Answer:
37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.
Explanation:
Equation for the reaction:
2 CO + 2 NO ------> N2 + 2 CO2
2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen
At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.
So therefore, we can say:
2 * 22.4 L of CO produces 22.4 L of N2
44.8 L of CO produces 22.4 L of N2
Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:
44.8 L of CO = 22.4 L of N
x L = 18.9 L
x L = 18.9 * 44.8 / 22.4
x L = 18.9 * 2
x = 37.8 L
The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L
The answer is A, because of the chemical reaction taking place color can change (as in this case). Hope it helps!
Answer is: volume will be 3.97 liters.
Boyle's Law: the pressure volume law - volume of a given amount of gas held varies inversely with the applied pressure when the temperature and mass are constant.
p₁V₁ = p₂V₂.
p₁ = 755 torr.
V₁ = 5.00 l.
p₂ = 1.25 atm · 760 torr/atm.
p₂ = 950 torr.
755 torr · 5 l = 950 torr · V₂.
V₂ = 755 torr · 5 l / 950 torr.
V₂ = 3.97 l.
When pressure goes up, volume goes down.
When volume goes up, pressure goes down.
Answer:
Explanation:
Iso-electronic species have same number of electrons . Positive charged ions will have smaller size . As electrons add , size increases due to electronic repulsion .
Following species are isoelectronic .
Al³⁺ < Mg²⁺ < Na¹⁺ < Ne < F⁻¹ < O⁻² < N⁻³
Answer:
1. C+ ---- O-
2. O+ ---- Cl-
3. O+ ----- F-
4. C+ ----- N-
5. Cl- ----- C+
6. S- ----- H+
7. S+ ----- Cl -
Explanation:
Electronegativity determines the polarity . There may be two atoms in a bond with high electronegativity, in such cases the positive charge is given to atom with comparatively lower electronegativity. Electronegativity determines the easiness with which an atom attract electrons in a chemical bond. A polar bond is formed when the difference in the electronegativity of two combining atoms is between 0.4 and 1.7. The correct direction is
1. C+ ---- O-
2. O+ ---- Cl-
3. O+ ----- F-
4. C+ ----- N-
5. Cl- ----- C+
6. S- ----- H+
7. S+ ----- Cl -
