<span>1. The correct option is ADENOSINE TRIPHOSPHATE, ATP. ATP is the basic unit of energy transfer in the living cells. ATP is the principal energy source for metabolic functions, in the cells, ATP are consumed by endothermic metabolic reactions and they are produced by exothermic metabolic reactions.
2. To form ATP, A THIRD PHOSPHATE GROUP HAS TO BE ADDED TO ADP. ADP has two phosphate groups while ATP has three phosphate group. ADP is usulally converted to ATP by the addition of a single phosphate group.
3. ADP and ATP work together and the two can be interconverted. ATP can be hydrolysed to ADP and ADP can be converted to ATP by the addition of a single phosphate group. When energy is needed inside a cell, the ATP will split off one of its phosphate group and become ADP. This split off process produce a high qunatity of energy which is then available for the cell to use.</span>
Answer:
15.95
Explanation:
This question is a modification of the calculation of the empirical formula of a compound given its percent composition and atomic weights of the elements in the compound.
Here we are given the formula and the percent composition, so we know that there are 4 atoms of E per 2 atoms of N so lets solve using the information given.
In 100 grams of the binary compound we have
30.46 g N
69.54 g E
The number of moles is the mass divided by atomic weight:
mol N = 30.46 g / A.W N = 30.46 g / 14.00 g/mol = 2.18 mol N
mol E = 65.54 g / A.W E
Thus,
4 mol E/ 2 mol N = ( 69.54 g/ A.W E ) / 2.18
2 A.E = 65.54 g / 2.18 ⇒ A.W E = 69.54 g / ( 2 x 2.18 ) = 15.94 g
So the A.W is 15.94 g/mol which is close the atomic weight of O.
Answer:
74.0 g/mol
Explanation:
Step 1: Write the generic neutralization reaction
HA + NaOH ⇒ NaA + H₂O
Step 2: Calculate the reacting moles of NaOH
At the equivalence point, 33.83 mL of 0.115 M NaOH react.
0.03383 L × 0.115 mol/L = 3.89 × 10⁻³ mol
Step 3: Calculate the moles of HA that completely react with 3.89 × 10⁻³ moles of NaOH
The molar ratio of HA to NaOH is 1:1. The reacting moles of HA is 1/1 × 3.89 × 10⁻³ mol = 3.89 × 10⁻³ mol.
Step 4: Calculate the molar mass of the acid
3.89 × 10⁻³ moles of HA have a mass of 0.288 g.
M = 0.288 g / 3.89 × 10⁻³ mol = 74.0 g/mol
