Explain how a common housecat gets “worms.” No google or else i report you.

Compare the amount of heat required to vaporize a 200.-gram sample of H2O(ℓ) at its boiling point to the amount of heat required

stira [4]

The enthalpy of vaporization of H2O is higher than the enthalpy of fusion of H2O, therefore vaporizing the same mass of H2O would require more heat/energy than melting the same mass of H2O.

6 0

3 years ago

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r

rodikova [14]

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

= 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

$K=Ae^{\frac{-E}{RT}}$

If $K_1=Ae^{\frac{-E_1}{RT}} -------equation 1$ &

$K_2=Ae^{\frac{-E_2}{RT}} -------equation 2$

$\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}$

$E_1= E_2-RT*In(\frac{K_1}{K_2})$

$E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})$

$E_1 = 28957.39292$ $J/mol$

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0

3 years ago

Calculate the mass of khc8h4o4 that reacts with 15 ml of the naoh solution

dlinn [17]

The balanced chemical reaction would be:

KHC8H4O4 (aq) + NaOH (aq) → NaKC8H4O4 (aq) + H2O.

The concentration of the NaOH is equal 0.1 M. We use this and the volume given above to determine the mass of KHC8H4O4. We do as follows:

0.1 mol / L NaOH (.015 L) ( 1 mol KHC8H4O4 / 1 mol NaOH) (204 g / 1 mol) = 0.306 g KHC8H4O4

6 0

3 years ago

How are the terms molar mass and atomic mass different from one another

Ronch [10]

1.Molar mass is the mass of one mole per single element while atomic mass is the mass of an atom at rest or is the number of protons and neutrons.

2.Molar mass is measured in grams per mole while atomic mass is “unitless.”

3.Atomic mass is measured via mass spectrometry while molar mass is computed via atomic weight.

8 0

3 years ago

How many liters of carbon dioxide can be produced if 37.8 grams of carbon disulfide react with excess oxygen gas at 28.85 degree

alekssr [168]

37.8 grams of CS2 equals to 37.8/76=0.5 mole. So the products have 0.5 mole CO2 which is 11.2 liters at STP. So according to the gas law, the volume at given condition is 12.4 liters. So the answer should be 12.2 liters.

8 0

3 years ago