<u>Full Question:</u>
The list below includes some of the properties of butane, a common fuel. Identify the chemical properties in the list. Check all of the boxes that apply.
denser than water
burns readily in air
boiling point of –1.1°C
odorless
does not react with water
burns readily in air
does not react with water
<h3><u>
Explanation:</u></h3>
The type of alkane that is used in many products includes Butane. It is found as a natural gas in the environment. It is found on the deeper part of ground. It can be obtained by drilling process and gets used up in many of the products that is used for commercial purposes.
Molecular mass that is associated with butane is 58.12 g/mol. The boiling point of butane is -1 degree Celsius and -140 degree Celsius is its melting point. It is a liquefied gas and does not react with water. It will burn in air more readily.
Answer:
H+/H3O , H2O
Explanation:
The ability to be a proton donor is the Bronsted-Lowry definition of acids. The Lewis definition of an acid is an electron pair acceptor, which covers molecules liKE BF3
The ability to accept a pair of electrons is what is common to all acids, not the ability to be a proton donor.
All acid solutions contain hydronium ions (H3O+), hydroxide ions (OH-) and water molecules. Each different acid solution will then have an anion that is exclusive to that acid. For example, hydrochloric acid solution will contain all of the above and chloride ions (Cl-).
All acids contain the acidic substance dissolved in water. Water naturally dissociates to a small amount, creating hydronium and hydroxide ions. But most of the water remains as water molecules.
Then when we add an acid, like HCl, the oxygen on the water attracts the hydrogen from the HCl. The electrons in the covalent bond remain with the chlorine, giving it a negative charge and thus it becomes the chloride ion (Cl-). The hydrogen now has a positive charge and as said before, is attracted to the water (specifically the lone pair of electrons on the oxygen) to create hydronium ions.
This creates extra hydronium ions, making the solution acidic. But remember, there are still water molecules, hydroxide ions and the negative ion all in solution for all acids.
FeSO₄*7H₂O(s) = FeSO₄(s) + 7H₂O(g)
M(FeSO₄*7H₂O)=278.0 g/mol
M(FeSO₄)=151.9 g/mol
m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)=m(FeSO₄)/M(FeSO₄)
m(FeSO₄)=M(FeSO₄)m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)
m(FeSO₄)=151.9*100.0/278.0=54.6 g
m(FeSO₄)=54.6 g
Assume that the amount needed from the 5% acid is x and that the amount needed from the 6.5% acid is y.
We are given that:
The volume of the final solution is 200 ml
This means that:
x + y = 200
This can be rewritten as:
x = 200 - y .......> equation I
We are also given that:
The concentration of the final solution is 6%
This means that:
5%x + 6.5%y = 6% (x+y)
This can be rewritten as:
0.05 x + 0.065 y = 0.06 (x+y) ............> equation II
Substitute with equation I in equation II and solve for y as follows:
0.05 x + 0.065 y = 0.06 (x+y)
0.05 (200-y) + 0.065 y = 0.06 (200-y+y)
10 - 0.05 y + 0.065 y = 12
0.015y = 12-10 = 2
y = 2/0.015
y = 133.3334 ml
Substitute with the y in equation I to get the x as follows:
x = 200 - y
x = 200 - 133.3334
x = 66.6667 ml
Based on the above calculations:
The amount required from the 5% acid = x = 66.6667 ml
The amount required from the 6.5% acid = y = 133.3334 ml
Hope this helps :)
Answer:
See explanation below for answers
Explanation:
We know that the balance is tared, so the innitial weight would be zero. Now, let's answer this by parts.
a) mass of displaced water.
In this case all we need to do is to substract the 0.70 with the 0.13 g. so:
mW = 0.70 - 0.13
mW = 0.57 g of water
b) Volume of water.
In this case, we have the density of water, so we use the formula for density and solve for volume:
d = m/V
V = m/d
Replacing:
Vw = 0.57/0.9982
Vw = 0.5710 mL of water
c) volume of the metal weight
In this case the volume would be the volume displaced of water, which would be 0.5710 mL
d) the mass of the metal weight.
In this case, it would be the mass when the metal weight hits the bottom which is 0.70 g
e) density.
using the above formula of density we calculate the density of the metal
d = 0.70 / 0.5710
d = 1.2259 g/mL
