This is a type of element or substance that is not a metal.

One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbon dioxid

natka813 [3]

Answer:

Enzyme is carbonic anhydrase

Substrate is $CO_2$

Turnover number is $10^{7}$

Explanation:

An enzyme is used by a living organism as a catalyst to perform a specific biochemical reaction.

A substrate is a molecule upon which an enzyme acts.

Turnover number refers to the number of substrate molecules transformed by a single enzyme molecule per minute. Here, the enzyme is the rate-limiting factor.

Here,

Enzyme is carbonic anhydrase

Substrate is $CO_2$

Turnover number is $10^{7}$

8 0

3 years ago

Amylose is a form of starch which has ________. only β-1,4-bonds between glucose units only α-1,4-links bonds glucose units hemi

forsale [732]

Answer: Amylose is a form of starch which has only α-1,4-links bonds glucose units.

Explanation:

Amylose is a polysaccharide made up of α(1-4) bound glucose molecules. The carbon atoms on glucose are numbered, starting at the aldehyde (C=O) carbon, so, in amylose, the 1-carbon on one glucose molecule is linked to the 4-carbon on the next glucose molecule.

5 0

3 years ago

In another experiment, a 0.150 M BF4^-(aq) solution is prepared by dissolving NaBF4(s) in distilled water. The BF4^-(aq) ions in

Ilia_Sergeevich [38]

Answer:

A) Forward rate = 1.1934 × 10^(-4) M/min

B) I disagree with the claim

Explanation:

A) We are told that [HF] reaches a constant value of 0.0174 M at equilibrium.

The reversible reaction given to us is;

BF4-(aq) +H20(l) → BF3OH-(aq) + HF(aq)

From this, we can see that the stoichiometric ratio is 1:1:1:1

Thus, concentration of [BF4-] is now;

[BF4-] = 0.150 - 0.0174

[BF4-] = 0.1326 M

From the rate law, we are told the forward rate is kf [BF4-].

We are given Kf = 9.00 × 10^(-4) /min

Thus;

Forward rate = 9.00 × 10^(-4) /min × (0.1326M)

Forward rate = 1.1934 × 10^(-4) M/min

(B) The student claims that the initial rate of the reverse reaction is equal to zero can't be true because at equilibrium, rates for the forward and reverse reactions are usually equal.

Thus, I disagree with the claim.

3 0

3 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

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 $CO_2=1.03g$

Mass of $H_2O=0.14g$

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

So, in 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

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 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

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 = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

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

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago

Balance this equation- _P4 + _02 --> _P203

ikadub [295]

Answer:

Explanation:

If you insist on filling in the first blank you can put a one.

3 0

3 years ago