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3 years ago

Which ionic compound is the most likely to dissolve in water?

Chemistry

2 answers:

harkovskaia [24]3 years ago

5 0

AC03 I am pretty sure

timurjin [86]3 years ago

5 0

Answer:

BaSO₄

Explanation:

Hello,

In this case, one could determine the compound most likely to be dissolved in water by locating the main nonmetal in the periodic table, thus, the farthest one from barium will lead to the formation of the most polar or ionic bond, which will be the most soluble, in such a way, since sulfur is farther than both carbon and phosphorous, BaSO₄ will be the compound most likely to be dissolved in water.

Best regards.

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Imagine two solutions with the same concentration and the same boiling point, but one has benzene as the solvent and the other h

aleksandrvk [35]

Answer:

Concentration = 1.32 m

Boiling point = $83.44^{0}C$

Explanation:

Let the boiling point of two solutions = T

Let the molal concentration of the two solutions= m

The relation between elevation in boiling point and molal concentration is:

$elevationinboiling point=K_{b}Xmolality$

i) For benzene

$T-80.1=2.53Xm$

$T=2.53m+80.1$ ....(1)

ii) For carbon tetrachloride

$T-76.8=5.03Xm$

$T=5.03Xm+76.8$ ...(2)

Equating the two equations:

$5.03Xm+76.8=2.53m+80.1\\m=1.32molal$

Putting value of "m" in equation "1"

$T=2.53(1.32)+80.1=83.44^{0}C$

5 0

3 years ago

A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO 3 in a coffee cup calorimeter. If both solution

uysha [10]

Answer:

THE STANDARD HEAT OF NEUTRALIZATION OF THE BASE SODIUM HYDROXIDE BY THE ACID HYDROGEN TRIOXONITRATE V ACID IS -56 kJ / mol.

Explanation:

Volume of 0.3 M NaOh = 100 mL

Volume of 0.3 M HNO3 = 100 mL

Initail temp of NaOH and HNO3 = 35 °C = 35 + 273 K = 308 K

Final temp. of mixture = 37 °C = 37 + 273 K = 310 K

We can make the following assumptions form the question given:

1. specific heat of the reaction mixture is the same as the specific heat of water = 4.2 J/g K

2. the toal mass of the reaction mixture is 200 mL = 200 g since no heat is lost to the calorimeter or surrounding.

3. initail temperature of the reaction mixture is equal to the average temperature of the two reactant solutions

= ( 308 + 308 /2) = 308 K

4. Rise in temeperature for the reaction = 310 -308 K = 2 K

Then the total heat evolved during the reaction = mass * specifc heat capacity * temperature change

Heat = 200 g * 4.2 J/g K * 2 K

Heat = 1680 J

EQUATION FOR THE REACTION

HNO3 + NaOH -------> NaNO3 + H20

From the equation, 1 mole of HNO3 reacts with 1 mole of NaOH to prouce mole of water.

100 mL of 0.5 M HNO3 contains 100 * 0.3 /1000 = 0.03 mole of acid

This result is same for the base NaOH = 0.03 mole of base

So therefore,

0.03 mole of acid will react with 0.03 mole of base to produce 0.03 mole of water to evolved 1680 J of heat energy.

The production of 1 mole of water will evolve 1680 / 0.03 J of heat

= 56 000 J or 56 kJ of heat energy per mole of water.

So therefore, 1the standard heat of neutralization of sodium hydroxide by trioxoxnitrate V acid is -56 kJ/mol.

5 0

3 years ago

Which statement best explains why atoms form chemical bonds with other atoms?

dedylja [7]

Sodium loses an electron and chlorine gains an electron. Simply put. It's a little more complicated, but that's what i think they're looking for.

6 0

4 years ago

A sample of nitrogen gas, stored in a 2.96-L container at 32.0°C, exerts a pressure of 4.27 atm. Calculate the number of moles o

inysia [295]

Answer: The number of moles of nitrogen gas is 0.505 moles.

Explanation:

To calculate the number of moles of nitrogen gas, we use ideal gas equation, which is:

$PV=nRT$

where,

P = pressure of the gas = 4.27 atm

V = Volume of the gas = 2.96 L

T = Temperature of the gas = $32.0^oC=[32.0+273]K=305K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of gas = ?

Putting values in above equation, we get:

$4.27atm\times 2.96L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 305K\\n=0.505mol$

Hence, the number of moles of nitrogen gas is 0.505 moles.

6 0

4 years ago

Read 2 more answers

Calculate the work, w, gained or lost by the system when a gas expands from 15 L to 45 L against a constant external pressure of

marta [7]

Answer:

Work done by the system = 4545 J

Explanation:

The expression for the calculation of work done is shown below as:

$w=-P\times \Delta V$

Where, P is the pressure

$\Delta V$ is the change in volume

From the question,

$\Delta V$ = 45 - 15 L = 30 L

P = 1.5 atm

$w=-1.5\times 30\ atmL$

Also, 1 atmL = 101 J

So,

$w=-1.0\times0.75\times 101\ J=-4545\ J$ (negative sign implies work is done by the system)

Work done by the system = 4545 J

6 0

3 years ago