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sladkih [1.3K]
3 years ago
13

Name the reactants and products of Na3PO4 + 3KOHàNaOH + K3PO4 (Question 2 and 3)

Chemistry
1 answer:
Leona [35]3 years ago
6 0

Answer:

Question 2: Na3PO4, KOH; Question 3: Na3PO4, KOH

Explanation:

Question 2

The reactants in a chemical equation are the species on the left side of the reaction arrow.

Thus the reactants are Na3PO4, KOH (sodium phosphate and potassium hydroxide).

Question 3.

The products in a chemical equation are the species on the right side of the reaction arrow.

Thus the products are NaOH, K3PO4 (sodium hydroxide and potassium phosphate).

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The rate constant for this first‑order reaction is
jenyasd209 [6]

Answer:

t = 5.7634 s

Explanation:

  • A → Pdts
  • - rA = K (CA)∧α = - δCA/δt

∴ T = 400°C

∴ α = 1 ....first-order

∴ CAo = 0.950 M

∴ CA = 0.300 M

⇒ t = ?

⇒ - δCA/δt = K*CA

⇒ - ∫δCA/CA = K*∫δt

⇒ Ln (CAo/CA) = K*t

⇒ t = Ln(CAo/CA) / K

⇒ t = (Ln(0.950/0.300)) / (0.200 s-1)

⇒ t = 1.1527 / 0.200 s-1

⇒ t = 5.7634 s

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Neon doesn't take part in chemical reaction why give reason​
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One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 43.3 mL of toluene (d = 0.867
Thepotemich [5.8K]

<u>Answer:</u>

(A)

Density = Mass / Volume

So  

Mass = Density × Volume

= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene

1C_6 H_5 CH_3  + 9 O_2  > 7 CO_2  + 4 H_2 O

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of CO_2 gas and 4 moles of H_2 O Vapour

So the mole ratio is 1 : 11

37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $

(C)

1mole contains 6.022\times10^{23} molecules

37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $

6 0
4 years ago
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