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3 years ago

Which of the following statements is correct?

Chemistry

2 answers:

marshall27 [118]3 years ago

8 0

Answer:

that was wrong

Explanation:

RoseWind [281]3 years ago

7 0

D. Blue litmus paper turns red when placed in a base.

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Fudgin [204]

For every 2 Mol NaOH you would get 1 Mol N2H4

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2 years ago

When iocnic solids dissolve in water, the break down into the ____ ions (cations) and ___ ions (anions) that they contain?

ddd [48]

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3 years ago

Why is a bannana curved?

Yanka [14]

Answer:

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2 years ago

Read 2 more answers

The radius of a vanadium atom is 130 pm. How many vanadium atoms would have to be laid side by side to span a distance of 1.30 m

monitta

Answer:

5 000 000 (5 million atoms)

Explanation:

Let us assume that a vanadium atom has a spherical shape.

diameter of a sphere = 2 x radius of the sphere

Thus,

Radius of a vanadium atom = 130 pm

= 130 x $10^{-12}$ m

The diameter of a vanadium atom = 2 x radius

= 2 x 130 x $10^{-12}$

= 260 x $10^{-12}$ m

Given a distance of 1.30 mm = 1.30 x $10^{-3}$ m,

The number of vanadium atoms required to span the distance = $\frac{1.3*10^{-3} }{260*10^{-12} }$

= 5000000

Therefore, the number of vanadium atom that would span a distance of 1.30 mm is 5 million.

3 0

3 years ago

Ne4ueva [31]

Answer:

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.

Explanation:

Volume of cylindrical reservoir = V

Radius of the cylindrical reservoir = r = d/2

d = diameter of the cylindrical reservoir = d = $4.50\times 10^1 m=45 m$

r = d/2 = 22.5 m

Depth of the reservoir = h = 10.0 m

$V=\pi r^2 h$

$=3.14\times (22.5 m)^2\times 10.0 m=15,896.25 m^3=15,896,250 L$

$1 m^3=1000 l$

Volume of water cylindrical reservoir : V

Density of water,d = 1 kg/L

Mass of water cylindrical reservoir = m

$m=d\times V=1 kg/L\times 15,896,250 L=15,896,250 kg$

1.6 kilogram of fluorine per million kilograms of water. (Given)

Concentration of fluorine in water = 1.6 kg/ 1000,000 kg of water

In 1000,000 kg of water = 1.6 kg of fluorine

Then $15,896,250 kg$ of water have x mass of fluorine:

$\frac{x}{15,896,250 kg\text{kg of water}}=\frac{1.6 kg}{1000,000 \text{kg of water}}$

$x=\frac{1.6 kg}{1000,000}\times 15,896,250 kg=25.434 kg$

15,896,250 kg water of contains mass 25.434 kg of fluorine.

25.434 kg = 25434 g

25,434 grams of fluorine should be added to give 1.6 ppm.

Percentage of fluorine in hydrogen hexafluorosilicate :

Molar mass hydrogen hexafluorosilicate = 144 g/mol

$F\%=\frac{6\times 19 g/mol}{144 g/mol}\times 100=79.16\%$

Total mass of hydrogen hexafluorosilicate = m'

$79.16\%=\frac{25,434 g}{m'}\times 100$

m' = 32,127.02 g

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.

5 0

2 years ago