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3 years ago

5

How many grams of NH4ClO4 would dissolve in 250 mL at 30 C

1 answer:

baherus [9]3 years ago

8 0

Ammonium perchlorate appears as a white, crystalline solid or powder. Classified as a division 1.1 explosive if powdered into particles smaller than 15 microns in diameter or if powdered into larger particles but thoroughly dried. Does not readily burn, but will burn if contaminated by combustible material. May explode under prolonged exposure to heat or fire. Used to make rocket propellants, explosives, pyrotechnics, as an etching and engraving agent, and in analytical chemistry.

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Balance the following skeleton ionic reaction, and calculate E°cell to decide whether the reaction is spontaneous: Ag(s) + Cu2+

tia_tia [17]

Answer:

2Ag(s) + Cu^2+(aq) ----------> 2Ag^+(aq) + Cu(s)

Explanation:

Ag(s)/Ag^+ (aq) is the anode as shown while Cu^2+(aq)/Cu^2(s) is the cathode.

E°cell= E°cathode -E°anode= 0.34 -0.80= -0.5V

The cell is not spontaneous as written because E°cell is negative. This implies that the electrodes of the cell must be interchanged to make the cell spontaneous.

5 0

3 years ago

Read 2 more answers

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

3 years ago

I need help ASAP

vova2212 [387]

Answer:

1.4 g/cm3

Explanation:

Density = Mass/Volume

Mass = 21g

Volume = 15cm3

Density = 21/15 = 1.4

8 0

2 years ago

True or false transuranium elements are synthetic elements with atomic #’s of 93 or higher.

devlian [24]

It's absolutely TRUE...........

8 0

3 years ago

If a horse is moving and then come to a stop is the horse accelerating or decelerating

quester [9]

Answer:

decelerating hope this helps❤️

7 0

3 years ago

Read 2 more answers