Answer:
1.71 × 10²¹ molecules
Explanation:
1 gram is equal to 1000 milligrams. The mass in grams corresponding to 500 mg is:
500 mg × (1 g / 1000 mg) = 0.500 g
The molar mass of ascorbic acid is 176.12 g/mol. The moles corresponding to 0.500 grams of ascorbic acid are:
0.500 g × (1 mol/ 176.12 g) = 0.00284 mol
In 1 mole of ascorbic acid, there are 6.02 × 10²³ molecules of ascorbic acid (Avogadro's number). The molecules in 0.00284 moles are:
0.00284 mol × (6.02 × 10²³ molecules/ 1 mol) = 1.71 × 10²¹ molecule
The name is Potassium bromide.
From the fact that oxygen is in group 16 and carbon is in group 14, the structure of CO2 must be O=C=O. In methane, there is no bond between any of the hydrogen atoms. The structure of H2O2 is H–O–O–H.
Carbon is in group 14 hence it has four valence electrons and oxygen is in group 16 hence it has six valence electrons. This implies that each oxygen atom will share four electrons with carbon in a covalent bond to form the structure O=C=O.
In CH4, we know that carbon is tetravalent so it forms for bonds. Therefore, there is no bond between hydrogen atoms so it bonds with each hydrogen atom; hydrogen only forms one bond.
In H2O2, there is the peroxide ion that has the structure O-O. Hence, the correct structure of H2O2 is H–O–O–H.
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<h3>Answer:</h3><h2>Chemical properties</h2><h3>Explanation:</h3>
By its very definition, a chemical property is one which is exhibited as a result of a chemical reaction. This may happen during or after the reaction. This is because in a chemical reaction there is a transformation in the physical composition of the components and this directly affects its chemical properties.
Answer:
M.Mass = 3.66 g/mol
Data Given:
M.Mass = M = ??
Density = d = 0.1633 g/L
Temperature = T = 273.15 K (Standard)
Pressure = P = 1 atm (standard)
Solution:
Let us suppose that the gas is an ideal gas. Therefore, we will apply Ideal Gas equation i.e.
P V = n R T ---- (1)
Also, we know that;
Moles = n = mass / M.Mass
Or, n = m / M
Substituting n in Eq. 1.
P V = m/M R T --- (2)
Rearranging Eq.2 i.e.
P M = m/V R T --- (3)
As,
Mass / Volume = m/V = Density = d
So, Eq. 3 can be written as,
P M = d R T
Solving for M.Mass i.e.
M = d R T / P
Putting values,
M = 0.1633 g/L × 0.08205 L.atm.K⁻¹.mol⁻¹ × 273.15 K / 1 atm
M = 3.66 g/mol
