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antoniya [11.8K]
3 years ago
11

Aluminum has a density of 2.70 g/mL. Calculate the mass (in grams) of a piece of aluminum having a volume of 381 mL . Express yo

ur answer to three significant figures.
Chemistry
2 answers:
laiz [17]3 years ago
4 0

Answer:

Mass ≈ 1030g (Three significant Figures)

Explanation:

We were given;

Density = 2.70 g/mL

Mass = ?

Volume = 381 mL

The relationship between all three parameters mentioned above is given by the formular below;

Density = Mass / Volume

Upon solving for Mass, we have;

Mass = Density * Volume

Substituting the respective values;

Mass = 2.7 * 381

Mass = 1028.7g

Mass ≈ 1030g (Three significant Figures)

BaLLatris [955]3 years ago
3 0
Multiply the volume by the density. 1 ml = 1 cm^3

1.11 g/cm^3 * 1 cm^3/mL * 386 mL = 428 g

All your units cancel out except for g which is what you would expect since you want units of mass.
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Without wasting much of our time, Here is the correct question.

Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen hemoglobin(aq) + O2(aq) -------> hemoglobin O2(aq) is first order in hemoglobin and first order in dissolved oxygen, with a rate constant of 4 × 10⁷ L mol⁻¹ s⁻¹. Calculate the initial rate at which oxygen will be bound to hemoglobin if the concentration of hemoglobin is 2 × 10⁻⁹ M and that of oxygen is 5 × 10⁻⁵M.

Answer:

4 × 10⁻⁶ M s⁻¹

Explanation:

The equation for the reaction between Hemoglobin molecules in blood that binds with oxygen molecule can be represent by:

hemoglobin_{(aq)  +  O_{2(aq)   ---------> hemoglobin.O_{2(aq)

Now, we are also being told to calculate only!, the  initial rate at which oxygen will be bound to hemoglobin.

So, If it is first order in hemoglobin and also first order in Oxygen molecule at the initial rate of the the reaction, therefore, the rate  for the reaction can be expressed as :

rate = k [hemoglobin_{(aq)}][O_{2(aq)}]

Let's not forget that we are so given some parameters;

where

k (rate constant) = 4 × 10⁷ L mol⁻¹ s⁻¹

[ hemoglobin_{(aq) ] = 2 × 10⁻⁹ M

[  O_{2(aq)  ]  =  5 × 10⁻⁵ M

Substituting our data given into the above rate formula, we have:

rate = (4 × 10⁷ L mol⁻¹ s⁻¹) × (2 × 10⁻⁹ M) × (5 × 10⁻⁵ M)

rate = 4 × 10⁻⁶ M s⁻¹     ( given that 1 M = 1 mol L⁻¹ )

∴ the initial rate at which oxygen will be bound to hemoglobin = 4 × 10⁻⁶ M s⁻¹

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