HELPPP

The vitamin ______ is one of the components of coenzyme a, which is involved in _____. pantothenic acid; carboxylation pantothen

timama [110]

From what I can recall, I believe the answer is pantothenic acid; acyl transfer. I'm a bio major, hope I helped. :)

8 0

3 years ago

Plants undergo photosynthesis to produce glucose according to the reaction below. What mass of water is required to produce 5.0g

solniwko [45]

Answer:

option a) 3 g

Explanation:

mass of Glucose = 5 g

Mass of H₂O = ?

Reaction Given:

6CO₂ + 6H₂O ----> C₆H₁₂O₆ + 6O₂

Solution:

First we have to find mass of glucose from balanced reaction.

So,

Look at the reaction

6CO₂ + 6H₂O -------> C₆H₁₂O₆ + 6O₂

6 mol 1 mol

As 6 mole of water (H₂O) give 1 mole of Glucose (C₆H₁₂O₆ )

Convert moles to mass

molar mass of C₆H₁₂O₆ = 6(12) + 12(1) + 6(16)

molar mass of C₆H₁₂O₆ = 72 + 12 + 96

molar mass of C₆H₁₂O₆= 180 g/mol

molar mass of H₂O = 2(1) + 16 = 18 g/mol

Now

6CO₂ + 6H₂O ---------> C₆H₁₂O₆ + 6O₂

6 mol (18 g/mol) 1 mol (180 g/mol)

108 g 180 g

108 g of water (H₂O) produce 180 g of glucose (C₆H₁₂O₆)

So

if 108 g of water (H₂O) produce 180 g of glucose (C₆H₁₂O₆) so how many grams of water (H₂O) will be required to produce 5 g of glucose (C₆H₁₂O₆).

Apply Unity Formula

108 g of water (H₂O) ≅ 180 g of glucose (C₆H₁₂O₆)

X g of water (H₂O) ≅ 5 g of glucose (C₆H₁₂O₆)

Do cross multiply

mass of water (H₂O) = 108 g x 5 g / 180 g

mass of water (H₂O) = 3 g

So 3 g of water is required to produce 5 g of glucose.

7 0

3 years ago

Mars2501 [29]

Answer:

Explanation: Carbon and silicon BOTH come from Group 14 of the Periodic Table , i.e. both formally have the same number of valence electrons, 4 such electrons.........

5 0

2 years ago

Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should as

UNO [17]

Answer: 0.100 m $K_2SO_4$

Explanation:

Elevation in boiling point is given by:

$\Delta T_b=i\times K_b\times m$

$\Delta T_b=T_b-T_b^0$ = Elevation in boiling point

i= vant hoff factor

$K_b$ = boiling point constant

m= molality

1. For 0.100 m $K_2SO_{4}$

$K_2SO_4\rightarrow 2K^{+}+SO_4^{2-}$

, i= 3 as it is a electrolyte and dissociate to give 3 ions. and concentration of ions will be $3\times 0.100=0.300$

2. For 0.100 m $LiNO_3$

$LiNO_3\rightarrow Li^{+}+NO_3^{-}$

, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be $2\times 0.100=0.200$

3. For 0.200 m $C_2H_8O_3$

, i= 1 as it is a non electrolyte and does not dissociate, concentration of ions will be $1\times 0.200=0.200$

4. For 0.060 m $Na_3PO_4$

$Na_3PO_4\rightarrow 3Na^{+}+PO_4^{3-}$

, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be $4\times 0.060=0.24m$

Thus as concentration of solute is highest for $K_2SO_4$ , the elevation in boiling point is highest and thus has the highest boiling point.

3 0

3 years ago

This reaction was at equilibrium when 0.2 atm of iodine gas was pumped into the container, what happened to the equilibrium and

n200080 [17]

Answer:

Q was < K. Partial pressure of hydrogen decreased, iodine increased

Explanation:

After iodine was added the Q was [Select] K so the reaction shifted toward the Products [Select] ,The partial pressure of hydrogen [Select], Iodine [Select] |,and hydrogen iodide Decreased

Based on the equilibrium:

H2(g) + I2(g) ⇄ 2HI(g)

K of equilibrium is:

K = [HI]² / [H2] [I2]

Where [] are concentrations at equilibrium

And Q is:

Q = [HI]² / [H2] [I2]

Where [] are actual concentrations of the reactants.

When the reaction is in equilibrium, K=Q.

But as [I2] is increased, Q decreases and Q was < K

The only concentration that increases is [I2], doing partial pressure of hydrogen decreased, iodine increased

7 0

3 years ago