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Alona [7]
3 years ago
5

A 0.75M solution of CH3OH is prepared in 0.500 kg of water. How many moles of CH3OH are needed?

Chemistry
1 answer:
4vir4ik [10]3 years ago
5 0

Answer:

We need 0.375 mol of CH3OH to prepare the solution

Explanation:

For the problem they give us the following data:

Solution concentration 0,75 M

Mass of Solvent is 0,5Kg

knowing that the density of water is 1g / mL,  we find the volume of water:

                           d = \frac{g}{mL} \\\\ V= \frac{g}{d}  = \frac{500g}{1 \frac{g}{mL} } = 500mL = 0,5 L

Now, find moles of CH_{3} OH are needed using the molarity equation:

                           M = \frac{ moles }{ V (L)} \\\\\\molesCH_{3}OH  = M . V(L) = 0,75 M . 0,5 L\\\\molesCH_{3}OH = 0,375 mol

therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH,  resulting in a concentration of 0,75M

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The standard free energy change for a reaction in an electrolytic cell is always:_________
sattari [20]

Answer: The standard free energy change for a reaction in an electrolytic cell is always positive.

Explanation:

Electrolytic cells use electric currents to drive a non-spontaneous reaction forward.

Relation of standard free energy change and emf of cell

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where,

\Delta G^o = standard free energy change

  n= no of electrons gained or lost

F= faraday's constant

E^0_{cell} = standard emf

E^0_{cell}   = standard emf = -ve  , for non spontaneous reaction

Thus  \Delta G^o=(-ve)(-ve)=+ve

Thus standard free energy change for a reaction in an electrolytic cell is always positive.

8 0
3 years ago
Identify the correct coefficients to balance the redox reaction with the lowest possible integer coefficients.
Monica [59]

Answer:

\rm 3\; Ag^{1+} + 1\; Al \to 1\; Al^{3+} + 3\; Ag.

Explanation:

Electrons are conserved in a chemical equation.

The superscript of \rm Ag^{1+} indicates that each of these ions carries a charge of +1. That corresponds to the shortage of one electron for each \rm Ag^{+} ion.

Similarly, the superscript +3 on each \rm Al^{3+} ion indicates a shortage of three electrons per such ion.

Assume that the coefficient of \rm Ag^{+} (among the reactants) is x, and that the coefficient of \rm Al^{3+} (among the reactants) is y.

\rm \mathnormal{x}\; Ag^{1+} + ?\; Al \to \mathnormal{y}\; Al^{3+} + ?\; Ag.

There would thus be x silver (\rm Ag) atoms and y aluminum (\rm Al) atoms on either side of the equation. Hence, the coefficient for \rm Al\! and \rm Ag\! would be y\! and x\!, respectively.

\rm \mathnormal{x}\; Ag^{1+} + \mathnormal{y}\; Al \to \mathnormal{y}\; Al^{3+} + \mathnormal{x}\; Ag.

The x \rm Ag^{1+} ions on the left-hand side of the equation would correspond to the shortage of x electrons. On the other hand, the y Al^{3+} ions on the right-hand side of this equation would correspond to the shortage of 3\, y electrons.

Just like atoms, electrons are also conserved in a chemical reaction. Therefore, if the left-hand side has a shortage of x electrons, the right-hand side should also be x\! electrons short of being neutral. On the other hand, it is already shown that the right-hand side would have a shortage of 3\, y electrons. These two expressions should have the same value. Therefore, x = 3\, y.

The smallest integer x and y that could satisfy this relation are x = 3 and y = 1. The equation becomes:

\rm 3\; Ag^{1+} + 1\; Al \to 1\; Al^{3+} + 3\; Ag.

6 0
3 years ago
What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 2.0 atm and the temperatu
tatyana61 [14]

Answer:

1) 6.0 atm.

2) 2.066 atm.

Explanation:

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<em>PV = nRT.</em>

where, P is the pressure of the gas.

V is the volume of the container.

n is the no. of moles of the gas.

R is the general gas constant.

T is the temperature of the gas (K).

<em>1) What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 2.0 atm and the temperature is held constant?</em>

  • At constant T and at two different (P, and V):

<em>P₁V₁ = P₂V₂.</em>

P₁ = 2.0 atm, V₁ = 150.0 mL.

P₂ = ??? atm, V₂ = 50.0 mL.

<em>∴ P₂ = P₁V₁/V₂</em> = (2.0 atm)(150.0 mL)/(50.0 mL) = <em>6.0 atm.</em>

<em>2. A sample of a gas in a rigid container at 30.0°C and 2.00 atm has its temperature increased to 40.0°C. What will be the new pressure?</em>

<em></em>

  • Since the container is rigid, so it has constant V.
  • At constant V and at two different (P, and T):

<em>P₁/T₁ = P₂/T₂.</em>

P₁ = 2.0 atm, T₁ = 30.0°C + 273 = 303 K.

P₂ = ??? atm, T₂ = 40.0°C + 273 = 313 K.

<em>∴ P₂ = P₁T₂/T₁ </em>= (2.0 atm)(313.0 K)/(303.0 K) =<em> 2.066 atm.</em>

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4 years ago
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