A solution is made by mixing 12.54 grams of NaNO3 with enough water to make 2.5 L of solution. What is the molarity?

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate

vovangra [49]

Answer:

K2 = 61.2 M^-1.S^-1

Explanation:

We complete the question fully:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mol . If the rate constant of this reaction is 6.7M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

Answer is as follows:

The question asks us to calculate the value of the rate constant at a certain temperature, given that it is at a particular value for a particular temperature. We solve the question as follows:

According to Arrhenius equation, the relationship between temperature and activation energy is as follows:

k = Ae^-(Ea/RT)

where, k = rate constant

A = pre-exponential factor

Ea = activation energy

R = gas constant

T = temperature in kelvin

From the equation, the following was derived for a double temperature problem:

ln(k2/k1) = (-Ea/R) * (1/T1 - 1/T2)

We list out the parameters as follows:

T1= (244 + 273.15) K = 517.15 K

T2= (324+ 273.15) K =597.15 K

K1 = 6.7 , K2 = ?

R = 8.314 J/mol K

Ea = 71.0 kJ/mol = 71000 J/mol

Putting the given values into the above formula as follows:

ln(k2/6.7) = (-71000/8.314) * (1/517.15 - 1/597.15)

lnk2 - 1.902 = 8539.8 * 0.000259

lnK2 = 1.902 + 2.21

lnK2 = 4.114

K2 = e^(4.114)

K2 = 61.2

Hence, K2 = 61.2 (M.S)^-1

7 0

4 years ago

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What is the mass of a liquid that has a volume of 50.0ml and a density of 2.40g/ml?

Iteru [2.4K]

Answer:

120g Using Density Equation.

Explanation:

Density =mass/volume

We need to the solve for the mass.

Mass=Density * Volume

Mass= 2.40 g/ml * 50.0 ml = 120g

5 0

3 years ago

How many atoms of oxygen are present in 7.51 grams of<br> glycine with formula C₂H5O2N?

Blizzard [7]

1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.

How to calculate number of atoms?

The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

However, the number of moles of oxygen in glycine can be calculated using the following expression:

Molar mass of C₂H5O2N = 75.07g/mol

Mass of oxygen in glycine = 32g/mol

Hence; 32/75.07 × 7.51 = 3.2grams of oxygen in glycine

Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles

Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms

Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.

Learn more about number of atoms at: brainly.com/question/8834373

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3 0

2 years ago

Iron is biologically important in the transport of oxygen by red blood cells from the lungs to the various organs of the body. I

Aneli [31]

Answer : The number of iron atoms present in each red blood cell are, $1.077\times 10^9$

Explanation :

First we have to calculate the moles of iron.

$\text{Moles of iron}=\frac{\text{Mass of iron}}{\text{Molar mass of iron}}=\frac{2.90g}{55.85g/mole}=0.0519moles$

Now we have to calculate the number of iron atoms.

As, 1 mole of iron contains $6.022\times 10^{23}$ number of iron atoms

So, 0.0519 mole of iron contains $0.0519\times 6.022\times 10^{23}=3.125\times 10^{22}$ number of iron atoms

Now we have to calculate the number of iron atoms are present in each red blood cell.

Number of iron atoms are present in each red blood cell = $\frac{\text{Number of iron atoms}}{\text{Total number of red blood cells}}$

Number of iron atoms are present in each red blood cell = $\frac{3.125\times 10^{22}}{2.90\times 10^{13}}$

Number of iron atoms are present in each red blood cell = $1.077\times 10^9$

Therefore, the number of iron atoms present in each red blood cell are, $1.077\times 10^9$

6 0

3 years ago

The combustion reaction of isopropyl alcohol is given below: C 3 H 7 O H ( l ) + 9 2 O 2 ( g ) → 3 C O 2 ( g ) + 4 H 2 O ( g ) T

jek_recluse [69]

Answer:

the heat of formation of isopropyl alcohol is -317.82 kJ/mol

Explanation:

The heat of combustion of isopropyl alcohol is given as follows;

C₃H₇OH (l) +(9/2)O₂ → 3CO₂(g) + 4H₂O (g)

The heat of combustion of CO₂ and H₂O are given as follows

C (s) + O₂ (g) → CO₂(g) = −393.50 kJ

H₂ (g) + 1/2·O₂(g) → H₂O (l) = −285.83 kJ

Therefore we have

3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ which we can write as

3C (s) + 3O₂ (g) → 3CO₂(g) = −393.50 kJ × 3 =

4H₂ (g) + 2·O₂(g) → 4H₂O (l) = −285.83 kJ × 4

3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ = +2006 kJ/mol

-1180.5 - 1143.32 +2006 = -317.82 kJ/mol

Therefore, the heat of formation of isopropyl alcohol = -317.82 kJ/mol.

3 0

4 years ago