Answer:
We need 12.26 grams H2SO4
Explanation:
Step 1: Data given
Volume of a H2SO4 solution = 500 mL = 0.500 L
Concentration of the H2SO4 solution = 0.250 M
Molar mass of H2SO4 = 98.08 g/mol
Step 2: Calculate moles H2SO4
Moles H2SO4 = concentration * volume
Moles H2SO4 = 0.250 M * 0.500 L
Moles H2SO4 = 0.125 moles
Step 3: Calculate mass of H2SO4
Mass of H2SO4 = moles * molar mass
Mass of H2SO4 = 0.125 moles * 98.08 g/mol
Mass of H2SO4 = 12.26 grams
We need 12.26 grams H2SO4
Answer:
When the surface water is warm, the storm sucks up heat energy from the water, just like a straw sucks up a liquid. This creates moisture in the air. If wind conditions are right, the storm becomes a hurricane. This heat energy is the fuel for the storm.
Explanation:
I hope this can help you!
Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation.
To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign.
Pb(s) --> Pb2+ +2e- E0 = +0.13 V
Ag+ + e- ---> Ag E0 = +0.80 V
Adding up the E0's would yield an overall electric cell potential of +0.93 V.
Let's assume that the gas has ideal gas behavior.
Then we can use ideal gas equation,
PV = nRT
Where, P is Pressure of the gas (Pa), V is volume of the gas (m³), n is the number of moles of gas (mol), R is the Universal gas constant (8.314 J mol⁻¹ K⁻¹) and T is the temperature in Kelvin (K)
The given data for the gas is,
P = 2.8 atm = 283710 Pa
V = 98 L = 98 x 10⁻³ m³
T = 292 K
R = 8.314 J mol⁻¹ K⁻¹
n = ?
By applying the formula,
283710 Pa x 98 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 292 K
n = 11.45 mol
Hence,moles of gas is 11.45 mol.
<span>The atoms or molecules attain enough kinetic energy to overcome any intermolecular attractions they have. Since there are no longer any attractive forces between the particles, they are free to drift away into space. The same sort of thing happens in ordinary evaporation, but only at the surface. </span>
