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3 years ago

how was it possible that Mendeleev was able to predict the properties of elements that no one knew about

1 answer:

7 0

<span>He used the patterns he found in properties of known elements to predict the properties of unknown elements.</span>

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How many grams of H2 will be produced, given 27.4 g of Al to start with?

pishuonlain [190]

3.07g H2

27.4/26.98/2x3x1.01x2=3.07

6 0

3 years ago

Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolv

Triss [41]

Answer:

1. $3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

2. $5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

Explanation:

The $pH$ value of Aspirin solution = 2.62

$pH=-\log[H^+]$

$[H^+]=10^{-2.62}=0.00240 M$

Moles of s asprin = $\frac{2.00 g}{180 g/mol}=0.01111 mol$

Volume of the solution = 0.600 L

The initial concentration of Aspirin = c = $\frac{0.01111 mol}{0.600 L}=0.0185 M$

$HAs\rightleftharpoons As^-+H^+$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_a=\frac{[As^-][H^+]}{[HAs]}$ :

$K_a=\frac{x\times x }{(c-x)}$

$=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}$

$K_a=3.57\times 10^{-4}$

$3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

The $pH$ value of ethylamine = 11.87

$pH+pOH=14$

$pOH=14-11.87=2.13$

$pOH=-\log[OH^-]$

$[OH^-]=10^{-2.13}=0.00741 M$

The initial concentration of ethylamine = c = 0.100 M

$C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}$ :

$K_b=\frac{x\times x}{(c-x)}$

$=\frac{0.00741\times 0.00741}{(0.100-0.00741)}$

$K_b=5.93\times 10^{-4}$

$5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

3 0

3 years ago

Does Kinetic energy depend only on an object’s speed and velocity?

BaLLatris [955]

yes .. . . . .. . . .. ....... . ..dlnv3r;'mw,c kc;oqc,

xkdnnd

4 0

3 years ago

Read 2 more answers

9. At equilibrium a 2 L vessel contains 0.360

klio [65]

Answer:

Ke = 34570.707

Explanation:

H2(g) + Br2(g) → 2 HBr(g)

equilibrium constant (Ke):

⇒ Ke = [HBr]² / [Br2] [H2]

∴ [HBr] = (37.0 mol) / (2 L) = 18.5 mol/L

∴ [Br2] = (0.110 mol) / (2 L) = 0.055 mol/L

∴ [H2] = (0.360 mol) / (2 L) = 0.18 mol/L

⇒ Ke = (18.5 mol/L)² / (0.055 mol/L)(0.18 mol/L)

⇒ Ke = 34570.707

3 0

3 years ago

How many moles of FeCl3 could be produced from 6.1 moles of Cly?<br> 2 Fe + 3Cl2 —> 2 FeCl2

Black_prince [1.1K]

Answer:

4.1 moles of FeCl₃

Explanation:

The reaction expression is given as shown below:

2Fe + 3Cl₂ → 2FeCl₃

Number of moles of Cl₂ = 6.1moles

So;

We know that from the balanced reaction expression:

3 moles of Cl₂ will produce 2 moles of FeCl₃

Therefore 6.1moles of Cl₂ will produce $\frac{6.1 x 2}{3}$ = 4.1 moles of FeCl₃

The number of moles is 4.1 moles of FeCl₃

6 0

2 years ago