3.01× 1024 particles are the number of particles are there in 5 grams of sodium carbonate.
<h3>
How many particles are there in 5 grams of sodium carbonate?</h3>
There are 6.022 × 1023 particles in one gram of a substance according to Avogadro's number. So when we find out for 5 grams, then we multiply 5 with 6.022 × 1023, we get 3.01 × 1024 particles. For one gram atomic weight of hydrogen, one mole of hydrogen contains 6.022 × 1023 hydrogen atoms.
So we can conclude that 3.01× 1024 particles are the number of particles are there in 5 grams of sodium carbonate.
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Answer:
32.76mole of P
Explanation:
The reaction expression is given as;
P₄O₁₀ → 4P + 5O₂
Given parameters:
Number of moles of P₄O₁₀ = 8.19mol
Unknown:
Number of moles of P produced = ?
Solution:
From the balanced reaction expression;
1 mole of P₄O₁₀ will produce 4 mole of P
8.19mole of P₄O₁₀ will therefore produce 4 x 8.19 = 32.76mole of P
Answer: I believe the 1st and 3rd reactions are better obtained through reference sources and the 2nd and 4th are easiest and safest to measure in the laboratory.
Explanation:
I am also working on this Pre-lab right now, and I looked back at the first question to help get my answer. In the first question (a), it is noted that ammonia gas and gaseous hydrochloric acid are both potentially dangerous in gaseous form. I saw that both the 1st and 3rd reactions contained noxious gases (I knew this because there was a (g) in both of these reactions). Using the knowledge from the first question that the noxious gases were potentially dangerous, I assumed that those reactions were the ones that are better obtained through the reference sources. The 2nd and 4th reactions did not contain any noxious gases, so I assumed those ones were easiest and safest to measure in the laboratory. Hope this helps!
For the problem presented, the correct Lewis structure for SrO is <span>[Sr]^+2 + [O]^<span>-2. </span>I am hoping that this answer has
satisfied your query about this specific question. Also, if you have further
questions, please don’t hesitate to ask away.</span>