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Cloud [144]
3 years ago
12

What is gluecose used for

Chemistry
1 answer:
Amanda [17]3 years ago
4 0
Glucose is used by intestinal cells and red blood cells, while the rest reaches the liver, adipose tissue and muscle cells, where it is absorbed and stored as glycogen.

(it is saved to be used later)
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How much thermal energy (q) is required to heat 15.2g of a metal (specific heat=0.397J/C) from 21.0C to 40.3C? Show steps please
Elena L [17]
Q = ?

Cp = 0.397 J/ºC

Δt =  40.3 - 21.0<span> => 19.3</span><span> ºC</span>

m = 15.2 g

Q = m x Cp x Δt

Q = 15.2 x 0.397 x 19.3

Q ≈ 116.46 J

<span>hope this helps! </span>
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Bacteria cells are different from plant and animal cells because they don't have what???!
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3 years ago
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Conversion factors are useful in solving problems in which a given measurement must be expressed in
spin [16.1K]

Conversion factors are useful in solving problems in which a given measurement must be expressed in some other units of measure.


Conversion factors
are helpful in solving issues where one unit of measurement needs to be expressed in another. In general, a measurement's numerical value changes when it is multiplied by a conversion factor, while the quantity being measured's real size stays the same.
A conversion factor is a number that is used to multiply or divide one set of units into another. If a conversion is required, it must be done using the correct conversion factor to get an identical value. For instance, 12 inches equals one foot when converting between inches and feet.

To learn more about Conversion factors please visit -brainly.com/question/23510660
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4 0
2 years ago
Consider the following reaction: CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at25 ∘C under these conditions: PCO2PCC
Salsk061 [2.6K]

<u>Answer:</u> The \Delta G for the reaction is 54.6 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

  • To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

  • The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.760atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.180atm

Putting values in above equation, we get:

K_p=\frac{(0.760)^2}{0.140\times 0.180}\\\\K_p=22.92

  • To calculate the Gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 22.92

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(22.92))\\\\\Delta G=54659.78J/mol=54.6kJ/mol

Hence, the \Delta G for the reaction is 54.6 kJ/mol

7 0
3 years ago
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