Answer:
when water changes from a liquid on earths surface to a gas in the atmosphere,this is known as evaporation
Antifreeze is an additive in water-based liquid to lower down the freezing point of such liquid. It is used to make use of the colligative properties of solutions specifically freezing-point depression for cold climate and boiling-point elevation to allow higher coolant temperature.
1 m = 0,001 km
1m³ = 0,000000001 km³
278 m³ = 0,000000278 km³ = 2,78×10^(-7) km³
:•)
Answer:
H_2O + 2CrO_4^2- + 3SO_3^2- -> 3SO_3^2- + 2CrO_2^- + 2OH^-
Explanation:
Reduction half reaction
2H_2O + CrO_4^2- + 3e -> CrO_2^- + 4OH^-
Oxidation half reaction
2OH^- + SO_3^2- -> SO_4^2- + H_2O + 2e
Balanced overall equation
H_2O + 2CrO_4^2- + 3SO_3^2- -> 3SO_3^2- + 2CrO_2^- + 2OH^-
Answer:
Explanation:
10 mL = .01 L .
25 mL = .025 mL .
10 mL of .1 M NaOH will contain .01 x .1 = .001 moles
25 mL of .1M HCl will contain .025 x .1 = .0025 moles
acid will neutralise and after neutralisation moles of acid remaining
= .0025 - .001 = .0015 moles .
Total volume = .01 + .025 = .035 L
concentration of remaining HCl = .0015 / .035
Option D is correct.
= .042857 M
= 42.857 x 10⁻³ M .
pH = - log [42.857 x 10⁻³]
= 3 - log 42.857
= 3 - 1.632
= 1.368 .
