Answer:
There is 17,114825 g of powdered drink mix needed
Explanation:
<u>Step 1 :</u> Calculate moles
As given, the concentration of the drink is 0.5 M, this means 0.5 mol / L
Since the volume is 100mL, we have to convert the concentration,
⇒0.5 / 1 = x /0.1 ⇒ 0.5* 0.1 = x = 0.05 M
This means there is 0.05 mol per 100mL
e
<u>Step 2 </u>: calculate mass of the powdered drink
here we use the formula n (mole) = m(mass) / M (Molar mass)
⇒ since powdered drink mix is usually made of sucrose (C12H22O11) and has a molar mass of 342.2965 g/mol.
0.05 mol = mass / 342.2965 g/mol
To find the mass, we isolate it ⇒0.05 mol * 342.2965 g/mol = 17,114825g
There is 17,114825 g of powdered drink mix needed
Answer:
benzamide
Explanation:
Compound melting Point ,ºC Melting Pont Mixture, ºC
X 131 - 133
trans-cinnamic 133 - 134 110 - 120
acid
benzamide 128 - 130 130-132
malic acid 131 -133 114 -124
Benzoin 135 - 137 108 - 116
The compound X is benzamide since the melting point range is the one closest to this compound ( 130-132 ºC)
The reason there is not an exact match is not due due to the presence of impurities. The presence of impurities always lower the melting point ( it is a coligative property such as the melting point depresion of salt and water )
The reason for the deviation must be be some other factors such as preparation of the sample in the capillary, errors in reading the thermometer, rate of heating, etc.
Answer:
Option D is correct.
H₂O + CO₂ → H₂CO₃
Explanation:
First of all we will get to know what law of conservation of mass states.
According to this law, mass can neither be created nor destroyed in a chemical equation.
This law was given by French chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
Example:
6CO₂ + 6H₂O + energy → C₆H₁₂O₆ + 6O₂
there are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass.
Now we will apply this law to given chemical equations:
A) H₂ + O₂ → H₂O
There are two hydrogen and two oxygen atoms present on left side while on right side only one oxygen and two hydrogen atoms are present so mass in not conserved. This equation not follow the law of conservation of mass.
B) Mg + HCl → H₂ + MgCl₂
In this equation one Mg, one H and one Cl atoms are present on left side while on right side two hydrogen, one Mg and two chlorine atoms are present. This equation also not follow the law of conservation of mass.
C) KClO₃ → KCl + O₂
There are one K, one Cl and three O atoms are present on left side of chemical equation while on right side one K one Cl and two oxygen atoms are present. This equation also not following the law of conservation of mass.
D) H₂O + CO₂ → H₂CO₃
There are two hydrogen, one carbon and three oxygen atoms are present on both side of equation thus, mass remain conserved. Thus is correct option.
Answer:
C₄F₈
Explanation:
Using their mole ratio to compute their mass
molar mass of carbon = 12.0107 g/mol
molar mass of fluorine gas = 37.99681
let x = mass of carbon
given mass of fluorine = 1.70 g
x / 12.01067 = 1.70 / 37.99687
cross multiply
x = ( 1.70 × 12) / 37.99687 = 20.4 / 37.99687 = 0.53688 g
mass of one mole of CF₂ = 0.53688 + 1.70 = 2.23688 g
number of mole of CF₂ = 8.93 g / 2.23688 = 3.992 approx 4
molecular formula of CF₂ = 4 (CF₂) = C₄F₈
