Answer:
3) The relative concentrations of each gas must remain constant.
4)The concentration of each gas will not change.
Explanation:
- For the equilibrium system:
<em>X₂ + Y₂ ⇄ 2XY,</em>
The no. of moles of gases in each side is constant; there is 2 moles of gases at reactants side and 2 moles of gases at products side.
So, changing the volume will not affect on the equilibrium system.
<em>So, the right choice is:</em>
3) The relative concentrations of each gas must remain constant.
4)The concentration of each gas will not change.
A homogeneous mixture<span> has the same uniform appearance and composition throughout. Many homogeneous </span>mixtures<span> are commonly referred to as </span>solutions<span>. A </span>heterogeneous mixture <span>consists of visibly </span>different<span> substances or phases. The three phases or states of matter are gas, liquid, and solid.
</span>
Phosphoenol pyruvate enzyme is not part of gluconeogenesis.
<h3>Phosphoenol pyruvate</h3>
The ester formed when pyruvate and phosphate are combined to form an enol results in phosphoenol pyruvate (2-phosphoenolpyruvate, or PEP). As an anion, it exists. In biochemistry, PEP is a crucial intermediary. Involved in glycolysis and gluconeogenesis, it boasts the highest-energy phosphate bond yet discovered in an organism (61.9 kJ/mol). It also plays a role in carbon fixation and the manufacture of a number of aromatic chemicals in plants. In bacteria, it provides energy for the phosphotransferase system. Enolase reacts with 2-phosphoglyceric acid to produce PEP as a result. Pyruvate kinase (PK) converts PEP to pyruvic acid, and this process produces adenosine triphosphate (ATP) via substrate-level phosphorylation. One of the main units of currency for chemical energy in cells is ATP.
Learn more about Phosphoenol pyruvate here:
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In Endothermic reactions, energy is absorbed.
Answer is: the percent by mass of NaHCO₃ is 2,43%.
m(NaHCO₃) = 10 g.
V(H₂O) = 400 ml.
d(H₂O) = 1 g/ml.
m(H₂O) = V(H₂O) · d(H₂O).
m(H₂O) = 400 ml · 1 g/ml.
m(H₂O) = 400 g.
m(solution) = m(H₂O) + m(NaHCO₃).
m(solution) = 400 g + 10 g.
m(solution) = 410 g.
ω(NaHCO₃) = 10 g ÷ 410 g · 100%.
ω(NaHCO₃) = 2,43 %
