Please answer. Brainliest!

Which of the following solutions contains the largest number of moles of dissolved particles? a.) 25 ml of sodium chloride b.) 2

kirza4 [7]

Answer:

b.) 25 ml of aluminum hydroxide

Explanation:

For this question, we have to assume that we have the same concentration for all the solutions, for example, 1 M. Additionally, we have to take into account the ionization reaction for each species:

a) $NaCl~->~Na^+~+~Cl^-$ we have to ions

b) $Al(OH)_3~->~Al^+^3~+~3OH^-$ we have fourth ions

c) $CH_3COOH~->~CH_3COO^-~+~H^+$ we have two ions

d) $C_6H_1_2O_6~->~C_6H_1_2O_6$ we have one ion

If we have the same volume and the same concentration the variables that will help us to answer the question would be the number of ions. If we have more ions we will have more particles dissolved. Therefore the answer would be b) (due to the fourth ions).

I hope it helps

5 0

2 years ago

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A 25.00 mL sample of 0.320 M KOH is titrated with 0.750 M HBr at 25 °C.

Fudgin [204]

Answer:

a. pH = 13.50

b. pH = 13.15

Explanation:

Hello!

In this case, since the undergoing chemical reaction between KOH and HBr is:

$HBr+KOH\rightarrow KBr+H_2O$

As they are both strong. In such a way, since the initial analyte is the 25.00 mL solution of 0.320-M KOH, we first compute the pOH it has, considering that all the KOH is ionized in potassium and hydroxide ions:

$pOH=-log([OH^-])=-log(0.320)=0.50$

Thus, the pH is:

$pH+pOH=14\\pH=14-pOH=14-0.50\\pH=13.50$

Which is the same answer for a and b as they ask the same.

Moreover, once 5.00 mL of the HBr is added, we need to compute the reacting moles of each substance:

$n_{KOH}=0.02500mL*0.320mol/L=0.00800mol\\\\n_{HBr}=0.005L*0.750mol/L=0.00375mol$

It means that since there are more moles of KOH, we need to compute the remaining moles after those 0.00375 moles of acid consume 0.00375 moles of base because they are in a 1:1 mole ratio:

$n_{KOH}^{remaining}=0.00425mol$

Next, we compute the resulting concentration of hydroxide ions (equal to the concentration of KOH) in the final solution of 30.00 mL (25.00 mL + 5.00 mL):

$[OH^-]=[KOH]=\frac{0.00425mol}{0.03000L}=0.142M$

So the pOH and the pH turn out:

$pOH=-log(0.142)=0.849\\pH+pOH=14\\pH=14-pOH=14-0.849\\pH=13.15$

Best regards!

7 0

3 years ago

Is sand a element, compound or a mixture?

Ainat [17]

Answer:s its a mixture

Explanation:

3 0

2 years ago

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What is the pressure of 0.60 moles of a gas if its volume is 10.0 liters at 35.0°C

eduard

Answer: 1.52 atm

Explanation:

Given that:

Volume of gas V = 10.0L

Temperature T = 35.0°C

Convert Celsius to Kelvin

(35.0°C + 273 = 308K)

Pressure P = ?

Number of moles = 0.6 moles

Molar gas constant R is a constant with a value of 0.0821 atm L K-1 mol-1

Then, apply ideal gas equation

pV = nRT

p x 10.0L = 0.6 moles x (0.0821 atm L K-1 mol-1 x 308K)

p x 10.0L = 15.17 atm L

p = 15.17 atm L / 10.0L

p = 1.517 atm (round to the nearest hundredth as 1.52 atm)

Thus, the pressure of the gas is 1.52 atm

3 0

2 years ago

In an experiment, a student was assigned to find the formula of an unknown hydrate that weighed 15.67g. The anhydrate, XCO3 (mol

meriva

Answer:

5.0 moles of water per one mole of anhydrate

Explanation:

To solve this question we must find the moles of the anhydrate. The difference in mass between the dry and the anhydrate gives the mass of water. Thus, we can find the moles of water and the moles of water per mole of anhydrate:

Moles Anhydrate:

7.58g * (1mol / 84.32g) = 0.0899 moles XCO3

Moles water:

15.67g - 7.58g = 8.09g * (1mol / 18.01g) = 0.449 moles H2O

Moles of water per mole of anhydrate:

0.449 moles H2O / 0.0899 moles XCO3 =

5.0 moles of water per one mole of anhydrate

6 0

2 years ago