corrected question:
Determining Density and Using Density to Determine Volume or Mass
(a) Calculate the density of mercury if 1.00 × 10 g occupies a volume of 7.36 cm³
(b) Calculate the volume of 65.0 g of liquid methanol (wood alcohol) if its density is 0.791 g/mL.
(c) What is the mass in grams of a cube of gold (density = 19.32 g/cm) if the length of the cube is 2.00 cm?
(d) Calculate the density of a 374.5-g sample of copper if it has a volume of 41.8 cm³ A student needs 15.0 g of ethanol for an experiment. If the density of ethanol is 0.789 g/mL, how many milliliters of ethanol are needed? What is the mass, in grams, of 25.0 mL of mercury (density = 13.6 g/mL)?
Answer:
density = 
ρ=m/v ,m=ρv, v=m/ρ
(a)m=1*10g , v=7.36cm³
ρ=10/7.36 =1.36g/cm³
(b) m=65g, ρ=0.791 g/mL.
v= 65/0.791 =82.17g/mL
(c) ρ=19.32g/cm³, l=2cm, v=l³=8cm³
m=19..32*8=154.56g/cm³
(d) mass of copper=374.5g , v=41.8cm³
ρ=374.5/41.8 =8.96g/cm³
mass of ethanol=15g, density of ethanol=0.789g/mL
v=15/0.789 =19.01mL
volume of mecury=25mL, density of mercury=13.6g/mL
m=25*13.6=340g
Answer:
Final concentration of NaOH = 0.75 M
Explanation:
For
:-
Given mass = 90.0 g
Molar mass of NaOH = 39.997 g/mol
The formula for the calculation of moles is shown below:
Thus,

Molarity is defined as the number of moles present in one liter of the solution. It is basically the ratio of the moles of the solute to the liters of the solution.
The expression for the molarity, according to its definition is shown below as:
Where, Volume must be in Liter.
It is denoted by M.
Given, Volume = 3.00 L
So,
<u>Final concentration of NaOH = 0.75 M</u>
Answer:
Initial rate of the reaction when concentration of hydrogen gas is doubled will be
.
Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
Initial rate of the reaction = R = 
![R = k\times [N_2][H_2]^3](https://tex.z-dn.net/?f=R%20%3D%20k%5Ctimes%20%5BN_2%5D%5BH_2%5D%5E3)
![4.0\times 10^5 M/s=k\times [N_2][H_2]^3](https://tex.z-dn.net/?f=4.0%5Ctimes%2010%5E5%20M%2Fs%3Dk%5Ctimes%20%5BN_2%5D%5BH_2%5D%5E3)
The initial rate of the reaction when concentration of hydrogen gas is doubled : R'
![[H_2]'=2[H_2]](https://tex.z-dn.net/?f=%5BH_2%5D%27%3D2%5BH_2%5D)
![R'=k\times [N_2][H_2]'^3=k\times [N_2][2H_2]^3](https://tex.z-dn.net/?f=R%27%3Dk%5Ctimes%20%5BN_2%5D%5BH_2%5D%27%5E3%3Dk%5Ctimes%20%5BN_2%5D%5B2H_2%5D%5E3)
![R'=8\times k\times [N_2][H_2]^3](https://tex.z-dn.net/?f=R%27%3D8%5Ctimes%20k%5Ctimes%20%5BN_2%5D%5BH_2%5D%5E3)

Initial rate of the reaction when concentration of hydrogen gas is doubled will be
.
Not sure what you are asking. I have two possible answers though...
It could either be more negatively charged, or valence electrons.
The more away from the nucleus a electron is, the more negatively charged it is.
The electrons on the outermost electron shell is valence electrons.
Again, I don't know what you were asking, but one of these answers may be correct.
This is seen in the first law of Thermodynamics stating that matter and energy cannot be destroyed nor created.