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2 years ago

13

What is the mass of a 1.71-l sample of a liquid that has a density of 0.921g/ml?

1 answer:

Temka [501]2 years ago

8 0

Hey there!:

Volume in mL :

1.71 L * 1000 => 1710 mL

Density = 0.921 g/mL

Therefore:

Mass = Density * Volume

Mass = 0.921 * 1710

Mass = 1574.91 g

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The electrons formed from the aerobic oxidation of glucose are

SVEN [57.7K]

Answer:

i and ii

Explanation:

In the aerobic oxidation of glucose, the electrons formed are transferred to O2 after several others transfer reactions like passing through coenzymes NAD+ and FAD

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2 years ago

4. What type of evidence do you think would be most difficult to collect

mars1129 [50]

I think it is trace evidence since it is really small and hard to find.

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3 years ago

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What is the molarity of a 2.4-liter solution containing 124 grams of HF?

ioda

Molar mass:

HF = 1 + 19 = 20.0 g/mol

Number of moles :

124 / 20.0 => 6.2 moles

Volume = 2.4 L

M = n / V

M = 6.2 / 2.4

M = 2.6 M

Answer A

hope this helps!

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3 years ago

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Titanium has an HCP crystal structure, a c/a ratio of 1.669, an atomic weight of 47.87 g/mol, and a density of 4.51 g/cm3. Compu

IrinaK [193]

Answer : The atomic radius for Ti is, $1.45\times 10^{-8}cm$

Explanation :

Atomic weight = 47.87 g/mole

Avogadro's number $(N_{A})=6.022\times 10^{23} mol^{-1}$

First we have to calculate the volume of HCP crystal structure.

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times V}$ .............(1)

where,

$\rho$ = density = $4.51g/cm^3$

Z = number of atom in unit cell (for HCP = 6)

M = atomic mass = 47.87 g/mole

$(N_{A})$ = Avogadro's number

V = volume of HCP crystal structure = ?

Now put all the values in above formula (1), we get

$4.51g/cm^3=\frac{6\times (47.87g/mol)}{(6.022\times 10^{23}mol^{-1}) \times V}$

$V=1.06\times 10^{-22}cm^3$

Now we have to calculate the atomic radius for Ti.

Formula used :

$V=6R^2c\sqrt{3}$

Given:

c/a ratio = 1.669 that means, c = 1.669 a

Now put (c = 1.669 a) and (a = 2R) in this formula, we get:

$V=6R^2\times (1.669a)\sqrt{3}$

$V=6R^2\times (1.669\times 2R)\sqrt{3}$

$V=(1.669)\times (12\sqrt{3})R^3$

Now put all the given values in this formula, we get:

$1.06\times 10^{-22}cm^3=(1.669)\times (12\sqrt{3})R^3$

$R=1.45\times 10^{-8}cm$

Therefore, the atomic radius for Ti is, $1.45\times 10^{-8}cm$

3 0

3 years ago

If the gram-formula mass of substance X is 180 g/mol, determine the molarity of the solution at Point E

Alina [70]

<h3><u>Answer;</u></h3>

<u>= 5 M or 5 moles/liter</u>

<h3><u>Explanation;</u></h3>

At point E, 90 g of substances X are dissolved in 100 g of the solvent.

100g of the solvent is equal to 100 ml

Molarity is the number of moles of a substance in one liter of a solvent.

90 g of X are in 100 ml

But; the RFM of X = 180 g/l

Therefore; the moles of X in 90 g = 90/180

= 0.5 moles

Therefore;

0.5 moles of X are contained in 100 ml of the solvent;

Thus, molarity = 0.5 × 1000/100

=<u> 5 M or 5 moles/liter</u>

7 0

3 years ago