the fraction of barium will yield mass of barium in barium sulfate. Then:
- if this number is equal to 6.595 g, we have a pure sample of barium chloride;
- if this number is lower than 6.595 g, this means we have an impure sample of barium chloride, as we were only able to precipitate a fraction of 6.595 g.
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:
The correct answer is d. Atoms share one or more pairs of electrons.
Explanation:
In a covalent bond, electrons of the last energy level are shared, to reach the octet (except for the Hydrogen that reaches its stability with 2 electrons). This type of bond occurs between two nonmetallic compounds. Example: H202 (hydrogen peroxide) formed between oxygen and hydrogen.