Answer:
B The water at the bottom of the lake will never freeze
Explanation:
Its simple
Answer:Artificial light from cities has created a permanent "skyglow" at night, obscuring our view of the stars. Here's their map of artificial sky brightness in North America, represented as a ratio of "natural" nighttime sky brightness. In the black areas, the natural night sky is still (mostly) visible.
Explanation:
The values of x represents that number of moles of water molecules that is present per mole of the salt magnesium sulfate. To determine the value for this, we need to know how much is the water that is lost after heating the sample assuming that all of the water molecules are evaporated leaving only the unhydrated form of the salt. We calculate as follows:
Mass of hydrated salt = 3.484 g
Mass after heating = 1.701 g
Mass lost = 3.484 g - 1.701 g = 1.783 g
The mass lost is equal to the mass of water lost.
Moles water lost = 1.783 g ( 1 mol / 18.02 g ) = 0.0989 mol H2O
Moles of unhydrated salt = 1.701 g ( 1 mol / 120.37 g ) = 0.0141 mol MgSO4
moles water / moles MgSO4 = 0.0989 mol H2O / 0.0141 mol MgSO4 = 7
Therefore, the value of x is 7.
Answer:
Cathode: Ag
Anode: Br₂
Explanation:
In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:
Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V
Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V
Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V
As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.
For the anode an oxidation must occurs, so the reactions for the nonmetals are:
F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V
Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V
Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V
For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.
A 10.0g
ok hope this helps
