To be able to answer this equations, we must set given information. Suppose the reaction to yield NO is:
N₂ + O₂ → 2 NO
Next, suppose you have 1 g of each of the reactants. Determine first which is the limiting reactant.
1 g N₂ (1 mol N₂/ 28 g)(2 mol NO/1 mol N₂)= 0.07154 mol NO present
Number of molecules = 0.07154 mol NO(6.022×10²³ molecules/mol)
<em>Number of molecules = 4.3×10²² molecules NO present</em>
Answer:
(1)=(A), (2)=(B), (3)=D, (4)=C, (5)=E, (6)=F
Explanation:
(1) Glassware used to accurately transfer small volumes = (A) Graduated pipette, that is basically a glass tube with graduation of different volumes to be dispensed.
(2) Glassware used to accurately transfer a small, single volume = (B) Volumetric pipette, that is a glass tube with a central glass bulb and is used to dispense accurately an unique volume of liquid everytime.
(3) Glassware to deliver a volume not known in advance = (D) Buret (or burette), that is used to dispense slowly a volume of liquid when a titration process is needed
(4) Glassware best used when greater access to the contents is needed = (C) Beaker, that is basically a very open glass cylinder with a spout
(5) Glassware used to prevent splashing or evaporation = (E) Erlenmeyer flask, that has a small open at the top and is useful when the liquid needs to be swirled as, for example, during a titration.
(6) Glassware used to make accurate solutions = (F) Volumetric flask, that has a long slim neck that provides a higher accuracy when a exact volume of liquid needs to be used for preparation of a solution.
Answer:
k= 1.925×10^-4 s^-1
1.2 ×10^20 atoms/s
Explanation:
From the information provided;
t1/2=Half life= 1.00 hour or 3600 seconds
Then;
t1/2= 0.693/k
Where k= rate constant
k= 0.693/t1/2 = 0.693/3600
k= 1.925×10^-4 s^-1
Since 1 mole of the nuclide contains 6.02×10^23 atoms
Rate of decay= rate constant × number of atoms
Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms
Rate of decay= 1.2 ×10^20 atoms/s
D is the answer, I believe.
An isolated system is one that allows neither heat or matter to enter or exit, and since the liquid remained the same temperature, one can conclude that neither energy nor matter is passing through.
Answer:
The answers are explained below
Explanation:
a)
Given: concentration of salt/base = 0.031
concentration of acid = 0.050
we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59
b)
we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O
Moles i............0.05...................0.01.................0.031.....................0
Moles r...........-0.01.................-0.01................0.01........................0.01
moles f...........0.04....................0....................0.041.....................0.01
c)
we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041
Hence, we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71
d)
pOH = -log (0.01/0.510) = 1.71
pH = 14 - 1.71 = 12.29
e)
Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.
in water. Used in salad dressing and while cooking food. It has a sour taste.
) is used in baking food like: cakes, cookies, breads. Baking soda is one of the ingredient while baking breads and cakes.