Answer:
0.133 mL
Explanation:
Given data
- Initial concentration (C₁): 15.0 M
- Initial volume (V₁): to be determined
- Final concentration (C₂): 0.001 M
- Final volume (V₂): 2.00 L
We can find the volume of the concentrated solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.001 M × 2.00 L / 15.0 M
V₁ = 1.33 × 10⁻⁴ L = 0.133 mL
Answer:
it is not correct I have an answer for ''why''
Explanation:
You need to balance it
s= 2
0=2
S+30•2 --------> SO•3
2 2 2 2
S +3O•2-------->SO•2......MULTIPLY
2 4 2 4......LCM IS 4
2 1 2 1....DIVIDE BY 4
SO NOW WE GOT THE BALANCE NO. WE CAN PUT IT IN THE EQUATIONS.
2S+3O•2 -----> 2SO•3
now you can check the both sides its balances the correct answer is this
Explanation:
every two moles of methanol will need three moles of oxygen gas and products 2 miles of carbon dioxide and 4 miles of water
Answer:
2 Atm; 2.016 g
Explanation:
Changing the volume without changing the temperature or mass only changes the pressure. Volume and pressure are inversely proportional so halving the volume will double the pressure.
P = 1 Atm, T = 0 °C are "standard" temperature and pressure (STP). The volume of 1 mole of gas is 22.4 L under these conditions. That means the amount of hydrogen gas in the cylinder is 1 mole, so has a mass of 2.016 g.
After the volume reduction, the pressure is 2 Atm, and the mass remains 2.016 g.
Answer: see figure attached and explanation below.
Explanation:
1) Chemical equation (given):
Fe + CuCl₂ → Cu + FeCl₂
2) ΔHf reactants: -256 kJ/mol (given)
3) ΔHf products: - 321 kJ/mol (given)
4) ΔH reaction = ΔHf products - ΔHf reactants = - 321 kJ/mol - (- 256 kJ/mol) = - 65 kJ/mol
5) Conclusion:
i) Since ΔHf of products is less (more negative) than ΔHf of reactants, the reaction is exhotermic: the reaction released energy, which is the reason why the products content less potential energy than the reactants.
ii) Then, the energy diagram is the typical one of an exothermic reaction: the products start a certain potential energy level, the energy incrases until reaching the activation energy (the energy barrier to form the activated complex) and then energy decreases until a level below the energy of the reactants.
iii) See the attached figure with such kind of diagram showing the products at a lower level than the reactans