Answer:
The correct answer is option C, that is, has an irritating odor and is colorless.
Explanation:
Sulfur dioxide refers to a gas. It has a nasty, sharp smell and is invisible. It associates with other components to produce harmful compounds like sulfurous acid, sulfuric acid, and sulfate particles. It is the chemical compound with the formula SO. At standard temperature, it is a toxic gas with an irritating and pungent smell. The gas is released naturally by the volcanic activity.
Answer:
Your missing the entire Transitional Metal group as well as all the Lanthanides and Actinides.
Explanation:
The groups are:
Alkali metals.
Alkaline earth metals.
Transitional metals.
Crystallogens.
Pnictogens.
Chalcogens.
Halogens.
Noble gases.
I believe the correct answers from the choices listed above are the second and the last option. At constant pressure, the systems 2A(g) + B(g) ---> 4C(g) and 2C(g) A(s) + B(s) ---> C(g) produces work to the surroundings. <span>When a gas is evolved during a chemical reaction, the gas can be imagined as displacing the atmosphere - pushing it back against the atmospheric pressure. The work done is therefore V*P where V is the volume of gas evolved, and P is the atmospheric pressure. </span>
Answer:
pH of the final solution = 9.15
Explanation:
Equation of the reaction: HCl + NH₃ ----> NH₄Cl
Number of moles of NH₃ = molarity * volume (L)
= 0.4 M * (300/1000) * 1 L = 0.12 moles
Number of moles of HCl = molarity * volume (L)
= 0.3 M * (175/1000) * 1 L = 0.0525 moles
Since all he acid is used up in the reaction, number of moles of acid used up equals number of moles of NH₄Cl produced
Number moles of NH₄Cl produced = 0.0525 moles
Number of moles of base left unreacted = 0.12 - 0.0525 = 0.0675
pOH = pKb + log([salt]/[base])
pKb = -logKb
pOH = -log (1.8 * 10⁻⁵) + log (0.0525/0.06755)
pOh = 4.744 + 0.109
pOH = 4.853
pH = 14 - pOH
pH = 14 - 4.853
pH = 9.15
Therefore, pH of the final solution = 9.15
Answer:
pH = 8.477
Explanation:
∴ ni 0.3 - -
nf - 0.3 0.3
∴ ni 0.50 - 0.3
nf 0.5 - X X 0.3 + X
∴ Ka = 2.0 E-9 = ([H3O+]*[OBr-]) / [HOBr]
⇒ Ka = 2.0 E-9 = ((X/1L)*(0.3 + X)/1 L) / ((0.5 - X)/1L)
⇒ 2.0 E-9 = ( 0.3X + X² ) / (0.5 - X)
⇒ X² + 0.3X - 1 E-9 = 0
⇒ X = 4.333 E-9 M
according Henderson-Hauselbach:
- pH = pk + Log [A-] / [HA]
∴ [OBr-] = 0.3 mol/ 1 L + 4.333 E-9 M = 0.300 M
∴ [HOBr] = 0.5 mol / 1 L - 4.33 E-9 M = 0.500 M
∴ pKa = - log Ka = - Log ( 2.0 E-9 ) = 8.6989
⇒ pH = 8.6989 + Log ( 0.300/0.500 )
⇒ pH = 8.477