The shape of the nucleus is maintained by

What are some examples of properties of water?

gavmur [86]

Examples : cohesion , polarity , high specific heat , adhesion.<span />

4 0

3 years ago

Draw the major byproduct of the reaction (not the minor species discussed in q2b). At which stage did you remove it from your sa

givi [52]

You need to do something like that your self so sorry can help.

3 0

3 years ago

Which of the following statements is true?

dybincka [34]

Answer:C

Explanation:

A process is called spontaneous if the process takes place on its own without the intervention of external factors.

The spontaneous processes are generally quick with observable rates of reaction.But processes can be spontaneous even with negligible rates of reaction.

Spontaneity of reactions depend on temperature.

This is because spontaneity is measured by gibbs energy or enthalpy.

Both of these measures are dependent on temperature.

So,temperature affects spontaneity.

7 0

3 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

3 years ago

Calculate the molar mass of N2O. Report your answer to 2 decimal places.

kvv77 [185]

Answer:

013 g/mol is the molar mass of N2O.

5 0

2 years ago