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Lera25 [3.4K]
3 years ago
13

Which polyatomic ion forms A neutral compound when combined with group a1 monotonic ion in a 1:1

Chemistry
1 answer:
kolezko [41]3 years ago
6 0

Answer:

"Nitrate" is a polyatomic ion forms A neutral compound when combined with group a1 monotonic ion in a 1:1

Explanation:

Nitrate is a polyatomic ion with the molecular formula NO⁻ ₃. Organic complexes that comprise the nitrate ester as a functional group are also named nitrates. Nitrates are shared apparatuses of fertilizers and explosives. Almost all nitrate salts are soluble in water. Usually nitrates that enter the body by eating or drinking leave the body without harm. These nitrites in the blood cause changes in hemoglobin, or the molecules that help move oxygen in the body. Nitrates can make it so that less oxygen is available for the body to function properly.

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Explanation:

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A certain compound is made up of one phosphorus (P) atom, three chlorine (Cl) atoms, and one oxygen (O) atom. What is the chemic
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3 years ago
A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base
yawa3891 [41]

Answer:

11.39

Explanation:

Given that:

pK_{b}=4.82

K_{b}=10^{-4.82}=1.5136\times 10^{-5}

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{1.805\ g}{82.0343\ g/mol}

Moles= 0.022\ moles

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.022}{0.055}

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

                                  B +   H₂O    ⇄     BH⁺ +        OH⁻

At t=0                        0.4                          -              -

At t =equilibrium     (0.4-x)                        x           x            

The expression for dissociation constant is:

K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}

1.5136\times 10^{-5}=\frac {x^2}{0.4-x}

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³  M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>

5 0
2 years ago
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