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arsen [322]
4 years ago
6

For a particular lab experiment, you need 150 mL of a 2.50 M solution of ammonium nitrate. How many grams of NH4NO3 will you nee

d for the solution?
8.0 g

10.02 g

25.0 g

875.0 g

30.0 g
Chemistry
1 answer:
Anna35 [415]4 years ago
4 0

Answer:

mass NH4NO3 = 30.0 g

Explanation:

∴ V = 150 mL

∴<em> C</em> NH4NO3 = 2.50 M

⇒ mass NH4NO3 = ?

∴ mol NH4NO3 = (0.150 L)×(2.50 mol/L) = 0.375 mol NH4NO3

∴ Mw NH4NO3 = 80.043 g/mol

⇒ mass NH4NO3 = (0.375 mol)×(80.043 g/mol) = 30.016 g

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The protein catalase catalyzes the reaction The Malcolm Baldrige National Quality Award aims to:
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The question is missing a part, so the complete question is as follows:

The protein catalase catalyzes the reaction The Malcolm Bladrigde National Quality Awards aims to: 2H2O2 (aq) ⟶ 2H2O (l) + O2 (g) and has a Michaelis-Menten constant of KM = 25mM and a turnover number of 4.0 × 10 7 s -1. The total enzyme concentration is 0.012 μM and the intial substrate concentration is 5.14 μM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Calculate the initial rate, R (often written as V0), of this reaction.

1) Calculate Rmax

The turnover number (Kcat) is a ratio of how many molecules of substrate can be converted into product per catalytic site of a given concentration of enzyme per unit of time:

Kcat = \frac{Vmax}{Et},

where:

Vmax is maximum rate of reaction when all the enzyme sites are saturated with substrate

Et is total enzyme concentration or concentration of total enzyme catalytic sites.

Calculating:

Kcat = \frac{Vmax}{Et}

Vmax = Kcat · Et

Vmax = 4×10^{7} · 1.2 × 10^{-8}

Vmax = 4.8 × 10^{-1} M

2) Calculate the initial rate of this reaction (R):

The Michaelis-Menten equation studies the dynamics of an enzymatic reaction. This model can explain how an enzyme enhances the rate of a reaction and how the reaction rate depends on the concentration of the enzyme and its substrate. The equation is:

V0 = \frac{[S].(Vmax)}{KM + [S]}, where:

[S] is the substrate's concentration

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The answers are Vmax = 4.8 × 10^{-1} M and V0 = 0.478 M.

 

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