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3 years ago

The rate of product formation by an enzyme-catalyzed reaction would be increased by

1 answer:

5 0

The rate of product formation by an enzyme-catalyzed reaction would be increased by temperature (until it reaches optimim temperature)

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Why is it difficult to accurately represent ammonia and methane on paper?

andrew-mc [135]

Answer:

The correct answer is because the molecular structure.

Explanation:

The difficulty of ammonia and methane to be represented on paper is due to the molecular structure. These compounds have a three-dimensional projection with defined angles. Ammonia presents angles of 109.5º between the atom of Nitrogen and those of Oxygen. The ammonia presents 107.8º between the oxygen atoms.

In the methane molecule, there is 109.5º between the hydrogen molecules and the carbon atom. This results in the need for a 3D representation of the molecule.

Have a nice day!

6 0

2 years ago

Why do noble gases have the least amount of electronegativity?

sattari [20]

<span>Their orbitals are completely filled</span>

4 0

2 years ago

The fuel used to power the booster rockets on space shuttles is a mixture of aluminum metal and ammonium perchlorate. The follow

pishuonlain [190]

Answer:

Therefore 373 mole of Al produce 746 mole of water.

Explanation:

Given reaction is

3 Al+3NH₄ClO₄→Al₂O₃+AlCl₃+3NO+6H₂O

From the above reaction it is clear that 3 mole of Al produce 6 mole of water.

Therefore

3 mole of Al produce 6 mole of water.

1 mole of Al produce $\frac{6}{3}$ mole of water.

373 mole of Al produce $\frac{6\times 373}{3}$ mole of water.

= 746 mole of water.

Therefore 373 mole of Al produce 746 mole of water.

7 0

2 years ago

A 298 lb person must receive Heparin, determine the number of units given every hour (Heparin 8.0 units/kg per hour). Enter your

yarga [219]

Answer:

1,081.1 units of heparin should be given to 298 lb person.

Explanation:

We are given:

Weight of the person = 298 lb = 135.143 kg

Conversion factor used:

1 lb = 0.4535 kg

Number of unit of heparin to be given in an hour = 8.0 units/kg

Number of units given to the patient weighing 135.143 kg :

$8.0 units/kg\times 135.143 kg=1,081.144 units\approx 1,081.1 units$

1,081.1 units of heparin should be given to 298 lb person.

4 0

2 years ago

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