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Svetach [21]
3 years ago
15

Examine this reaction:

Chemistry
1 answer:
Alina [70]3 years ago
8 0

Answer:

The normal amount of disaccharide would be produced, but fewer monosaccharides would be produced.

Explanation:

The first reaction, the conversion of starch into disaccharides, is catalyzed by the enzyme amylase. <u>Since amylase is present in a normal amount, a normal amount of disaccharides will be produced.</u>

In the second reaction, these disaccharides will be transformed into monosaccharides by a disaccharidase. However, since t<u>here is less disaccharidase, there will be fewer monosaccharides produced than if it was a normal amount of amylase.</u>

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How many total atoms are in 0.380 g of P2O5
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Consider the following generic reaction for which Kp = 5.51 × 105 at 25°C:
lora16 [44]

Answer:

K_c=1.35\times 10^7

Explanation:

The relation between Kp and Kc is given below:

K_p= K_c\times (RT)^{\Delta n}

Where,  

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

2R_{(g)}+A_{(g)}\rightleftharpoons2Z_{(g)}

Given: Kp = 5.51\times 10^5  

Temperature =  25°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

T = (25 + 273.15) K = 298.15 K  

R = 0.082057 L atm.mol⁻¹K⁻¹

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5.51\times 10^5= K_c\times (0.082057\times 299)^{-1}

K_c\frac{1}{24.535043}=551000

K_c=13518808.69=1.35\times 10^7

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3 years ago
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Answer:

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