Answer:
Molarity of the solution = 3.000 M
Volume of the solution = 250.0 mL = 0.25 L
moles in 250.0 mL = molarity x volume of the solution
= 3.000 M x 0.25 L
= 0.75 mol
Hence, 0.75 mol of NaCl is needed to prepare 250.0 mL of 3.000 M NaCl solution.
Moles (mol) = mass (g) / molar mass (g/mol)
Moles of NaCl in 250.0 mL = 0.75 mol
Molar mass of NaCl = 58.44 g/mol
Mass of NaCl in 250.0 mL = Moles x Molar mass
= 0.75 mol x 58.44 g/mol
= 43.83 g
Hence, 43.83 g of NaCl is needed to prepare 250.0 mL of 3.000 M solution.
Explanation:
I believe the third one if i am not mistaken
Answer:
a. 1.23 V
b. No maximum
Explanation:
Required:
a. Is there a minimum standard reduction potential that the half-reaction used at the cathode of this cell can have?
b. Is there a maximum standard reduction potential that the half-reaction used at the cathode of this cell can have?
The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E°cell = E°red, cat - E°red, an
If E°cell must be at least 1.10 V (E°cell > 1.10 V),
E°red, cat - E°red, an > 1.10 V
E°red, cat - 0.13V > 1.10 V
E°red, cat > 1.23 V
The minimum standard reduction potential is 1.23 V while there is no maximum standard reduction potential.
<span> endothermic is the answer
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<u>Answer:</u> The volume of oxygen gas at STP is 446 mL
<u>Explanation:</u>
STP conditions are:
Pressure of the gas = 1 atm
Temperature of the gas = 273 K
To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law. The equation follows:
where,
are the initial pressure, volume and temperature of the gas
are the final pressure, volume and temperature of the gas
We are given:
Putting values in above equation, we get:
Hence, the volume of oxygen gas at STP is 446 mL
