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shusha [124]
3 years ago
10

According to the periodic table, the neutral atom of what element has 4 protons,5 neutrons,and 4 electrons

Chemistry
1 answer:
Kaylis [27]3 years ago
7 0

Since it has a 4 protons and 5 neutrons, it has a mass number of 9. The number of electrons is irrelevant for this question. Atomic number is the number of protons, so it is 4. So it is Beryllium

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Aqueous potassium sulfide and aqueous cobalt (II) chloride are mixed, and a double replacement reaction occurs. What is the corr
alekssr [168]
K2S (aq) + CoCl2( aq) ----->   2KCl (aq) + CoS (s)
potassium   +  cobalt                potassium chloride  + carbonyl  sulfide 
sulfide          chloride 

carbonyl sulfide :-  it is chemical compound with linear formula (OCS ) normally written as (CoS)  .it does not show its structure . its is colorless flammable gas with an unpleasant odour.
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7 0
3 years ago
Select the correct answer.
frozen [14]

Answer:

1.79x10^-14

Explanation:

pOH + pH = 14

H+=10^-pH

- Hope that helps! Please let me know if you need further explanation.

3 0
3 years ago
1.2 L sample of gas is determined to contain 0.5 moles of nitrogen. At the same temperature and pressure, how many moles of gas
hram777 [196]

A 20 L sample of the gas contains 8.3 mol N₂.

According to <em>Avogadro’s Law,</em> if <em>p</em> and <em>T</em> are constant

<em>V</em>₂/<em>V</em>₁ = <em>n</em>₂/<em>n</em>₁

<em>n</em>₂ = <em>n</em>₁ × <em>V</em>₂/<em>V</em>₁

___________

<em>n</em>₁ = 0.5 mol; <em>V</em>₁ = 1.2 L

<em>n</em>₂ = ?;           <em>V</em>₂ = 20 L

∴<em>n</em>₂ = 0.5 mol × (20 L/1.2 L) = 8.3 mol

4 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
3 years ago
What product of an acid base reaction is an ionic compound apex?
charle [14.2K]
When acids react with bases they produce salt and water such as:
HCl + NaOH → NaCl + H₂O
According to strength of acid and base, we have 4 types of salts:
salt of strong acid and strong base like: NaCl
salt of weak acid and strong base like: CH₃COONa
salt of strong acid and weak base like: NH₄Cl
salt of weak acid and weak base like: CH₃COONH₄
8 0
3 years ago
Read 2 more answers
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