Answer:
add 7.5L of water
Explanation:
M1×V1=M2×V2
M is molarity, V is volume
0.7 × 10 = 0.4 × V2
V2= 17.5L
vol. of water to add= 17.5 - 10 = 7.5L
True
Explanation:
It is correct to say that the biosphere includes all of the fish that are in the ocean. Even though the ocean is part of the hydrosphere.
The biosphere includes all life form on earth wherever they may dwell from deep abyss to the high atmosphere.
The different spheres on earth are interacting with one another and they depend on each other.
Just like the earth systems, the spheres of the earth relates with each other. Even the ocean is contained in the geosphere.
learn more:
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Answer:
Mass = 42.8g
Explanation:
4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )
Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.
Step 1: Determine the balanced chemical equation for the chemical reaction.
The balanced chemical equation is already given.
Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).
Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol
Oxygen = 63.4g × 1mol / 32g = 1.9813mol
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.
Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.
5 moles of O2 = 6 moles of H2O
1.9831 moles = x
x = (1.9831 * 6 ) / 5
x = 2.37972 moles
Mass of H2O = Molar mass * Molar mass
Mass = 2.7972 * 18
Mass = 42.8g
Answer:
2.01V ( To three significant digits)
Explanation:
First we show the standard reduction potentials of Cu2+(aq)/Cu(s) system and Al3+(aq)/Al(s) system. We can clearly see from the balanced redox reaction equation that aluminium is the anode and was the oxidized specie while copper is the cathode and was the reduced specie. This observation is necessary when substituting values of concentration into the Nernst equation.
The next thing to do is to obtain the standard cell potential as shown in the image attached and subsequently substitute values of concentration and standard cell potential into the Nernst equation as shown. This gives the cell potential under the given conditions.
