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3 years ago

What is the atom glucose made of?

Chemistry

2 answers:

PIT_PIT [208]3 years ago

5 0

Glucose is made up of "CARBON".........

defon3 years ago

4 0

Glucose contains (C) carbon, (O) oxygen, and (H) hydrogen atoms.
the formula for glucose is C6H12O6
hope this helps..
brianliest..?

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Describe the energy of the products in comparison to the energy the atoms had when the were wandering around the room free.

Over [174]

Answer: The atoms of the solid products are tightly packed whereas those in gas are loose.

Explanation:

Atoms in the solid structure are linked to each other by strong force of attraction and they collectively form an three-dimensional structure. The binding of the atoms form the shape of the solid product. The gas exhibit free atoms in it. The atoms are not linked to one another by strong force of attraction. The atoms as a part of gas have higher energy as those present in the solid products. So, the atoms of the gas wander freely as compared to those in the solid products.

7 0

3 years ago

The Henry's law constant for helium gas in water at 30 °C is 3.70 × 10-4 M/atm. When the partial pressure of helium above a samp

igor_vitrenko [27]

Answer:

The concentration of helium in the water is 2.405×10^-4 M

Explanation:

Concentration = Henry's law constant × partial pressure of helium

Henry's law constant = 3.7×10^-4 M/atm

Partial pressure of helium = 0.65 atm

Concentration = 3.7×10^-4 × 0.65 = 2.405×10^-4 M

4 0

3 years ago

Ethanol (C2H5OH) melts a - 144 oC and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vapo

hammer [34]

Answer:

For a: The total heat required is 36621.5 J

For b: The total heat required is 58944.5 J

Explanation:

For a:

To calculate the heat required at different temperature, we use the equation:

$q=mc\Delta T$ .........(1)

where,

q = heat absorbed

m = mass of substance

$c$ = specific heat capacity of substance

$\Delta T$ = change in temperature

To calculate the amount of heat required at same temperature, we use the equation:

$q=m\times \Delta H$ ........(2)

where,

q = heat absorbed

m = mass of substance

$\Delta H$ = enthalpy of the reaction

The processes involved in the given problem are:

$1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

For process 1:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K$

Putting values in equation 1, we get:

$q_1=42.0g\times 2.3J/g.K\times 43K\\\\q_1=4153.8J$

For process 2:

We are given:

Conversion factor: 1 kJ = 1000 J

Molar mass of ethanol = 46 g/mol

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{35.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 773.04J/g\\\\q_2=32467.7J$

Total heat required = $[q_1+q_2]$

Total heat required = $[4153.8J+32467.7J]=36621.5J$

Hence, the total heat required is 36621.5 J

For b:

The processes involved in the given problem are:

$1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

For process 1:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_s=0.97J/g.K\\T_2=-144^oC\\T_1=-155^oC\\\Delta T=[T_2-T_1]=[-144-(-155)]^oC=11^oC=11K$

Putting values in equation 1, we get:

$q_1=42.0g\times 0.97J/g.K\times 11K\\\\q_1=448.14J$

For process 2:

We are given:

$m=42.0g\\\Delta H_{fusion}=5.02kJ/mol=\frac{5.02kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=109.13J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 109.13J/g\\\\q_2=4583.5J$

For process 3:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=-144^oC\\\Delta T=[T_2-T_1]=[78-(-144)]^oC=222^oC=222K$

Putting values in equation 1, we get:

$q_3=42.0g\times 2.3J/g.K\times 222K\\\\q_3=21445.2J$

For process 4:

We are given:

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{38.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_4=42.0g\times 773.04J/g\\\\q_4=32467.7J$

Total heat required = $[q_1+q_2+q_3+q_4]$

Total heat required = $[448.14+4583.5+21445.2+32467.7]J=58944.5J$

Hence, the total heat required is 58944.5 J

8 0

3 years ago

A 2.6 liter container of nitrogen had a pressure of 3.2 atm. What volume will the gas occupy at a pressure of

Dominik [7]

Answer:

A sample of oxygen gas occupies a volume of 250. ... volume will it occupy at 800. torr pressure? ... A 2.0 liter container of nitrogen had a pressure of 3.2 atm. ... A sample of hydrogen at 1.5 atm had its pressure decreased to 0.50 atm producing

Explanation:

8 0

3 years ago

Kostya randomly chooses a number from 1 to 10. What is the probability he chooses a number less than 5?

Ghella [55]

A. 1/2

Explanation- There is a 5/10 chance of choosing on of the numbers which simplifies to 1/2

3 0

3 years ago

Read 2 more answers