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Alexxandr [17]
4 years ago
11

A titanium cube contains 2.86×1023 atoms. What is the edge length l of the cube? The density of titanium is 4.50 g/cm3 . (The v

olume of a cube is V=l3.) Express the length in centimeters to three significant figures.
Chemistry
1 answer:
elena55 [62]4 years ago
4 0

Answer:

I = 1.716 cm

Explanation:

We want the length of the cube. In this case, we know that the volume of a cube is:

V = I³

From here we can solve for I:

I = ∛V  (1)

Now, the volume can be obtained using data of density. The expression for density is:

d = m/V

Solvinf for V:

V = m/d (2)

The mass of the cube, can be obtained with the moles and molar mass of titanium:

m = n * MM (3)

And finally, the moles (n) can be obtained using avogadro number:

1 mol ----> 6.02x10²³ atoms (4)

Therefore, to get the length we need to calculate with the atoms of titanium, the moles, then mass, volume and finally the length. Let's calculate the moles of titanium using the ratio of (4):

nTi = 2.86x10²³ / 6.02x10²³ = 0.475 moles

Now, let's calculate the mass using (3) and the molecular mass of titanium which is 47.867 g/mol:

m = 0.475 * 47.867 = 22.737 g

Let's replace this value in (2) to get the volume:

V = 22.737 / 4.5 = 5.053 cm³

Finally, the length using (1):

I = ∛5.053

I = 1.716 cm

This is the length of the cube

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8 0
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The discharge of chromate ions (CrO42-) to sewers or natural waters is of concern because of both its ecological impacts and its
muminat

Answer:

45727g

Explanation:

So, have the overall ionic equation given as the following;

CrO42^- + 3 Fe2^+ + 8 H2O ------> Cr(OH)^3(s) + 3 Fe(OH)^3(s) + 4 H^+.

So, we have (from the question) that the amount or quantity of the waste stream daily = 60m^3/h, and the waste stream daily contains waste stream containing = 4.0 mg/L Cr, and the discharge limit = 0.1 mg/L.

Step one: convert m^3/ h to L/h. Therefore, 60 m^3/h × 1000dm^3 = 60000 L/h .

Step two: Determine or calculate the the value of Cr used up.

The value of Car used up ={ 60,000 × ( 4.0 - 0.1) } ÷ 1000 = 234 g.

Step three: Determine or calculate the mass of Cr(OH)3 and the mass of Fe(OH)3.

The number of moles of Cr = 234/52 = 4.5 moles.

Molar mass of Cr(OH)3 = 103 g/mol and the molar mass of Fe(OH)3 = 106.8 g/mol.

Thus, the mass of Cr(OH)3 = 4.5 × 103 = 463.5 g.

And the mass of Fe(OH)3 = 13.5 × 106.8 = 1441.8 g.

Hence, the total = 463.5 g + 1441.8 g = 1905.3 g.

Step four: Determine or calculate the How much particulate matter would be generated daily.

The amount of the particulate that would be generated daily = 24 × 1905.3 = 45727g.

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Name for this compound Ag2O
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it is silver oxide my friend mark as brainlest t

tthanxxxxx

silver oxide

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