CHEMISTRY COLLEGE

4. Oxygenating myoglobin. The myoglobin content of some human muscles is about 8 g kg 1. In sperm whale, the myoglobin content of muscle is about 80 g kg 1 . (a) How much O 2 is bound to myoglobin in human muscle and in sperm whale muscle? Assume that the myoglobin is saturated with O 2, and that the molecular weights of human and sperm whale myoglobin are the same. Berg, Jeremy M.. Biochemistry (p. 213). W. H. Freeman. Kindle Edition.

Answers

Answer 1

Answer:

Answer:

0.01454 grams of oxygen is present per kilogram of human muscle.

0.1454 grams of oxygen is present per kilogram of sperm whale muscle.

Explanation:

Given : The myoglobin is fully saturated with oxygen gas.

Moles of myoglobin = Moles of oxygen gas

Molecular mass of myoglobin = 17,600 g/mol

(assumed same fro whale and human )

Myoglobin content in humans = 8 g/kg

Mole of molyoglobin in human muscles per kg : $(8 g/kg)/(17,600 g/mol)$

Mass of oxygen present in per kg of human muscle:

Moles of oxygen gas × 32g/mol :

$(8 g/kg)/(17,600 g/mol)* 32 g/mol=0.01454$ g/kg of human muscle

Myoglobin content in whales= 80 g/kg

Mole of molyoglobin in whale muscles per kg : $(80 g/kg)/(17,600 g/mol)$

Moles of myoglobin = Moles of oxygen gas

Mass of oxygen present in per kg of sperm whale muscle:

Moles of oxygen gas × 32g/mol :

$(80 g/kg)/(17,600 g/mol)* 32 g/mol=0.1454$ g/kg of whale muscle

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Which is the most valid reason a group of people might oppose the storage of spent fuel rods in their community?

Answers

Answer:

The most likely reason is that spent fuel bars can infect the community with radiation causing numerous health problems.

Explanation:

A spent fuel is a fuel that has been exposed to irradiation in a nuclear reactor. This fuel expels radiation and this is one of the main reasons why a group of people oppose the storage of spent fuel bars in their community. This is because through this fuel can occur a leak of radiation contaminating the community and generating major health problems.

The fuel rods will remain radioactive for a long time....hope this helps

Balance the equation: KL+Pb(NO3)2→PbL2+KNO3

Answers

is that Lr? I'm not sure what the L one stand for?
2:1:1:2

A 230 kg steel crate is being pushed along a cement floor. The force of friction is 480 N to the left and the applied force is 1860 N to the right. The forces acting on the crate are

Answers

are radioactive waves , hope this helps

How much heat is transferred when 7.19 grams of H2 reacts with excess nitrogen, according to the following equation: N2(g) + 3 H2 (g) --> 2 NH3 (g) \DeltaΔH = +46.2 kJ

Answers

Answer:

$Q=54.8kJ$

Explanation:

Hello there!

In this case, according to the given chemical reaction, it is possible for us to realize that the 46.2 kJ of energy are given per mole of reaction, which are related to 3 moles of hydrogen; Thus, we can calculate the energy per mole of hydrogen as shown below:

$\Delta H=(46.2kJ)/(mol) *(1mol)/(3molH_2)\n\n \Delta H=15.4(kJ)/(molH_2)$

Now, to calculate the total energy, we convert the grams to moles of hydrogen as shown below:

$Q=7.19gH_2*(1molH_2)/(2.02gmolH_2)*15.4(kJ)/(molH_2) \n\nQ=54.8kJ$

Best regards!

What will be the theoretical yield of tungsten(is) ,W, if 45.0 g of WO3 combines as completely as possible with 1.50 g of H2

Answers

Answer:

35.6 g of W, is the theoretical yield

Explanation:

This is the reaction

WO₃ + 3H₂ → 3H₂O + W

Let's determine the limiting reactant:

Mass / molar mass = moles

45 g / 231.84 g/mol = 0.194 moles

1.50 g / 2 g/mol = 0.75 moles

Ratio is 1:3. 1 mol of tungsten(VI) oxide needs 3 moles of hydrogen to react.

Let's make rules of three:

1 mol of tungsten(VI) oxide needs 3 moles of H₂

Then 0.194 moles of tungsten(VI) oxide would need (0.194 .3) /1 = 0.582 moles (I have 0.75 moles of H₂, so the H₂ is my excess.. Then, the limiting is the tungsten(VI) oxide)

3 moles of H₂ need 1 mol of WO₃ to react

0.75 moles of H₂ would need (0.75 . 1)/3 = 0.25 moles

It's ok. I do not have enough WO₃.

Finally, the ratio is 1:1 (WO₃ - W), so 0.194 moles of WO₃ will produce the same amount of W.

Let's convert the moles to mass (molar mass . mol)

0.194 mol . 183.84 g/mol = 35.6 g

Exactly 1.5 g of a fuel burns under conditions of constant pressure and then again under conditions of constant volume. In measurement A the reaction produces 25.9 kJ of heat , and in measurement B the reaction produces 23.3 kJ of heat. Which measurement (A or B) corresponds to conditions of constant pressure? Which one corresponds to conditions of constant volume? Explain.

Answers

Answer:

Process B : constant pressure condition

Process A : constant volume condition

Explanation:

In case of constant pressure, some of the energy is used to do work on the surrounding to keep pressure constant. Due to this, the total heat energy is less than in case of constant volume. In Case of constant Volume all of heat is available, produced in reaction because no work is done.

If we look at our data,we will find that process B has energy 23.3 KJ which is less than process A, the energy of which is 25.9 KJ. It means Process B is occurred at constant pressure condition and Process A has occurred at constant volume condition

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