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3 years ago

More than three fourths of the air we breath

Chemistry

1 answer:

o-na [289]3 years ago

5 0

Answer:

Is made up of nitrogen oxygen carbon dioxide water vapor and many other gases as well as particles of liquids and solids. is the most abundant gas in the atmosphere. It makes up more than three fourths of the air we breathe.

Explanation:

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Suppose you start with one liter of vinegar and repeatedly remove 0.08 L, replace with water, mix, and repeat. a. Find a formula

Bumek [7]

Answer:

0.92^n

Explanation:

Given that :

Initial amount of vinegar = 1 Litre

Number of litres removed repeatedly = 0.08 Litre

Since the amount removed each time is constant, then ;

Initial % = 100% = 100/100 = 1

. Using the relation :

Amount of vinegar in mixture :

Initial * (1 - amount removed / initial amount)^n

n = number of times repeated

1 * (1 - 0.08/1)^n

1 * (1 - 0.08)^n

1 * 0.92^n

Hence,

For nth removal,

Concentration will be :

0.92^n ; for n ≥ 1

7 0

2 years ago

Problem 12.002 the molar analysis of a gas mixture at 30°c, 2 bar is 40% n2, 50% co2, 10% ch4. determine

andreev551 [17]

The problem is incomplete. However, there can only be two probable questions for this problem. First, you can be asked the individual partial pressures of each gas. Second, you can be asked the volume occupied by each gas. I can answer both cases for you.

1.

Let's assume ideal gas.
Pressure for N₂: 2 bar*0.4 = 0.8 bar
Pressure for CO₂: 2 bar*0.5 = 1 bar
Pressure for CH₄: 2 bar*0.1 = 0.2 bar

2. For the volume, let's find the total volume first.

V = nRT/P = (1 mol)(8.314 J/mol-K)(30 +273 K)/(2 bar*10⁵ Pa/1 bar)
V = 0.0126 m³
Hence,
Volume for N₂: 0.0126 bar*0.4 = 0.00504 m³
Volume for CO₂: 0.0126*0.5 = 0.0063 m³
Volume for CH₄: 0.0126*0.1 = 0.00126 m³

3 0

3 years ago

In glycolysis, if glucose is labeled at the carbon 6 position (see page 1 for numbering of carbons in glucose) A) the carbon wit

Oliga [24]

Answer:

D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled.

Explanation:

Glycolysis has 2 phase (1) preparatory phase (2) pay-off phase.

<u>(1) Preparatory phase</u>

During preparatory phase glucose is converted into fructose-1,6-bisphosphate. Till this time the carbon numbering remains the same i.e. if we will label carbon at 6th position of glucose, its position will remian the same in fructose-1,6-bisphosphate that means the labeled carbon will still remain at 6th position.

When fructose-1,6-bisphosphate is further catalyzed with the help of enzyme aldolase it is cleaved into two 3 carbon intermediates which are glyceraldehyde 3-phosphate (GAP) and dihyroxyacetone phosphate (DHAP). In this conversion, the first three carbons of fructose-1,6-bisphosphate become carbons of DHAP while the last three carbons of fructose-1,6-bisphosphate will become carbons of GAP. It simply means that GAP will acquire the last carbon of fructose-1,6-bisphosphate which is labeled. Now the last carbon of GAP which has phosphate will be labeled.

<u>(2) Pay-off phase</u>

During this phase, GAP is dehydrogenated into 1,3-bisphosphoglycerate (BPG) with the help of enzyme glyceraldehyde 3-phosphate dehydrogenase. This oxidation is coupled to phosphorylation of C1 of GAP and this is the reason why 1,3-bisphosphoglycerate has phosphates at 2 positions i.e. at position 1 in which phosphate is newly added and position 3rd which already had labeled carbon.

It is pertinent to mention here that<u> BPG has a mixed anhydride and the bond at C1 is a very high energy bond.</u> In the next step, this high energy bond is hydrolyzed into a carboxylic acid with the help of enzyme phosphoglycerate kinase and the final product is 3-phosphoglycerate. Hence, the carbon with low energy phosphate i.e. the carbon at 3rd position remains labeled.

3 0

3 years ago

For the following reaction, 6.94 grams of water are mixed with excess sulfur dioxide . Assume that the percent yield of sulfurou

Alexxx [7]

<h3>Answer:</h3>

#a. Theoretical yield = 31.6 g

#b. Actual yield = 25.72 g

<h3>Explanation:</h3>

The equation for the reaction between sulfur dioxide and water to form sulfurous acid is given by the equation;

SO₂(g) + H₂O(l) → H₂SO₃(aq)

The percent yield of H₂SO₃ is 81.4%

Mass of water that reacted is 6.94 g

#a. To get the theoretical yield of H₂SO₃ we need to follow the following steps

Step 1: Calculate the moles of water

Molar mass of water = 18.02 g/mol

Mass of water = 6.94 g

But, moles = Mass/molar mass

Moles of water = 6.94 g ÷ 18.02 g/mol

= 0.385 mol

Step 2: Calculate moles of H₂SO₃

From the equation, the mole ratio of water to H₂SO₃ is 1 : 1

Therefore, moles of water = moles of H₂SO₃

Hence, moles of H₂SO₃ = 0.385 mol

Step 3: Theoretical mass of H₂SO₃

Mass = moles × Molar mass

Molar mass of H₂SO₃ = 82.08 g/mol

Number of moles of H₂SO₃ = 0.385 mol

Therefore;

Theoretical mass of H₂SO₃ = 0.385 mol × 82.08 g/mol

= 31.60 g

Thus, the theoretical yield of H₂SO₃ is 31.6 g

<h3>#b. Calculating the actual yield</h3>

We need to calculate the actual yield

Percent yield of H₂SO₃ is 81.4%

Theoretical yield is 31.60 g

But; Percent yield = (Actual yield/theoretical yield)×100

Therefore;

Actual yield = Percent yield × theoretical yield)÷ 100

= (81.4 % × 31.6) ÷ 100

= 25.72 g

The percent yield of H₂SO₃ is 25.72 g

6 0

3 years ago

Please helppp asaappppppp

Zanzabum

Answer:

B. s, p, d, f

Explanation:

These things are often referred to as suborbitals and you normally have s,p,d,f.

S has 1 two orbitals

P has 3 orbitals

D has 5 orbitals

F has 7 orbitals

and each orbital can house 2 electrons

8 0

3 years ago