What is the name of this hydrocarbon? 2-butane 2-butene 2-butyne 4-butyne 4-propyne

Plz help :(

ziro4ka [17]

<h2>Answer:0.5 moles,1 mole</h2>

Explanation:

$Sr(NO_{3})_{2}$ is readily soluble in water.

$Sr(NO_{3})_{2}$ dissolves in water to give $Sr^{2+}$ ions and $NO_{3}^{-}$ ions.

The reaction is

$Sr(NO_{3})_{2}$ → $Sr^{2+}$ $+2NO_{3}^{-}$

So, $1$ mole of $Sr(NO_{3})_{2}$ gives $1$ mole of $Sr^{2+}$ and $2$ moles of $NO_{3}^{-}$ .

So, $0.5$ moles of $Sr(NO_{3})_{2}$ gives $0.5$ moles of $Sr^{2+}$ and $1$ mole of $NO_{3}^{-}$ .

5 0

3 years ago

Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result

Murljashka [212]

Answer:

(a) 2.7

(b) 4.44

(c) 4.886

(d) 5.363

(e) 5.570

(f) 12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

(c) = mol HA reacted = 0.100 L x 0.14 mol/L = 0.014 mol

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) = 4.886

(d) mol HA reacted = 150 x 10⁻³ L x x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) = 5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH = √(kb x (A⁻) where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA, 200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒ Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

4 0

2 years ago

Hwlp meeeeeeeeeeeeeeeeeeee

Vanyuwa [196]

The most appropriate answer is C !!

4 0

3 years ago

How do atoms form a new substance?

lubasha [3.4K]

By sharing electeons with each other
if they lose or gain electrons then they only form ions
they cannot lose neutrons as they are locked inside the nucleus

8 0

3 years ago

How does the comic strip define a polar molecule

zmey [24]

The answer is: A molecule with a difference in electrical charge between two ends.

Electronegativity (χ) is a property that describes the tendency of an atom to attract a shared pair of electrons.

Atoms with higher electronegativity attracts more electrons towards it, electrons are closer to that atom.

For example fluorine has electronegativity approximately χ = 4 and oxygen χ = 3,5, fluorine attracts electron and he has negative charge and oxygen has positive charge.

4 0

3 years ago