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4 years ago

What type of solution forms when lithium chloride dissolved in water?

1 answer:

6 0

Answer: strong electrolyte.

Justification:

1) Lithium chloride is an ionic compound, highly soluble in water.

2) This is the chemical equation:

LiCl(aq) → Li⁺ (aq) + Cl⁻ (aq)

3) Due to its ionic character and the small size of the ion Li⁺, LiCl dissociates almost completely, meaning the the solution is a strong electrolyte.

4) Electrolytes are solutions that carry electricity (the iones, charged species, are the carrierr of the electricity)

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In general, non-spontaneous reactions __________.

Rom4ik [11]

Very low gives free energy

3 0

3 years ago

What is the formula mass for sodium chloride?

Zigmanuir [339]

Atomic mass of sodium is 22.9 g/mol
atomic mass of chloride is 35.45 g/mol
sodium chloride is 22.99 + 35.45= 58.44 g/mol

3 0

3 years ago

Which phrases describe sedimentary rock? check all that apply.

yuradex [85]

Answer:

Sedimentary rock, rock formed at or near Earth's surface by the accumulation and lithification of sediment by the precipitation from solution at normal surface temperatures

Explanation:

7 0

3 years ago

ΔG o for the reaction H2(g) + I2(g) ⇌ 2HI(g) is 2.60 kJ/mol at 25°C. Calculate ΔG o , and predict the direction in which the rea

kondaur [170]

Answer:

The reaction is not spontaneous in the forward direction, but in the reverse direction.

Explanation:

Step 1: Data given

H2(g) + I2(g) ⇌ 2HI(g) ΔG° = 2.60 kJ/mol

Temperature = 25°C = 25+273 = 298 Kelvin

The initial pressures are:

pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

Step 2: Calculate ΔG

ΔG = ΔG° + RTln Q

with ΔG° = 2.60 kJ/mol

with R = 8.3145 J/K*mol

with T = 298 Kelvin

Q = the reaction quotient → has the same expression as equilibrium constant → in this case Kp = [p(HI)]²/ [p(H2)] [p(I2)]

with pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

Q = (3.10²)/(1.5*1.75)

Q = 3.661

ΔG = ΔG° + RTln Q

ΔG = 2600 J/mol + 8.3145 J/K*mol * 298 K * ln(3.661)

ΔG =5815.43 J/mol = 5.815 kJ/mol

To be spontaneous, ΔG should be <0.

ΔG >>0 so the reaction is not spontaneous in the forward direction, but in the reverse direction.

4 0

4 years ago

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

5 0

3 years ago