Answer:
All flowering is regulated by the integration of environmental cues into an internal sequence of processes. These processes regulate the ability of plant organs to produce and respond to an array of signals. The numerous regulatory switches permit precise control over the time of flowering.
Explanation:
Answer:
6.4 g BaSO₄
Explanation:
You have been given the molarity and the volume of the solution. To find the mass of the solution, you need to (1) find the moles BaSO₄ (via the molarity ratio) and then (2) convert moles BaSO₄ to grams BaSO₄ (via the molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs to reflect the sig figs of the given values.
Molarity (mol/L) = moles / volume (L)
(Step 1)
55 mL / 1,000 = 0.055 L
Molarity = moles / volume <----- Molarity ratio
0.5 (mol/L) = moles / 0.055 L <----- Insert values
0.0275 = moles <----- Multiply both sides by 0.055
(Step 2)
Molar Mass (BaSO₄): 137.33 g/mol + 32.065 g/mol + 4(15.998 g/mol)
Molar Mass (BaSO₄): 233.387 g/mol
0.0275 moles BaSO₄ 233.387 g
--------------------------------- x ------------------- = 6.4 g BaSO₄
1 mole
a. volume of NO : 41.785 L
b. mass of H2O : 18 g
c. volume of O2 : 9.52 L
<h3>Further explanation</h3>
Given
Reaction
4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)
Required
a. volume of NO
b. mass of H2O
c. volume of O2
Solution
Assume reactants at STP(0 C, 1 atm)
Products at 1000 C (1273 K)and 1 atm
a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :
volume NO at 1273 K and 1 atm
b. 15 L NH3 at STP ( 1mol = 22.4 L)
mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :
mass H2O(MW = 18 g/mol) :
c. mol NO at 1273 K and 1 atm :
mol ratio of NO : O2 = 4 : 5, so mol O2 :
Volume O2 at STP :
the answer is A. I (iodine)
no worries
