Approximately what percentage of a human's DNA is used for coding Information for the synthesis of proteins?

Naddika [18.5K]

Answer:

63.25 grams of CO₂

Explanation:

To convert from liters to grams, we first need to convert from liters to moles. To do this, we divide the liters by 22.4, the amount of liters of a gas per mole.

32.2 / 22.4

= 1.4375 moles of CO₂

Now we want to convert from moles to grams. To do this, we multiply the moles by the molar mass of CO₂. The total molar mass can be found on the periodic table by adding up the molar mass of carbon (12) and two oxygen (32).

12 + 32 = 44

Now we want to multiply the moles by the molar mass.

1.4375 • 44

= 63.25 grams of CO₂

This is your answer.

Hope this helps!

7 0

3 years ago

In a certain chemical reaction Compound A combines with Compound B to produce Compound C (and no other products). Measurements w

Karo-lina-s [1.5K]

Answer:

Theoretical yield of C = 7.9 g

Actual amount of C isolated = 5.0 g

Explanation:

Given that in the reaction, Compound A combines with Compound B t o produce Compound C and no other products, the reaction is a combination reaction. The equation of the reaction is given below:

A + B ---> C

Part A:

From the measurement of the masses of A and B present before and after the reaction, the following values were obtained:

Initial mass of A = 4.0 g

Final mass of A = 0.0 g

Mass of A used up + 4.0 - 0.0 = 4.0 g

Initial mass of B = 9.0 g

Final mass of B = 5.1 g

Mass of B used up = 9.0 - 5.1 = 3.9 g

According to the law of conservation of mass, matter can neither be created nor destroyed but can change from one form to another. Therefore, theoretical yield of C equal the sum of the masses of A and B used up.

Theoretical yield of C = 4.0 + 3.9 = 7.9 g

Part B:

Given that the percent yield is 63.0 % = 0.63

Actual amount of C isolated = 0.63 * 7.9 = 5.0 g

5 0

3 years ago

H2(g) + F2(g)2HF(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.20 moles

abruzzese [7]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$H_2(g)+F_2(g)\rightarrow 2HF(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})]$

We are given:

$\Delta S^o_{(HF(g))}=173.78J/K.mol\\\Delta S^o_{(H_2)}=130.68J/K.mol\\\Delta S^o_{(F_2)}=202.78J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K

We are given:

Moles of hydrogen gas reacted = 2.20 moles

By Stoichiometry of the reaction:

When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K

So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = $\frac{-14.1}{1}\times 2.20=-31.02J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

7 0

3 years ago

Can anyone please tell me the answer?

cricket20 [7]

Answer:

1=4

2=2

3=hydrogen bonding

4=SO2 is reduced to Sulphur

Explanation:

<h2>I am trying my best okay.</h2>

8 0

3 years ago

In addition to mass balance, oxidation-reduction reactions must be balanced such that the number of electrons lost in the oxidat

Fiesta28 [93]

Answer:

Part A: (1, 1, 4, 1, 1, 1)

Part B: (2, 6, 4, 2, 3, 8)

Explanation:

Redox reactions can be balanced using the half-reaction method. It has the following steps:

We write both half-reactions (reduction and oxidation)
We balance the masses using H⁺ and H₂O in acidic media or OH⁻ and H₂O in basic media.
We add electrons to balance electrically the half-reaction
We multiply the half-reaction by numbers to make sure the number of electrons gained and lost are the same.
We add both half-reactions and take the numbers to the general equation.

<em>Acidic solution</em>

SO₄²⁻(aq) + Sn²⁺(aq) + X ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + Y

1.

Reduction: SO₄²⁻ ⇒ SO₃²⁻

Oxidation: Sn²⁺ ⇒ Sn⁴⁺

2.

2 H⁺ + SO₄²⁻ ⇒ SO₃²⁻ + H₂O

Sn²⁺ ⇒ Sn⁴⁺

3.

2 H⁺ + SO₄²⁻ + 2 e⁻ ⇒ SO₃²⁻ + H₂O

Sn²⁺ ⇒ Sn⁴⁺ + 2 e⁻

4.

1 x [2 H⁺ + SO₄²⁻ + 2 e⁻ ⇒ SO₃²⁻ + H₂O]

1 x [Sn²⁺ ⇒ Sn⁴⁺ + 2 e⁻]

5.

2 H⁺ + SO₄²⁻ + 2 e⁻ + Sn²⁺ ⇄ SO₃²⁻ + H₂O + Sn⁴⁺ + 2 e⁻

2 H⁺ + SO₄²⁻ + Sn²⁺ ⇄ SO₃²⁻ + H₂O + Sn⁴⁺

Taking this to the general equation:

SO₄²⁻(aq) + Sn²⁺(aq) + 2 H⁺(aq) ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O(l)

Since H⁺ are spectator ions, they are not balanced automatically through this method and we have to balance them manually. In this case, we need to add 2 more H⁺ to the left.

SO₄²⁻(aq) + Sn²⁺(aq) + 4 H⁺(aq) ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O(l)

<em>Basic solution</em>

MnO₄⁻(aq) + F⁻(aq) + X ⇄ MnO₂(s) + F₂(aq) + Y

1.

Reduction: MnO₄⁻ ⇒ MnO₂

Oxidation: F⁻ ⇒ F₂

2.

2 H₂O + MnO₄⁻ ⇒ MnO₂ + 4 OH⁻

2 F⁻ ⇒ F₂

3.

2 H₂O + MnO₄⁻ + 3 e⁻ ⇒ MnO₂ + 4 OH⁻

2 F⁻ ⇒ F₂ + 2 e⁻

4.

2 × (2 H₂O + MnO₄⁻ + 3 e⁻ ⇒ MnO₂ + 4 OH⁻)

3 × (2 F⁻ ⇒ F₂ + 2 e⁻)

5.

4 H₂O + 2 MnO₄⁻ + 6 e⁻ + 6 F⁻ ⇄ 2 MnO₂ + 8 OH⁻ + 3 F₂ + 6 e⁻

4 H₂O + 2 MnO₄⁻ + 6 F⁻ ⇄ 2 MnO₂ + 8 OH⁻ + 3 F₂

Taking this to the general equation:

2 MnO₄⁻(aq) + 6 F⁻(aq) + 4 H₂O ⇄ 2 MnO₂(s) + 3 F₂(aq) + 8 OH⁻

This equation is balanced.

6 0

3 years ago