Density= mass/volume. volume= 12 x 6 x 10. 415/720=.57g/cm^3
Answer:
The correct option is C
Explanation:
From the question we are told that
The reaction is
Generally 
Here 
is the change in enthalpy
is the change in the internal energy
is the difference between that number of moles of product and the number of moles of reactant
Looking at the reaction we can discover that the elements that was consumed and the element that was formed is 
and 
and this are both gases so the change would occur in the number of moles
So
![\Delta H = \Delta U + [16 -23]* RT](https://tex.z-dn.net/?f=%5CDelta%20%20H%20%20%3D%20%20%5CDelta%20%20U%20%2B%20%5B16%20-23%5D%2A%20%20RT)
The negative sign in the equation tell us that the enthalpy
would be less than the Internal energy 
Answer:
a) Aqueous LiBr = Hydrogen Gas
b) Aqueous AgBr = solid Ag
c) Molten LiBr = solid Li
c) Molten AgBr = Solid Ag
Explanation:
a) Aqueous LiBr
This sample produces Hydrogen gas, because the H+ (conteined in the water) has a reduction potential higher than the Li+ from the salt. Therefore the hydrogen cation will reduce instead of the lithium one and form the gas.
b) Aqueous AgBr
This sample produces Solid Ag, because the Ag+ has a reduction potential higher than the H+ from the water. Therefore the silver cation will reduce instead of the hydrogen one and form the solid.
c) Molten LiBr
In a molten binary salt like LiBr there is only one cation present in the cathod. In this case the Li+, so it will reduce and form solid Li.
c) Molten AgBr
The same as the item above: there is only one cation present in the cathod. In this case the Ag+, so it will reduce and form solid Ag.
Answer:
That would be helium, with a melting point of 0.95 K (-272.20 °C)—although this happens only under considerable pressure (~25 atmospheres). At ordinary pressure, helium would remain liquid even if it could be chilled to absolute zero.
Answer: 226 Subscript 88 Baseline Upper R a right arrow Superscript 222 Subscript 86 Baseline Upper R n + Superscript 4 Subscript 2 Baseline Upper H e
Explanation:
Alpha decay : When a larger nuclei decays into smaller nuclei by releasing alpha particle. In this process, the mass number and atomic number is reduced by 4 and 2 units respectively.
The general representation of alpha decay reaction is:
Representation of Radium decays to form Radon
Thus 226 Subscript 88 Baseline Upper R a right arrow Superscript 222 Subscript 86 Baseline Upper R n + Superscript 4 Subscript 2 Baseline Upper H e represents alpha decay.
