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3 years ago

How much potassium chloride is needed to make 0.500 m solution with 1.50 L of water?

Chemistry

1 answer:

Andrew [12]3 years ago

4 0

Answer:

55.9 g KCl.

Explanation:

Hello there!

In this case, according to the definition of molality for the 0.500-molar solution, we need to divide the moles of solute (potassium chloride) over the kilograms of solvent as shown below:

$m=\frac{mol}{kilograms}$

Thus, solving for the moles of solute, we obtain:

$mol=m*kilograms$

Since the density of water is 1 kg/L, we obtain the following moles:

$mol=0.500mol/kg*1.50kg\\\\mol=0.75mol$

Next, since the molar mass of KCl is 74.5513 g/mol, the mass would be:

$0.75mol*\frac{74.5513g}{1mol}\\\\55.9g \ KCl$

Regards!

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Who much the velocity of a body when it travels 600m in 5 min

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Answer:

2 m/s

Explanation:

Applying the formulae of velocity,

V = d/t............. Equation 1

Where V = Velocity of the body, d = distance, t = time

From the question,

Given: d = 600 m, t = 5 minutes = (5×60) = 300 seconds.

Substitute these values into equation 1

V = 600/300

V = 2 m/s.

Hence the velocity of the body when it travels is 2 m/s

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2 years ago

2-phosphoglycerate(2PG) is converted to phosphoenolpyruvate (PEP) by the enzyme enolase. The standard free energy change(deltaGo

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Answer:

The correct option is: (D) -2.4 kJ/mol

Explanation:

Chemical reaction involved: 2PG ↔ PEP

Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol

Temperature: T = 37° C = 37 + 273.15 = 310.15 K (∵ 0°C = 273.15K)

Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol) (∵ 1 kJ = 1000 J)

Reactant concentration: 2PG = 0.5 mM

Product concentration: PEP = 0.1 mM

Reaction quotient: $Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2$

To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:

$\Delta G = \Delta G^{\circ } + 2.303 R T log Q_{r}$

$\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)$

$\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)$

Therefore, the Gibb's free energy change at 37° C (310.15 K): ΔG = (-2.45 kJ/mol)

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3 years ago

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2 years ago

Read 2 more answers

A sample of oxygen gas is compressed from 30.6 L to 1.8 L at constant temperature pressure of 1.8 atm. Calculate the amount of e

ad-work [718]

Answer:

the change in the internal energy of the system is 3,752.67 J

Explanation:

Given;

initial volume of the gas, V₁ = 30.6 L

final volume of the gas, V₂ = 1.8 L

constant pressure of the gas, P = 1.8 atm

Energy released by the system, Q = 1.5 kJ = 1,500 J

Apply pressure-volume work equation, to determine the work done on the gas;

w = -PΔV

w = -P(V₂ - V₁)

w = - 1.8 atm(1.8 L - 30.6 L)

w = 51.84 L.atm

w = 51.84 L.atm x 101.325 J/L.atm

w = 5,252.67 J

The change in the internal energy of the system is calculated as;

ΔU = Q + w

Since the heat is given out, Q = - 1,500 J

ΔU = -1,500 J + 5,252.67 J

ΔU = 3,752.67 J

Therefore, the change in the internal energy of the system is 3,752.67 J

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2 years ago

A block of has a length of 7cm, a width of 4cm, and a height of 10cm. Its mass measures 840g. What is it's density?

andriy [413]

Answer:

Explanation:

Density is

mass / volume = d

Mass:

840g

Volume:

7 cm x 4 cm x 10 cm = 280 cm^3

840g / 280 cm^3 = 3 g/cm^3

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2 years ago