What is the mass of 1.84 mol NaCl? Give your

What enzyme catalyses the reaction of PRPP transfer to hypoxanthine with formation of IMP?

Alona [7]

Answer:

HPRT

Explanation:

HPRT catalyzes the salvage reactions of hypoxanthine and guanine with PRPP to form IMP and GMP

The formation of GMP from IMP requires oxidation at C-2 of the purine ring, followed by a glutamine-dependent amidotransferase reaction that replaces the oxygen on C-2 with an amino group to yield 2-amino,6-oxy purine nucleoside monophosphate, or as this compound is commonly known, guanosine monophosphate.

8 0

3 years ago

Drag the description to the category

viktelen [127]

Answer:

ok

Explanation:

5 0

3 years ago

An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +

Maurinko [17]

Answer:

The simplified expression for the fraction is $\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$

Explanation:

From the given information:

O3* → O3 (1) fluorescence

O + O2 (2) decomposition

O3* + M → O3 + M (3) deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

$\text {X} = \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) } } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{ {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3 \times cM) }$

$\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$ since cM is the concentration of the inert molecule

7 0

4 years ago

A standard gold bar has a mass of 12.4 kg. When the gold bar is placed into a container of water the volume changes by 642 mL. W

AlekseyPX

Answer:

19.3 g/mL

Explanation:

The following data were obtained from the question:

Mass (m) = 12.4 kg

Volume (V) = 642 mL.

Density (D) =.?

Next, we shall convert 12.4 kg to grams. This can be obtained as follow:

1 kg = 1000 g

Therefore,

12.4 kg = 12.4 × 1000

12.4 kg = 12400 g

Therefore, 12.4 kg is equivalent to 12400 g.

Finally, we shall determine the density of the gold as follow:

Density is simply defined as the mass of the substance per unit volume of the substance. It can be represented mathematically as:

Density (D) = mass (m) / volume (V)

D = m/V

With the above formula, the density of gold can easily be obtained as follow:

Mass (m) = 12400 g

Volume (V) = 642 mL.

Density (D) =.?

D = m/V

D = 12400/642

D = 19.3 g/mL

Therefore, the density of hold is 19.3 g/mL

6 0

3 years ago

Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide. Ca) + H2O --> Ca(OH)2 In a particular

mars1129 [50]

Answer:

Percent yield = 92.5%

Explanation:

The question asks for the percent yield which can be defined as:

$\frac{actual yield}{theoretical yield} .100$

Where the actual yield is how much product was obtained, in this case 6.11 g of Ca(OH)₂, and the theoretical yield is how much product could be obtained with the given reactants theoretically, that is if the reaction would work perfectly. So we need to calculate first the theoretical yield.

1. First lets write the chemical equation reaction correctly and check that it is balanced:

CaO + H₂O → Ca(OH)₂

2. Calculate the amount of product Ca(OH)₂ that can be obtained with the given reactants (theoretical yield), which are 5.00g of CaO and excess of water. So the amount of CaO will determined how much Ca(OH)₂ we can obtained.

For this we'll use the molar ratio between CaO and Ca(OH)₂ which we see it is 1:1. For every mol of CaO we'll obtain a mol of Ca(OH)₂. So lets convert the 5.00 g of CaO to moles:

Molar Mass of CaO: 40.078 + 15.999 = 56.077 g/mol

moles of CaO = 5.00 g / 56.077 g/mol = 0.08916 moles

As we said before from the molar ratio moles of Ca(OH)₂ = moles of CaO

So the moles of Ca(OH)₂ that can be obtained are 56.077 g/mol

We need to convert this value to grams:

Molar Mass of Ca(OH)₂ = 40.078 + (15.999 + 1.008)*2 = 74.092 g/mol

Theoretical yield of Ca(OH)₂ = 0.08916 moles x 74.092 g/mol = 6.606 g

3. Calculate the percent yield:

$\frac{actual yield}{theoretical yield} .100$

Percent yield = (6.11 g / 6.606g) x 100 = 92.5 %

5 0

3 years ago

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