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4 years ago

What sugar provides energy for all cells of the body? Help me plz

Chemistry

2 answers:

Aliun [14]4 years ago

5 0

The answer is glucose

spin [16.1K]4 years ago

4 0

The answer should be glucose

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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

Would the car be able to travel all the way through the track of the roller coaster shown below? Why or Why not?

xxMikexx [17]

Answer:

No because there is not enough Potential Energy at Point 1 to make it all the way through point 5.

Explanation:

8 0

3 years ago

Read 2 more answers

How many oxygen atoms are in 163 g of aluminum sulfate?

djverab [1.8K]

Answer:

3.44 × 10²⁴ atoms of oxygen

Explanation:

Step 1: Calculate the moles corresponding to 163 g of Al₂(SO₄)₃

The molar mass of Al₂(SO₄)₃ is 342.15 g/mol.

163 g × 1 mol/342.15 g = 0.476 mol

Step 2: Calculate the molecules in 0.476 moles of Al₂(SO₄)₃

We will use Avogadro's number: there are 6.02 × 10²³ molecules of Al₂(SO₄)₃ in 1 mole of Al₂(SO₄)₃.

0.476 mol × 6.02 × 10²³ molecules/1 mol = 2.87 × 10²³ molecules

Step 3: Calculate the atoms of O in 2.87 × 10²³ molecules of Al₂(SO₄)₃

According to the chemical equation, the molar ratio of Al₂(SO₄)₃ to O is 1:12. The atoms of O are 12/1 × 2.87 × 10²³ = 3.44 × 10²⁴.

8 0

3 years ago

What is the density of carbon dioxide gas if 0.297 g occupies of 150.0 mL?

Helga [31]

P = m/v
P = 0.297g/150.0ml
P = 1.98x10^-3 g/ml

3 0

3 years ago

How many particles are in 0.0075 moles of aluminum

elixir [45]

Answer:

i dont know i dont know i dont know ind i hope you understand love ya

6 0

3 years ago