Answer:

Explanation:
Hello,
In this case, for the given information, we can compute the rate of disappearance of NO₂ by using the following rate relationship:

Whereas it is multiplied by the the inverse of the stoichiometric coefficient of NO₂ in the reaction that is 2. Moreover, the subscript <em>f</em> is referred to the final condition and the subscript <em>0</em> to the initial condition, thus, we obtain:

Clearly, it turns out negative since the concentration is diminishing due to its consumption.
Regards.
Answer:
Number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.
Number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.
Total number of ions = 1.49 × 10²³ + 1.49 × 10²³ = 2.98 × 10²³.
Explanation:
1 mole of any compound contains 6.023 × 10²³ molecules.
molecular weight of NaCl is 23 + 35.5 = 58.5 g.
so, 58.5 grams of NaCl makes 1 mole
⇒ 14.5 g of NaCl =
= 0.248 moles.
⇒ 0.248 mole contains 0.248 × 6.023×10²³ molecules
= 1.49 × 10²³ molecules.
And 1 molecule contains 1 Na ion and 1 Cl ion.
⇒ number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.
⇒ number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.
Total number of ions = 1.49 × 10²³ + 1.49 × 10²³ = 2.98 × 10²³.
Answer:
If this trend continues, the following week will be cooler, and a large amount of rain will fall.
Explanation:
Patterns and trends can often be found in data sets. During the week that Cho recorded the weather, the temperatures consistently dropped by one to four degrees each day. At the end of the week, the amount of precipitation increased daily.
Ca(OH)₂: strong base
pOH = a . M
a = valence ( amount of OH⁻)
M = concentration
Ca(OH)₂ ⇒ Ca²⁺ + 2OH⁻ (2 valence)
so:
pOH = 2 x 0.005
pOH = 0.01
pH = 14 - 0.01 = 13.99