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ycow [4]
2 years ago
8

The placement of carbonyl group of a kerose sugar is at second-carbon only (b) fir + carbon only () first and fast carbon (d) li

st carbon only Chirulit,​
Chemistry
1 answer:
olasank [31]2 years ago
6 0

Answer:

(d)

Explanation:

Carbonyl group can be the placement of kerosene sugar

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Rutherford, geiger, and marsden’s experiment demonstrated that the volume of the nucleus is roughly what fraction of the volume
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Rutherford, Geiger and Marsden's experiment proved that every atom has a nucleus and that this nucleus is of positive charge and contains the most of the mass of the atom. 0.005% of the volume occupied by the electrons is the volume of the nucleus.
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Calculate the number of moles of KOH in 6.70 mL of a 0.480 M KOH solution. (Enter your answer in scientific notation.)
andrew-mc [135]

Answer:

3.2×10^-3 mol

Explanation:

The equation for molarity is M= n/L. Where "M" is Molarity, "n" is the number of moles of solute, and "L" is the total liters in solution.

The question gives you the volume in mL, so to convert "mL" to "L" you need to divide by 1000. (6.70mL/ 1000L)= 0.0067L.

Now you can plug the numbers into the equation. 0.480M= n/ 0.0067L), multiply (0.480M×0.0067L)= 0.003216 mol. The scientific notation is 3.2×10^-3, 10^-3 because you move the decimal back three times and 3.2 because there are 2 sig figs.

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The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C
avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

q=c\times (T_{final}-T_{initial})

where,

q = heat gained = ?

c = specific heat = 5.20kJ/^oC

T_{final} = final temperature = 27.43^oC

T_{initial} = initial temperature = 22.93^oC

Now put all the given values in the above formula, we get:

q=5.20kJ/^oC\times (27.43-22.93)^oC

q=23.4kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = \frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole

\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

\Delta H=\Delta U+\Delta n_gRT

or,

\Delta U=\Delta H-\Delta n_gRT

where,

\Delta H = change in enthalpy = -2805.8kJ/mol

\Delta U = change in internal energy = ?

\Delta n_g = change in moles = 0   (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = 27.43^oC=273+27.43=300.43K

Now put all the given values in the above formula, we get:

\Delta U=\Delta H-\Delta n_gRT

\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K

\Delta U=-2805.8kJ/mol-0

\Delta U=-2805.8kJ/mol

Therefore, the internal energy change is -2805.8 kJ/mol

5 0
3 years ago
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