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3 years ago

Answer =

Chemistry

1 answer:

Anna [14]3 years ago

6 0

Answer:

37.83 moles

Explanation:

moles = grams / formula weight = 454 grams / 12g/mol = 37.83 moles

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Inessa [10]

Answer:

The heat would flow from the hot solid to the cool solid until all temperatures are near equal.

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2 years ago

What is the answer to question 7, NH4C2H3O2

Montano1993 [528]

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3 years ago

A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-

Komok [63]

Answer: The cell potential of the cell is +0.118 V

Explanation:

The half reactions for the cell is:

Oxidation half reaction (anode): $Ni(s)\rightarrow Ni^{2+}+2e^-$

Reduction half reaction (cathode): $Ni^{2+}+2e^-\rightarrow Ni(s)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

$[Ni^{2+}_{diluted}]$ = $1.00\times 10^{-4}M$

$[Ni^{2+}_{concentrated}]$ = 1.0 M

Putting values in above equation, we get:

$E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}$

$E_{cell}=0.118V$

Hence, the cell potential of the cell is +0.118 V

5 0

3 years ago

16. The concentration of a solution of potassium hydroxide is determined by titration with nitric

____ [38]

Answer:

$M_{base}=0.709M$

Explanation:

Hello,

In this case, since the reaction between potassium hydroxide and nitric acid is:

$KOH+HNO_3\rightarrow KNO_3+H_2O$

We can see a 1:1 mole ratio between the acid and base, therefore, for the titration analysis, we find the following equality at the equivalence point:

$n_{acid}=n_{base}$

That in terms of molarities and volumes is:

$M_{acid}V_{acid}=M_{base}V_{base}$

Thus, solving the molarity of the base (KOH), we obtain:

$M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} =\frac{0.498M*42.7mL}{30.0mL}\\ \\M_{base}=0.709M$

Regards.

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2 years ago

Do you think the amount of fat will affect its melting point?

antoniya [11.8K]

The amount of fat will affect its melting point

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2 years ago