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3 years ago

What net ionic equation describes the reaction when these solutions are mixed?

Chemistry

2 answers:

Aleksandr-060686 [28]3 years ago

8 0

Answer:

$2PO_{4}^{-3} (aq) + 3Ca^{+2}(aq)-->Ca_{3}(PO_{4})_{2}(s)$

Explanation:

In net ionic equation we remove the spectator ions. The ions which are present on both the side and are not forming any solid compound.

Let us write the balanced reaction first:

$2Na_{3}PO_{4}(aq)+3CaCl_{2}(aq)-->3Ca_{3}(PO_{4})_{2}+6NaCl(aq)$

The ionic reaction is:

$6Na^{+}(aq)++6Cl^{-}(aq)+2PO_{4}^{-3} (aq) + 3Ca^{+2}(aq)-->Ca_{3}(PO_{4})_{2}(s)+6Na^{+}(aq)+6Cl^{-}(aq)$

Thus net ionic reaction is

$2PO_{4}^{-3} (aq) + 3Ca^{+2}(aq)-->Ca_{3}(PO_{4})_{2}(s)$

almond37 [142]3 years ago

7 0

I think the correct answer from the choices listed above is option D. The net ionic equation that will describe the reaction of the reactants when mixed is expressed as:

<span>2PO43–(aq) + 3Ca2+(aq) → Ca3(PO4)2(s)
</span>
Hope this answers the question. Have a nice day.

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Wave length measures the width of the wave

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2 years ago

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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

If a gas occupies 79.5 mL at -1.4°C, what temperature, in Kelvin, would it

Anna35 [415]

Answer:

121 K

Explanation:

Step 1: Given data

Initial volume (V₁): 79.5 mL

Initial temperature (T₁): -1.4°C

Final volume (V₂): 35.3 mL

Step 2: Convert "-1.4°C" to Kelvin

We will use the following expression.

K = °C + 273.15 = -1.4°C + 273.15 = 271.8 K

Step 3: Calculate the final temperature of the gas (T₂)

Assuming ideal behavior and constant pressure, we can calculate the final temperature of the gas using Charles' law.

V₁/T₁ = V₂/T₂

T₂ = V₂ × T₁/V₁

T₂ = 35.3 mL × 271.8 K/79.5 mL = 121 K

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3 years ago

if i add 25 ml of water to 135 ml of a 0.25 M NaOH solution what will the molarity of the diluted solution be

DENIUS [597]

Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

$n = c \cdot V$ ,

where

$n$ is the number of moles of the solute in the solution.
$c$ is the concentration of the solution. $c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1}$ for the initial solution.
$V$ is the volume of the solution. For the initial solution, $V = 135\;\textbf{mL} = 0.135\;\textbf{L}$ for the initial solution.

$n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}$ .

What's the concentration of the diluted solution?

$\displaystyle c = \frac{n}{V}$ .

$n$ is the number of solute in the solution. Diluting the solution does not influence the value of $n$ . $n = 0.03375\;\text{mol}$ for the diluted solution.
Volume of the diluted solution: $25\;\text{mL} + 135\;\text{mL} = 160\;\textbf{mL} = 0.160\;\textbf{L}$ .

Concentration of the diluted solution:

$\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}$ .

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

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3 years ago