Answer:
2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.
12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution
Explanation:
First, by definition of solubility, in 100 g of water there are 0.0016 g of CaF₂. So, to know how many moles are 0.0016 g, you must know the molar mass of the compound. For that you know:
- Ca: 40 g/mole
- F: 19 g/mole
So the molar mass of CaF₂ is:
CaF₂= 40 g/mole + 2*19 g/mole= 78 g/mole
Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 0.0016 grams of the compound how many moles are there?
moles=2.05*10⁻⁵
<u><em>2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.</em></u>
Now, to answer the following question, you can apply the following rule of three: if by definition of density in 1 mL there is 1 g of CaF₂, in 1000 mL (where 1L = 1000mL) how much mass of the compound is there?
mass of CaF₂= 1000 g
Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 1000 grams of the compound how many moles are there?
moles=12.82
<u><em>12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution</em></u>
Answer:
d) cut the large sized Cu solid into smaller sized pieces
Explanation:
The aim of the question is to select the right condition for that would increases the rate of the reaction.
a) use a large sized piece of the solid Cu
This option is wrong. Reducing the surface area decreases the reaction rate.
b) lower the initial temperature below 25 °C for the liquid reactant, HNO3
Hugher temperatures leads to faster reactions hence this option is wrong.
c) use a 0.5 M HNO3 instead of 2.0 M HNO3
Higher concentration leads to increased rate of reaction. Hence this option is wrong.
d) cut the large sized Cu solid into smaller sized pieces
This leads to an increased surface area of the reactants, which leads to an increased rate of the reaction. This is the correct option.
Answer:
Explanation:
turn over number = R max / [E]t = K2
From given , R max = 249 * 10 ^ -6 mol. L^-1
T [E]t = 2.23 n mol. L^-1
= 2.23 * 10^-9 mol. L^-1
Putting values in above equation,
= 111.65 * 10^3 S^-1
Turn over number is maximum no of substrate molecule that can be converted into product molecules for unit time by enzyme molecule.
Answer : The fuel value and the fuel density of pentane is, 49.09 kJ/g and 
respectively.
Explanation :
Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).
As we are given that:
Fuel value = 
Molar mass of pentane = 72 g/mol
Fuel value = 
Fuel value = 49.09 kJ/g
Now we have to calculate the fuel density of pentane.
Fuel density = Fuel value × Density
Fuel density = (49.09 kJ/g) × (0.626g/mL)
Fuel density = 30.73 kJ/mL =
Thus, the fuel density of pentane is
Answer:
Yes it is a mineral
Explanation:
Susanna has found a mineral because the five characteristics of a mineral are naturally occurring solid, crystal structure, forms by inorganic process, and definite chemical composition. Everything that was listed in the question describes the fact that its a mineral.
Brainliest would be appreciated :)
