Which is a diatomic molecule? A)He B)O2 C) NaCl D)CuF2

The turnover number is defined as the maximum number of substrate molecules that can be converted into product molecules per uni

Nataliya [291]

Answer:

Explanation:

turn over number = R max / [E]t = K2

From given , R max = 249 * 10 ^ -6 mol. L^-1

T [E]t = 2.23 n mol. L^-1

= 2.23 * 10^-9 mol. L^-1

Putting values in above equation,

= 111.65 * 10^3 S^-1

Turn over number is maximum no of substrate molecule that can be converted into product molecules for unit time by enzyme molecule.

5 0

3 years ago

There are three bulbs powered by a battery. What will happen if the middle light burns out.

Vadim26 [7]

Answer:

nuclear war ;P

Explanation:

8 0

3 years ago

Which of the following is a statement of Hess's law?

attashe74 [19]

Answer:

C

Explanation:

the enthalpy of reaction is independent of the reaction path

7 0

3 years ago

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

3 years ago

The following diagrams represent mixtures of NO(g) and O2(g). These two substances react as follows: 2NO(g)+O2(g)→2NO2(g) It has

Alja [10]

This is an incomplete question, here is a complete question and an image is attached below.

The following diagrams represent mixtures of NO(g) and O₂(g). These two substances react as follows:

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

It has been determined experimentally that the rate is second order in NO and first order in O₂.

Based on this fact, which of the following mixtures will have the fastest initial rate?

The mixture (1). The mixture (2). The mixture (3).

Answer : The mixture 1 has the fastest initial rate.

Explanation :

The given chemical reaction is:

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The rate law expression is:

$Rate=k[NO]^2[O_2]$

Now we have to determine the number of molecules of $NO\text{ and }O_2$

In mixture 1 : There are 5 $NO$ and 4 $O_2$ molecules.

In mixture 2 : There are 7 $NO$ and 2 $O_2$ molecules.

In mixture 3 : There are 3 $NO$ and 5 $O_2$ molecules.

Now we have to determine the rate law expression for mixture 1, 2 and 3.

The rate law expression for mixture 1 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(5)^2\times (4)$

$Rate=k(100)$

The rate law expression for mixture 2 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(7)^2\times (2)$

$Rate=k(98)$

The rate law expression for mixture 3 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(3)^2\times (5)$

$Rate=k(45)$

Hence, the mixture 1 has the fastest initial rate.

4 0

3 years ago