Answer:
48.32 g of anhydrous MnSO4.
Explanation:
Equation of dehydration reaction:
MnSO4 •4H2O --> MnSO4 + 4H2O
Molar mass = 55 + 32 + (4*16) + 4((1*2) + 16)
= 223 g/mol
Mass of MnSO4 • 4H2O = 71.6 g
Number of moles = mass/molar mass
= 71.6/223
= 0.32 mol.
By stoichiometry, since 1 mole of MnSO4 •4H2O is dehydrated to give 1 mole of anhydrous MnSO4
Number of moles of MnSO4 = 0.32 mol.
Molar mass = 55 + 32 + (4*16)
= 151 g/mol.
Mass = 151 * 0.32
= 48.32 g of anhydrous MnSO4.
Answer:
It depends on the objects mass, the gravitational pull when up or down slopes, and the height of the reference point
Solid, Liquid, and Gas
Solid is the state in which matter maintains a fixed volume and shape; liquid is the state in which matter adapts to the shape of its container but varies only slightly in volume; and gas is the state in which matter expands to occupy the volume and shape of its container.
Answer:
The elements in Group 2 (beryllium, magnesium, calcium, strontium, barium, and radium) are called the alkaline earth metals (see Figure below). These elements have two valence electrons, both of which reside in the outermost s sublevel. The general electron configuration of all alkaline earth metals is ns
The question is incomplete. The complete question is :
Hydrogen is manufactured on an industrial scale by this sequence of reactions:
The net reaction is :
Write an equation that gives the overall equilibrium constant 
in terms of the equilibrium constants 
and 
. If you need to include any physical constants, be sure you use their standard symbols, which you'll find in the ALEKS Calculator.
Solution :
![$K_1 = \frac{[CO][H_2]^3}{[CH_4][H_2O]}$](https://tex.z-dn.net/?f=%24K_1%20%3D%20%5Cfrac%7B%5BCO%5D%5BH_2%5D%5E3%7D%7B%5BCH_4%5D%5BH_2O%5D%7D%24)
...............(1)
![$K_2 = \frac{[CO_2][H_2]}{[CO][H_2O]}$](https://tex.z-dn.net/?f=%24K_2%20%3D%20%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D%24)
...................(2)
![$K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$](https://tex.z-dn.net/?f=%24K%3D%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%5E4%7D%7B%5BCH_4%5D%5BH_2O%5D%5E2%7D%24)
On multiplication of equation (1) and (2), we get
![$K_1 \times K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \times \frac{[CO_2][H_2]}{[CO][H_2O]}$](https://tex.z-dn.net/?f=%24K_1%20%5Ctimes%20K_2%3D%5Cfrac%7B%5BCO%5D%5BH_2%5D%5E3%7D%7B%5BCH_4%5D%5BH_2O%5D%7D%20%5Ctimes%20%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D%24)
![$K_1K_2=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$](https://tex.z-dn.net/?f=%24K_1K_2%3D%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%5E4%7D%7B%5BCH_4%5D%5BH_2O%5D%5E2%7D%24)
.................(4)
Comparing equation (3) and equation (4), we get
