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3 years ago

True or false during reactions or light is needed carbon dioxide is converted to glucose

Chemistry

1 answer:

Helen [10]3 years ago

7 0

Answer:

True

Explanation:

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A solution of sodium hydroxide was titrated against a solution of sulfuric acid. How many moles of sodium hydroxide would react

jeka57 [31]

Answer:

2 mole of Sodium hydroxide reacts with 1 mole of Sulfuric acid

Explanation:

Write down the equation in the beginning with reactants and products:

NaOH + H₂SO₄ → Na₂SO₄ + H₂0

Now try to balance it. Try with Na first:

2NaOH + H₂SO₄ → Na₂SO₄ + H₂0

Na atoms are balanced. There are 6 Oxygen atoms on the right and 5 on the left. Balance by increasing the H₂O moles:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0

Check if H atoms are also balanced. They are. That means our final reaction is:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0

2 Moles of NaOH reacts with 1 mole of H₂SO₄

5 0

3 years ago

Identify whether each element is a halogen, a noble gas, or nonmetal only.

OLga [1]

Answer:

Astatine: Halogen

Nitrogen: Non-Metal

Krypton: Non-Metal, Noble Gas

Chlorine: Non-Metal

Sulfur: Non-metal

Explanation:

5 0

3 years ago

What do the colors of light combine to make _______________ light.

Karo-lina-s [1.5K]

the answer is dure light I'm not sure of this answer ,but it might be to go with .

8 0

2 years ago

What are the complete Ionic and Net Ionic equations for the following

zhenek [66]

Answer:

Check photo

Explanation:

5 0

3 years ago

The Michaelis constant for pancreatic lipase is 5 mM. At 60 C, lipase is subject to deactivation with a half-life of 8 min. Fat

Dmitriy789 [7]

Explanation:

According to the given data, we will calculate the following.

Half life of lipase $t_{1/2}$ = 8 min x 60 s/min

= 480 s

Rate constant for first order reaction is as follows.

$k_{d} = \frac{0.6932}{480}$

= $1.44 \times 10^{-3}s^{-1}
$

Initial fat concentration $S_{o}$ = 45 $mol/m^{3}$

= 45 mmol/L

Rate of hydrolysis $V_m_{o}$ = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration (S) = $S_{o} (1 - X)$

= 45 (1 - 0.80)

= 9 $mol/m^{3}$

or, = 9 mmol/L

It is given that $K_{m}$ = 5mmol/L

Therefore, time taken will be calculated as follows.

t = $-\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]$

Now, putting the given values into the above formula as follows.

t = $-\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]$

= $-\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s
}{K_{M} ln (\frac{45 mmol/L
}{9 mmol/L
}) + (45 mmol/L - 9 mmol/L
)]$

= $1642.83 s \times \frac{1 min}{60 sec}$

= 27.38 min

Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.

6 0

3 years ago