What is the mass of a neutron???? PLEASE HELP

A chemical reaction produces 653 550 kj of energy as heat in 142.3min. calculate the rate of energy transferred in kj per minute

miskamm [114]

We have the value of

Total energy produced in the chemical reaction=653 550 KJ

Time needed=142.3min

To calculate the rate of energy transfer, that is the amount of energy produced per minute.

Rate of energy transfer= $\frac{Total energy produced}{Time needed}$

= $\frac{653 550}{142.3}$

=4592.76 KJ min⁻¹

So, the rate of energy transfer is 4592.76 KJ min⁻¹.

6 0

3 years ago

Calculate the mass percent of copper in cus, copper(ii) sulfide. if you wish to obtain 10.0g of copper metal from copper(ii) sul

Aleksandr [31]

Using the relative atomic weights of both copper and sulfur ie copper = 63.55 and sulfur is 32.06 so 63.55+32.06=95.56 total mass and so of this, copper = 63.55/95.56=66.4%. So to get 10 grams of copper, use the formula 10g=66.4%xCuS so CuS=10/0.664=15.06 grams of CuS.

4 0

3 years ago

A covalent bond formed when 2 atoms share 6 valence electrons between them is a ______ bond

sleet_krkn [62]

COVALENT BOND IS THE BOND EXISTING BETWEEN 2 ATOMS THAT SHARE 6 ELECTRONS

5 0

2 years ago

There is a gas at 780 mm of Hg, in a volume of 5 liters and a temperature of 37 C, the volume is changed to 5.5 liters and the

HACTEHA [7]

Answer:

32.8 C

Explanation:

- Use combined gas law formula and rearrange.

- Hope that helped! Please let me know if you need further explanation.

6 0

3 years ago

The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal

spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

5 0

3 years ago