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dalvyx [7]
3 years ago
13

Volume of HCl used 25.0mL 4 l

Chemistry
1 answer:
borishaifa [10]3 years ago
8 0

Answer:

1.0 M

Explanation:

Reaction equation;

KOH(aq) + HCl(aq) -----> KCl(aq) + H2O(l)

Concentration of acid CA = ?

Concentration of base CB = 1.0 M

Volume of base VB = 25.60 - 0.50 = 25.1 ml

Volume of acid VB =  25.0 ml

Number of moles of acid NA = 1

Number of moles of base NB =2

CAVA/CBVB =NA/NB

CAVANB = CBVBNA

CA = CBVBNA/VANB

CA = 1 * 25.1 * 1/25.0 *1

CA = 1.0 M

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myrzilka [38]

Answer:

About 67 grams or 67.39 grams

Explanation:

First you would have to remember a few things:

 enthalpy to melt ice is called enthalpy of fusion.  this value is 6.02kJ/mol

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 water boils at 100 degrees C and water melts above 0 degrees C

 1 kilojoules is 1000 joules

  water's enthalpy of vaporization (steam) is 40.68 kJ/mol

  a mole of water is 18.02 grams

  we also have to assume the ice is at 0 degrees C

Step 1

Now start with your ice.  The enthalpy of fusion for ice is calculated with this formula:

q = n x ΔH    q= energy, n = moles of water, ΔH=enthalpy of fusion

Calculate how many moles of ice you have:

150g x (1 mol / 18.02 g) = 8.32 moles

Put that into the equation:

q = 8.32 mol x 6.02 = 50.09 kJ of energy to melt 150g of ice

Step 2

To raise 1 gram of water to the boiling point, it would take 4.18 joules times 100 (degrees C)  or 418 joules.

So if it takes 418 joules for just 1 gram of water, it would take 150 times that amount to raise 150g to 100 degrees C.  418 x 150 = 62,700 joules or 62.7 kilojoules.

So far you have already used 50.09 kJ to melt the ice and another 62.7 kJ to bring the water to boiling.  That's a total of 112.79 kJ.

Step 3

The final step is to see how much energy is left to vaporize the water.

Subtract the energy you used so far from what you were told you have.

265 kJ - 112.79 kJ = 152.21 kJ

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You know you only have 152.21 kJ left so find out how many moles that will vaporize.

152.21 kJ = mol x 40.68  or   mol = 152.21 / 40.68  = 3.74 moles

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3.74 mol   x  ( 18.02 g / 1 mol ) = 67.39 grams

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2 years ago
A weather balloon is filled with helium that occupies a volume of 5.57 104 L at 0.995 atm and 32.0°C. After it is released, it r
Alchen [17]

6.52 × 10⁴ L. (3 sig. fig.)

<h3>Explanation</h3>

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P\cdot V = n\cdot R\cdot T,

where

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  • R is the ideal gas constant, and
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The question is asking for the final volume V of the gas. Rearrange the ideal gas equation for volume:

V = \dfrac{n \cdot R \cdot T}{P}.

Both the temperature of the gas, T, and the pressure on the gas changed in this process. To find the new volume of the gas, change one variable at a time.

Start with the absolute temperature of the gas:

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The volume of the gas is proportional to its temperature if both n and P stay constant.

  • n won't change unless the balloon leaks, and
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V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}.

Now, keep the temperature at T_1 =258.65\;\text{K} and change the pressure on the gas:

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The volume of the gas is proportional to the reciprocal of its absolute temperature \dfrac{1}{T} if both n and T stays constant. In other words,

V_2 = V_1 \cdot\dfrac{\dfrac{1}{P_2}}{\dfrac{1}{P_1}} = V_1\cdot\dfrac{P_1}{P_2} = 4.72122\times 10^{4}\;\text{L}\times\dfrac{0.995\;\text{atm}}{0.720\;\text{atm}}=6.52\times 10^{4}\;\text{L}

(3 sig. fig. as in the question.).

See if you get the same result if you hold T constant, change P, and then move on to change T.

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