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3 years ago

Which of the following salts is the least soluble in water?

2 answers:

7 0

A. Strontium Phosphate

Solubility product constant is an equilibrium constant for the maximum amount of solute that can dissolve to form an aqueous solution. The value of the constant describes a solution which is saturated.

The greater the solubility product constant, the more soluble a solute is in the liquid. Consequently, the smaller the constant, the less soluble the solute is. The following list shows the solubility product constants of the given compounds in aqueous solutions at 298K in decreasing order:

Thallium Bromide - 3.71×10–6
Copper Iodide - 1.27×10<span>–12
</span>Silver Bromide - 5.35×10–13
Silver Arsenate - 1.03×10–22
Mercury Bromide - 6.40×10–23
Strontium Phosphate - 1×10–31
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The salt with the smallest solubility product constant is strontium phosphate, and therefore is the one which is least soluble in water.

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Mariana [72]3 years ago

4 0

Answer: I think it is Strontium Phosphate

Explanation:

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Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
$\text{NH}_3\; (g)$ : $-2 \; b = -1$
$\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

3 0

3 years ago

Combustion analysis of 0.300 g of an unknown compound containing carbon, hydrogen, and oxygen produced 0.5213 g of co2 and 0.283

Lyrx [107]

First, we have to get how many grams of C & H & O in the compound:
- the mass of C on CO2 = mass of CO2*molar mass of C /molar mass of CO2
= 0.5213 * 12 / 44 = 0.142 g
- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O
=0.2835 * 2 / 18 = 0.0315 g
- the mass of O = the total mass - the mass of C atom - the mass of H atom
= 0.3 - 0.142 - 0.0315 = 0.1265 g
Convert the mass to mole by divided by molar mass
C(0.142/12) H(0.0315/2) O(0.1265/16)
C(0.0118) H(0.01575) O(0.0079) by dividing by the smallest value 0.0079
C1.504 H3.99 O1 by rounding to the nearst fraction
C3/2 H4/1 )1/1 multiply by 2
∴ the emprical formula C3H8O2

6 0

3 years ago

PLEASE ANSWER I WILL MAKE BRAINLIEST!!!

tangare [24]

Using an example like Christmas lights, I would say yes because normally a lot of them would go out if one light is broken.

3 0

2 years ago

Read 2 more answers

Someone help me with chemistry on my account/page please :))

Gemiola [76]

Answer:

Explanation:

5 0

2 years ago

Formed when an amine is combined with a carboxyl group

alekssr [168]

Hello.

The answer is C.Amine

When an amine is combined (reacted) with a carboxyl group, an AMIDE + water is formed, and if you carry on heating under a vacuum, an imidazoline is formed.

Have a nice day

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3 years ago