6.02 x 10^6
Hoped that helped.
Explanation:
Bromine is a chemical element with the symbol Br and atomic number 35. It is the third-lightest halogen, and is a fuming red-brown liquid at room temperature that evaporates readily to form a similarly coloured vapour. Its properties are intermediate between those of chlorine and iodine. Isolated independently by two chemists, Carl Jacob Löwig (in 1825) and Antoine Jérôme Balard (in 1826), its name was derived from the Ancient Greek βρῶμος ("stench"), referring to its sharp and disagreeable smell.
Bromine, 35Br
Answer:
1.03 M
Explanation:
Step 1: Write the balanced equation
NaOH + HCl ⇒ NaCl + H₂O
Step 2: Calculate the reacting moles of HCl
30.0 mL (0.0300 L) of 0.500 M HCl react.
0.0300 L × 0.500 mol/L = 0.0150 mol
Step 3: Calculate the moles of NaOH that react with 0.0150 moles of HCl
The molar ratio of NaOH to HCl is 1:1. The moles of NaOH that react are 1/1 × 0.0150 mol = 0.0150 mol.
Step 4: Calculate the molar concentration of NaOH
0.0150 moles of NaOH are in 14.5 mL (0.0145 L).
M = 0.0150 mol/0.0145 L = 1.03 M
From Avogadro we obtained a physical constant of matter which is Avogadro's number, and from both scientists we understand that elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.
<h3>What is Avogadro's number?</h3>
Avogadro's number, or Avogadro's constant, is the number of particles found in one mole of a substance.
The Avogadro's number is given as 6.02 x 10²³.
Summary of Josef Loschmidt and Amedeo Avogadro Contribution to chemistry.
- Equal volumes of gas contain equal numbers of molecules,
- Elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.
Thus, from Avogadro we obtained a physical constant of matter which is Avogadro's number, and from both scientists we understand that elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.
Learn more about Avogadro's here: brainly.com/question/1581342
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Answer:
The six member ring and the position of the -OH group on the carbon (#4) identifies glucose from the -OH on C # 4 in a down projection in the Haworth structure). Fructose is recognized by having a five member ring and having six carbons, a hexose.
