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Vikentia [17]
3 years ago
6

Consider the reaction, C2H4(g) + H2(g) - C2H6(8), where AH = -137 kJ. How many kilojoules are released when 3.5 mol of CH4

Chemistry
1 answer:
antiseptic1488 [7]3 years ago
7 0

Answer: 480 kJ of energy is released when 3.5 mol of C_2H_4 reacts.

Explanation:

The balanced chemical reaction is:

C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)  \Delta H=-137kJ

Thus it is given that the reaction is exothermic (heat energy is released) as enthalpy change for the reaction is negative.

1 mole of C_2H_4 on reacting gives = 137 kJ of energy

Thus 3.5 moles  of C_2H_4 on reacting gives = \frac{137}{1}\times 3.5=480 kJ of energy

Thus 480 kJ of energy is released when 3.5 mol of C_2H_4 reacts.

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Calnexin and calreticulin catalyze the removal of the final glucose residue from glycoproteins during the folding process. True
zloy xaker [14]

Answer:

Explanation:

A) False.

Glucosidase (not calnexin nor calreticulin) helps to remove glucose residue.

Both calnexin and calreticulin rather have an affinity for last glucose residue of misfolded protein (Only misfolded proteins are marked by glycosyltransferase by attaching glucose residue). They attach with misfolded protein and with the help of other proteins like ERp57 (a type of protein disulfide isomerase) and try to fold it properly. If protein is properly folded then glucosidase removes the glucose residue thereby releasing the properly folded protein from calnexin or calreticulin. and now protein is transported to the Golgi body. If folding is still not proper then the same cycle of glycosylation -binding of calnexin/calreticulin and effort to fold it properly is repeated.

B) True.

Transketolase is a key enzyme of the pentose phosphate pathway. It contains thiamine diphosphate (TPP) as a cofactor. it does transfer 2 carbon residue from a ketose to aldose. So, effectively it converts one ketose sugar to aldose with 2 carbonless and aldose to ketose with 2 carbon more.

C) True.

Theoretically, for the evolution of one molecule of oxygen, only 8 photons are required. But in practice, it is known that there are many variants like wavelength and the energy of the photon. The larger the wavelength, like the one which is used in PS1 (more than 700nM), the lesser the energy. Secondly, the energy of the photon is also wasted as heat energy. Because of these factors, more than 8 photons are needed in reality.

D) Wrong.

Fructose 2,6 bisphosphate is a key substrate and affects both the enzymes- phosphofructokinase and fructose bisphosphatase allosterically during gluconeogenesis. It strongly favors the breakdown of glucose during glycolysis by activating phosphofructokinase but it inhibits fructose bisphosphatase. Hence it activates the kinase enzyme while inhibiting the phosphatase and maintains a huge supply of glucose in the system.

E) Wrong.

The Calvin cycle shares similarity with the pentose phosphate pathway as both are involved in the synthesis of sugar (Triose and Ribose). However, it does not share similarity with enzymes of glycolysis (which is primarily focused on the breakdown of glucose) and gluconeogenesis.

8 0
3 years ago
A solution contains 6.21 g of ethylene glycol dissolved in 25.0 g of water. If water has a boiling point elevation constant of 0
timurjin [86]
Calculate first the number of moles of ethylene glycol by dividing the mass by the molar mass.
                           n = (6.21 g ethylene glycol) / 62.1 g/mol
                              n = 0.1 mol
Then, calculate the molality by dividing the number of moles by the mass of water (in kg).
                           m = 0.1 mol/ (0.025 kg) = 4m
Then, use the equation,
                       Tb,f = Tb,i + (kb)(m)
Substituting the known values,
                       Tb,f = 100°C + (0.512°C.kg/mol)(4 mol/kg)
                          <em>Tb,f = 102.048°C</em>
5 0
3 years ago
Read 2 more answers
A 50.0 mL graduated cylinder has a mass of 65.1 g. When it is filled with an unknown liquid to the 49.3 mL mark, the cylinder an
Delvig [45]

<u>Answer:</u> The density of liquid is 1.12g/cm^3

<u>Explanation:</u>

We are given:

Mass of cylinder, m_1 = 65.1 g

Mass of liquid and cylinder combined, M = 120.5 g

Mass of liquid, m_2 = (M-m_1)=(120.5-65.1)g=55.4g

To calculate density of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

We are given:

Mass of liquid = 55.4 g

Volume of liquid = 49.3 mL = 49.3 cm^3    (Conversion factor:  1mL=1cm^3 )

Putting values in above equation, we get:

\text{Density of liquid}=\frac{55.4g}{49.3cm^3}\\\\\text{Density of liquid}=1.12g/cm^3

Hence, the density of liquid is 1.12g/cm^3

3 0
3 years ago
Write the equilibrium constant expression for this reaction: nh4
pochemuha
Kc= (nh4)
--------
(nh3) + (h2o)
4 0
3 years ago
What is the oxidation number of tin (Sn) in the compound Na2SnO2? A. -2 B. 0 C. +2 D. +3
Allisa [31]

O.N. of Na = +1

O.N. of O = -2

Let, O.N. of Tin = x

1*2 + x + -2*2 = 0

2+x-4 = 0

x-2 = 0

x = 2

SO OPTION C IS YOUR ANSWER......

7 0
3 years ago
Read 2 more answers
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