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3 years ago

The plant food contains nh4)3po4 what tests would you run to verify the presence of the nh4 ion and the po4 ion

2 answers:

algol133 years ago

8 0

For the presence of ammonium ion, there is a need to add sodium hydroxide solution to the water and warm the mixture. Test any vapor that gets produced with damp red litmus paper. It should turn blue as ammonia gas is discharged, which is alkaline. The ionic equation for the reaction is:

NH₄⁺ + OH⁻ ⇒ NH₃ + H₂O

For the presence of phosphate ions, the addition of barium ions is done. The ionic equation is:

3Ba₂⁺ + 2PO4³⁻ ⇒ Ba₃ (PO₄)₂ (precipitate)

exis [7]3 years ago

7 0

<h2>Answer with Explanation </h2>

To Check The Presence Of NH4+ Ion, by making essential with NaOH, and testing the exhaust radiated with wet red litmus paper, the alkali radiated will change its shading to blue.

(you ought to likewise have the capacity to smell the alkali)

NH+ and OH - > H2O and NH3 (g)

To Check The Presence Of The PO43-Ion, by including AgNO3 and seeing the generation of the splendid yellow Ag3PO4 accelerate, which at that point will "disintegrate" when by making the arrangement acidic with abundance HNO3,

(AgI will likewise ppt yellow, yet does not break down when fermented)

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Alborosie

Answer:

the drops of liquid are coming from the decreases. they are formed as the motion of the water particles in the air gas. this change in motion cause air in the air to change from a liquid to a water

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3 years ago

Read 2 more answers

Can someone explain this to me, please?

zimovet [89]

Answer:

Isotopes have same atomic numbers, no. of protons and no. of electrons. Only their no. of neutrons and atomic mass are changed.

Atomic Mass = 24

Atomic No. = 11

Hence,

No. of protons in Na-24 = 11

No. of neutrons = Atomic Mass - Atomic Number

No. of neutrons = 24 - 11

No. of neutrons = 13

Atomic Number = 11

$\rule[225]{225}{2}$

Hope this helped!

<h3>~AH1807</h3><h3>Peace!</h3>

5 0

2 years ago

Given the following data:N2(g) + O2(g)→ 2NO(g), ΔH=+180.7kJ2NO(g) + O2(g)→ 2NO2(g), ΔH=−113.1kJ2N2O(g) → 2N2(g) + O2(g), ΔH=−163

statuscvo [17]

Answer:

ΔH = +155.6 kJ

Explanation:

The Hess' Law states that the enthalpy of the overall reaction is the sum of the enthalpy of the step reactions. To do the addition of the reaction, we first must reorganize them, to disappear with the intermediaries (substances that are not presented in the overall reaction).

If the reaction is inverted, the signal of the enthalpy changes, and if its multiplied by a constant, the enthalpy must be multiplied by the same constant. Thus:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

2NO(g) + O₂(g) → 2NO₂(g) ΔH = -113.1 kJ

2N₂O(g) → 2N₂(g) + O₂(g) ΔH = -163.2 kJ

The intermediares are N₂ and O₂, thus, reorganizing the reactions:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

NO₂(g) → NO(g) + (1/2)O₂(g) ΔH = +56.55 kJ (inverted and multiplied by 1/2)

N₂O(g) → N₂(g) + (1/2)O₂(g) ΔH = -81.6 kJ (multiplied by 1/2)

------------------------------------------------------------------------------------

N₂O(g) + NO₂(g) → 3NO(g)

ΔH = +180.7 + 56.55 - 81.6

ΔH = +155.6 kJ

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3 years ago

Recycling aluminum cans and taking shorter showers are two ways to practice the blank of natural resources

enyata [817]

Conservation, because you are conserving natural resources(water) and reusing aluminum cans which help the environment. Hope this helps!

3 0

3 years ago

Read 2 more answers

EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

$\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)$ .

The equilibrium constant for this reaction is called the self-ionization constant of water:

$K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]$ .

Note that water isn't part of this constant.

The value of $K_w$ at 25 °C is $10^{-14}$ . How to memorize this value?

The pH of pure water at 25 °C is 7.
$[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$
However, $[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$ for pure water.
As a result, $K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}$ at 25 °C.

Back to this question. $[\text{H}^{+}]$ is given. 25 °C implies that $K_w = 10^{-14}$ . As a result,

$\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}$ .

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3 years ago