The lines in an element's atomic spectrum result from

4. Finally, convert hemoglobin concentration from g/dL to millimoles per Liter (milliMolar or mM). Hemoglobin has a molecular we

prohojiy [21]

To make 1 Molar solution of hemoglobin ; 1600 grams of hemoglobin will be dissolved in 1 liter of water

The molecular weight of Hemoglobin is approximately 16,000 Daltons, when hemoglobin is converted to mM

16000 Dalton = 16000 ( g/mol )

given that 1 Dalton = 1 g/mol

To make 1 molar solution of hemoglobin using 1 liter of water

1 liter = 1000 grams

16000 Dalton = 16000 g/mol

Hence 16,000 grams of Hemoglobin is required to make 1 Molar solution of hemoglobin using 1 liter of water.

learn more : brainly.com/question/23517096

8 0

2 years ago

For the reaction, 2A + B → 3C, rate of disappearance of A is 0.3 M/s in a certain time interval.What is the average rate of appe

andrezito [222]

Answer:

E) rate of appearance of C = 0.45 M/s

rate of the reaction = 0.15 M/s

Explanation:

2A + B → 3C

Writing rate law for the reaction:

Rate of reaction = - $\frac{1}{2}$ $\frac{d[A]}{dt}$ = - $\frac{d[B]}{dt}$ = $\frac{1}{3}$ $\frac{d[C]}{dt}$ → equation 1

Given that the rate of disappearance of A is 0.3 M/s

⇒ - $\frac{d[A]}{dt}$ = 0.3 M/s

⇒Rate of reaction = - $\frac{1}{2}$ $\frac{d[A]}{dt}$ = $\frac{1}{2}$ ×0.3 M/s

⇒Rate of reaction = 0.15 M/s

From equation 1, $\frac{d[C]}{dt}$ = - $\frac{3}{2}$ $\frac{d[A]}{dt}$ = $\frac{3}{2}$ ×0.3 M/s

⇒ $\frac{d[C]}{dt}$ = 0.45 M/s

or the rate of appearance of C = 0.45 M/s

7 0

3 years ago

Part B Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution

MissTica

Answer:

0,508g of H₂O₂

Explanation:

For the reaction:

2KMnO₄(aq) + H₂O₂(aq) + 3H₂SO₄(aq) → 3O₂(g) + 2MnSO₄(aq) + K₂SO₄(aq) + 4H₂O(l)

2 moles of KMnO₄ react with 1 mol of H₂O₂.

In the titration, moles of KMnO₄ required were:

1,68M×0,0178L = 0,0299 moles of KMnO₄. Moles of H₂O₂ are:

0,0299 moles of KMnO₄× $\frac{1molH_{2}O_{2}}{2molKMnO_{4}}$ = 0,01495 moles of H₂O₂. As molar mass of H₂O₂ is 34,01g/mol, mass of H₂O₂ was dissolved is:

0,01495 moles of H₂O₂× $\frac{34,01g}{1molH_{2}O_{2}}$ = 0,508g of H₂O₂

5 0

2 years ago

Caffeine, a stimulant found in coffee and soda, hasthe mass percent composition: C. 49.48%, H, 5.19%. N. 28.85% 0. 16.48% The mo

borishaifa [10]

We have the next % composition:

C. 49.48%

H, 5.19%.

N. 28.85%

0. 16.48%

We assume 100 g of sample

1) As we have 100 g of sample of Caffeine, we calculate the mass of each element involved here.

C. 49.48 g

H, 5.19 g

N. 28.85 g

0. 16.48 g

2) We calculate the number of moles of each element (we need the mass per mole of each element)

For C) 12.01 g/mol

49.48 g x (1 mol/12.01 g) = 4.120 moles

For H) 1.007 g/mol

5.19 g x (1 mol/1.007 g) = 5.154 moles

For O) 15.99 g/mol

16.48 g x (1 mol/15.99 g) = 1.030 moles

For N) 14.00 g/mol

28.85 g x (1 mol/14.00 g) = 2.060 moles

3) We choose the smallest number from 2) and divide the rest of them by it.

For C) 4.120 moles/1.030 moles= 4

For H) 5.154 moles/1.030 moles= 5

For O) 1.030 moles/1.030 moles= 1

For N) 2.060 moles/1.030 moles= 2

4) The numbers in 3) represents the subindex from the empirical formula of caffeine:

$C_4H_5O_1N_2$

5) We calculate the molar mass of our empirical formula, 97.06 g/mol.

We already have the molar mass of the molecular formula, so we proceed like this:

n= the molar mass of the molecular formula/the molar mass of the empirical formula

n = 194.19 g/mol/97.06 g/mol = 2 approx.

We use "n" and we multiply our empirical formula by n = 2:

Therefore, our molecular formula:

$C_8H_{10}O_2N_4$

8 0

1 year ago

How many significant figures are in 820 400.0 L

raketka [301]

Answer:

7 significant numbers

7 0

2 years ago