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3 years ago

A certain protein was found to contain 0.292% manganese by mass. Determine the minimum molecular mass of the protein.

1 answer:

4 0

$55u \ \ \ \ \ \ \Rightarrow \ \ \ \ \ 0,292\%\\
x \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ 100\%\\\\
x = \frac{55u*100\%}{0,292\%}\approx18835,62u$

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Use this equation for the following problems: 2NaN3 --> 2Na+3N2

olchik [2.2K]

Answer:

1) 65.0

2) 16.434 L = 16434 mL.

Explanation:

2NaN₃ → 2Na + 3N₂,

It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂.

23.6 L of N₂ contains → ??? g of N₂.

∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.

We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.

We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

using cross multiplication:

2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

??? mol of NaN₃ produce → 0.775 moles of N₂.

∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.

Finally, we can get the grams of NaN₃ needed:

mass = no. of moles x molar mass = (0.517 mol)(65.0 g/mol) = 33.6 g.

Q2: How many mL of N₂ result if 8.3 g Na are also produced?

We need to get the no. of moles of 8.3 g Na using the relation:

n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.

We can get the no. of moles of N₂ produced with 0.36 mol of Na:

using cross multiplication:

2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

0.36 moles of Na produced with → ??? moles of N₂.

∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.

We can get the mass of 0.54 mol of N₂:

mass = no. of moles x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.

Now, we can get the mL of 15.12 g of N₂:

using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

??? L of N₂ contains → 15.12 g of N₂.

∴ The volume of N₂ result = (1.0 L)(15.12 g)/(0.92 g) = 16.434 L = 16434 mL.

4 0

2 years ago

Balance the equations

slavikrds [6]

4. 4 C + S8 = 4 CS2

5. H2 + O2 = H2O2 is already balanced.

6. 4 Na + O2 = 2 Na2O

7 0

3 years ago

Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. Precision pertains

Eduardwww [97]

D.) closeness of a measurement to an accepted value
That is accuracy not precision.

6 0

3 years ago

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Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions:

cupoosta [38]

MgO
MgI2
PbO2
PbI4

These are the possible compounds

4 0

3 years ago

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Effective nuclear charge, Zeff, is defined as:

Alborosie

Answer:

The effective nuclear charge for a valence electron in oxygen atom: $Z_{eff} = 4.55$

Explanation:

Effective nuclear charge $[Z_{eff}]$ is the net nuclear charge experienced by the electron in a given atom. It is always less than the actual charge of the nucleus [Z], due to shielding by electrons in the inner shells.

It is equal to the difference between the actual nuclear charge or the atomic number (Z) and the shielding constant (s).

$\Rightarrow Z_{eff} = Z - s$

For an oxygen atom-

Electron configuration: (1s²) (2s² 2p⁴)

The atomic number (actual nuclear charge): Z = 8

The shielding constant (s) for a valence electron can be calculated by using the Slater's rules:

⇒ s = 5 × 0.35 + 2 × 0.85 = 1.75 + 1.7 = 3.45

Therefore, the effective nuclear charge for a valence electron in oxygen atom is:

$Z_{eff} = Z - s = 8 - 3.45 = 4.55$

Therefore, the effective nuclear charge for a valence electron in oxygen atom: $Z_{eff} = 4.55$

6 0

2 years ago