Answer:
The answer is (e) : phosphoglucomutase, UDP-glucose pyrophosphorylase, glycogen synthase then amylo-(1,4-1,6)-transglycosylase.
Explanation:
Phosphoglucomutase: Convert glucose-6-phosphate to glucose-1-phosphate.
UDP-glucose pyrophosphorylase: Form UDP-glucose from glucose-1-phosphate.
Glycogen synthase: Add the new glucose from UDP-glucose to the growing glycogen chain.
Amylo-(1,4-1,6)-transglycosylase: This is a branching enzyme, it initiates formation of branches evolving from the main chain.
Answer:
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Explanation:
<u>1. Balanced molecular equation</u>

<u>2. Mole ratio</u>

<u>3. Moles of HNO₃</u>
- Number of moles = Molarity × Volume in liters
- n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃
<u>4. Moles Ba(OH)₂</u>
- n = 0.700M × 0.0310 liter = 0.0217 mol
<u>5. Limiting reactant</u>
Actual ratio:

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.
Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.
<u>6. Final molarity of Ba(OH)₂</u>
- Molarity = number of moles / volume in liters
- Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M
Here, we are required to determine the volume of the earth which is 1.08326 × 10¹² km³ in liters.
<em>The volume of the earth is approximately</em>,
, 1.08326 × 10²⁴ liters
By conversion factors;
- <em>1dm³ = 1liter</em>
- However; <em>1km = 10000dm = 10⁴ </em><em>dm</em>
- Therefore, 1km³ = (10⁴)³ dm³.
Consequently, 1km³ = 10¹²dm³ = 10¹²liters.
The conversion factor from 1km³ to liters is therefore, c.f = 10¹²liters/km³
Therefore, the volume of the earth which is approximately, 1.08326 × 10¹² km³ can be expressed in liters as;
<em>1.08326 × 10¹² km³ × 10¹²liters/km³ </em>
The volume of the earth is approximately,
1.08326 × 10²⁴ liters.
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