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3 years ago

Specialized proteins which function as catalysts for organic reactions are:

Chemistry

2 answers:

Talja [164]3 years ago

5 0

Answer: enzymes

Explanation:

Enzymes are the special protein molecules present in the cells of the living organism. Enzymes do not participate as reactants in a chemical reaction instead they speed up the chemical reaction. This is the catalytic property of the enzyme. A substrate binds to the active site of the enzyme and gets converted into product. They play important role in respiration, digesting food, muscle and nerve function.

On the basis of the above description, Specialized proteins which function as catalysts for organic reactions are: enzymes.

grigory [225]3 years ago

4 0

Specialized proteins which function as catalysts for organic reactions are enzymes.

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A solution is 40.00% by volume benzene (C6H6) in carbon tetrachloride at 20°C. The vapor pressure of pure benzene at this temper

finlep [7]

Answer:

The total vapor pressure is 84.29 mmHg

Explanation:

Step 1: Data given

Solution = 40.00 (v/v) % benzene in CCl4

Temperature = 20.00 °C

The vapor pressure of pure benzene at 20.00 °C = 74.61 mmHg

Density of benzene is 0.87865 g/cm3

The vapor pressure of pure carbon tetrachloride is 91.32 mmHg

We suppose the total volume = 100 mL

Step 2: Calculate volume benzene and CCl4

40 % benzene = 40 mL

60 % mL CCl4 = 60 mL

Step 3: Calculate mass benzene

Mass = density * volume

Mass of benzene = 40.00 mL * 0.87865 g/mL = 35.146 g

Step 4: Calculate moles of benzene

Moles = mass / molar mass

Number of moles of benzene = 35.146 grams / 78 g/mol = 0.45059 mol

Step 5: Calculate mass of CCl4

Mass of CCl4 = 60 mL * 1.5940 g/mL = 95.64 g

Step 6: Calculate moles CCl4

Number of moles of CCl4 = 95.64 grams / 154g/mol = 0.62104 mol

Step 7: Calculate total number of moles

Total number of moles = moles benzene + moles CCl4

0.45059 moles + 0.62104 moles = 1.07163 mol

Step 8: Calculate mole fraction benzene and CCl4

Mole fraction = moles benzene / total moles

Mole fraction of benzene = 0.45059 / 1.07163 = 0.4205

Mole fraction of CCl4 = 0.62104 / 1.07163 = 0.5795

Step 9: Calculate partial pressure

Partial pressure of benzene = 0.4205 * 74.61 = 31.37 mmHg

Partial pressure of CCl4 = 0.5795 * 91.32 = 52.92 mmHg

Total vapor pressure = 31.37 + 52.92 = 84.29 mmHg

The total vapor pressure is 84.29 mmHg

7 0

3 years ago

How many significant figures are in 195.01

Masja [62]

Answer:

difficult to answer, need a bit more detail

5 0

3 years ago

When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin

loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.2^0C$ = Depression in freezing point

$K_f$ = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality = $\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9$

$8.2^0C=1\times K_f\times 1.9$

$K_f=4.32^0C/m$

Now Depression in freezing point for sodium chloride is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=20.0^0C$ = Depression in freezing point

$K_f$ = freezing point constant

m= molality = $\frac{136g\times 1000}{950g\times 58.5g/mol}=2.45$

$20.0^0C=i\times 4.32^0C\times 2.45$

$i=1.9$

Thus vant hoff factor for sodium chloride in X is 1.9

3 0

2 years ago

A 42.0g sample of compound containing only C and H was analyzed. The results showed that the sample contained 36.0g of C and 6.0

user100 [1]

Answer:

What is the empirical formula of the compound?

Explanation:

When the relative masses of elements in a hydrocarbon are given, it is possible to use this information to obtain the empirical formula by dividing the given masses of each element by the relative atomic masses of the element. The lowest ratio is now used to divide through to obtain the empirical formula of the compound.

The empirical formula only shows that ratio of atoms of each element present in the compound. From the information provided, the empirical formula of the compound is CH2. Hence the answer.

5 0

3 years ago

How many grams of hydrogen chloride can be produced from 1.00g of hydrogen and 55.0g of chlorine? what is the limiting reactant?

zloy xaker [14]

Answer:

$m_{HCl}=36.1gHCl$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the required grams of HCl by firstly identifying the limiting reactant via the moles of each reactant as they are in a 1:1 mole ratio:

$n_{H_2}=1.00gH_2*\frac{1molH_2}{2.02gH_2}=0.500molH_2\\\\ n_{Cl_2}=55.0gCl_2*\frac{1molCl_2}{70.9gCl_2}=0.776molCl_2$

Thus, we infer the hydrogen is the limiting reactant and therefore we use its 1:2 mole ratio with HCl whose molar mass is 36.46 g/mol:

$m_{HCl}=0.500molH_2*\frac{2molHCl}{1molH_2}*\frac{36.46gHCl}{1molHCl}\\\\m_{HCl}=36.1gHCl$

Regards!

4 0

2 years ago