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4 years ago

Describe the difference between the normal and mutated hemoglobin amino acid sequences:

Chemistry

1 answer:

astraxan [27]4 years ago

3 0

Answer:

The chain of colored boxes represent the first eight amino acids in the beta chain of hemoglobin. The sixth position in the normal beta chain has glutamic acid, while sickle beta chain has valine. This is the sole difference between the two

Explanation:

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A box has a volume of 45m3 and is filled with air held at 25∘C and 3.65atm. What will be the pressure (in atmospheres) if the sa

Marina CMI [18]

Answer:

Given:

Initial pressure: $3.65\; \rm atm$ .
Volume was reduced from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ .
Temperature was raised from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ .

New pressure: approximately $3.4\times 10\; \rm atm$ ( $34\; \rm atm$ .) (Assuming that the gas is an ideal gas.)

Explanation:

Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:

Reduce the volume of the gas from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ . Calculate the new pressure, $P_1$ .
Raise the temperature of the gas from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ . Calculate the final pressure, $P_2$ .

By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)

For this gas, $V_0 = 45\; \rm m^{3}$ while $V_1 = 5.0\; \rm m^{3}$ .

Let $P_0$ denote the pressure of this gas before the volume change ( $P_0 = 3.65\; \rm atm$ .) Let $P_1$ denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between $P_1\!$ and $P_0\!$ :

$\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0$ .

In other words, because the final volume is $(1/9)$ of the initial volume, the final pressure is $9$ times the initial pressure. Therefore:

$\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm$ .

On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)

Convert the unit of the temperature of this gas to degrees Kelvins:

$T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K$ .

$T_2 = (35 + 273.15)\; \rm K = 308.15\; \rm K$ .

Let $P_1$ denote the pressure of this gas before this temperature change ( $P_1 = 32.85\; \rm atm$ .) Let $P_2$ denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at $V_2 = V_1 = 5.0\; \rm m^{3}$ .

Apply Amonton's Law to find the ratio between $P_2$ and $P_1$ :

$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}$ .

Calculate $P_2$ , the final pressure of this gas:

$\begin{aligned} P_2 &= \frac{308.15\; \rm K}{298.15\; \rm K} \times P_1 \\ &= \frac{308.15\; \rm K}{298.15\; \rm K} \times 32.85\; \rm atm \approx 3.4 \times 10\; \rm atm\end{aligned}$ .

In other words, the pressure of this gas after the volume and the temperature changes would be approximately $3.4\times 10\; \rm atm$ .

8 0

3 years ago

How many moles of mercury are produced if 74g of oxygen are produced

Maurinko [17]

<span>4.6 moles Hg should be the answer. Hope this helps!</span>

6 0

3 years ago

The overall cell reaction occurring in an alkaline battery isZn(s) + MnO₂(s) + H₂O(l) → ZnO(s) + Mn(OH)₂(s) (e) In practice, vol

zubka84 [21]

b) Mass of MnO₂ = 5.981 g

Mass of H₂O = 1.2384 g

c) Total Mass of Reactant consumed = 11.708 g

b) Given Reaction

Zn(s) + MnO₂(s) + H₂O(l) → ZnO(s) + Mn(OH)₂(s)

Mass of Zn = 4.50 g

Moles of Zn = 0.0688 moles

Now,

Moles of Zn = moles of MnO₂ = moles of H₂O = moles of ZnO = moles of Mn(OH)₂

Hence ,

Moles of MnO₂ = 0.0688 moles

Mass of MnO₂ = 0.0688 × 86.9368 g

= 5.981 g

Similarly,

Moles of H₂O = 0.0688 moles

Mass of H₂O = 0.0688 × 18 g

= 1.2384 g

c) now ,

Moles of ZnO = 0.0688 moles

Mass of ZnO = 0.06880× 81.3794 g

= 5.598 g

Moles of Mn(OH)₂ = 0.0688 moles

Mass of Mn(OH)₂ =0.0688 × 88.952 g

= 6.11 g

Total mass of Product = 11.708 g

Total Mass of Reactant = 11.715 g

Hence,

Total mass of reactant consumed = 11.708 g

c) As total mass of reactant is more than that of mass of reactant consumed , Hence G is more than that of mass of reactant consumed .

G = - nFEcell

and no. of moles of reactant is greater than that of number of moles of reactant consumed .

Hence voltaic cell of given Capacity are heavier than that of mass of reactant consumed .

Thus from above conclusion we can say that , Mass of the reactant consumed is 11.708 g.

Learn more about Galvanic Cell here : brainly.com/question/19340007

#SPJ4

3 0

2 years ago

Which person do you agree with the most?

ki77a [65]

I agree with Allayna

3 0

3 years ago

Read 2 more answers

what is an empirical formulaWhat is the percent composition by mass of nitrogen in (NH4)2CO3 (gram-formula mass = 96.0 g/mol)?

Margarita [4]

Answer:

Percentage composition = 14.583%

Explanation:

In chemistry, the emprical formular of a compound is the simplest formular a compound can have. It shows the simplest ratio in which the elements are combined in the compound.

Percentage composition by mass of Nitrogen

Nitrogen = 14g/mol

In one mole of the compound;

Mass of Nitrogen = 1 mol * 14g/mol = 14g

Mass of compound = 1 mol * 96.0 g/mol = 96

Percentage composition of Nitrogen = (Mass of Nitrogen / Mass of compound) * 100

percentage composition = 14/96 * 100

Percentage composition = 0.14583 * 100

Percentage composition = 14.583%

3 0

3 years ago