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Svet_ta [14]
3 years ago
7

Will the final amount of biodiesel change if you actually have more or less amount of catalyst​

Chemistry
1 answer:
DerKrebs [107]3 years ago
4 0

Answer

I don know sorry

Explanation:

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Both are mainly composed of droplets of condensed water
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What does homogeneous mixture of elements
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a homogeneous mixture is a mixture that appears to be one thing. for example air would be one.

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What is the number of moles of NaOH in 16.5 mL of 0.750 Molar NaOH?
Alchen [17]

Answer:

0.0165 (L) * 0.750 (mol/L) = 0.0124 mol NaOH in 16.5 mL

Explanation:

5 0
3 years ago
What is the chemical equation for the alpha decay of erbium-144
iren [92.7K]

Answer:

Er-144 -------> Dy-140 + He-4

Explanation:

Alpha decay is the release of a hydrogen nucleus.  So the original atom will decrease the mass by 4 and the atomic number by 2.

8 0
3 years ago
One mole of an ideal gas doubles its volume in a reversible isothermal expansion. (a) What is the change in entropy of the gas?
Andre45 [30]

Explanation:

The given data is as follows.

       n = 1 mol,     V_{f} = 2V_{i}

       Q = 1500 J,      R = 8.314 J/mol k

(a)    \Delta S = \frac{dQ}{dT}

And, according to the first law of thermodynamics

                \Delta E_{int} = Q - W

And, in an isothermal process the change in internal energy of the gas is zero.

Hence,    0 = Q - W

or,             W = Q

Expression for work done in an isothermal process is as follows.

                   W = nRT ln \frac{V_{f}}{V_{i}}

As W = Q, Hence expression for Q will also be given as follows.

            Q = nRT ln \frac{V_{f}}{V_{i}}

Now,  

        \Delta S = \frac{nRT ln \frac{V_{f}}{V_{i}}}{T}

        [/tex]\Delta S = nR ln \frac{V_{f}}{V_{i}}[/tex]

                      = nR ln \frac{2V_{i}}{V_{i}}

                       = nR ln 2

                        = 1 \times 8.314 \times 0.693

                        = 5.76 J/K

Therefore, change in entropy is 5.76 J/K.

(b)    As,  Q = nRT ln \frac{V_{f}}{V_{i}}

                   = nRT ln \frac{2V_{i}}{V_{i}}

                   = nRT ln 2

           T = \frac{Q}{nR ln 2}

              = \frac{1500}{1 \times 8.314 ln 2}

              = 260.4 K

Therefore, temperature of the gas is 260.4 K.

7 0
3 years ago
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