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3 years ago

B. What are the units for the atomic radius?

1 answer:

5 0

Answer:Units used to measure atomic radius: Angstroms (Å): This the most common unit used. Equivalent to 1.0 x 10-10 meters. Nanometer (nm): Equivalent to 1.0 x 10-9 meters.

Explanation:

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A chemist wants a 0. 100 M solution of NaOH. She has 0. 0550 mol NaOH available. What volume of solution is needed to make the m

yuradex [85]

Answer:the answer is 0.0550L

Explanation:

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2 years ago

Under certain conditions Argon gas diffuses at a rate of 3.2 cm per second under the same conditions an unknown gas diffuses at

kozerog [31]

Answer:

20 g/mol

Explanation:

We can use <em>Graham’s Law of diffusion</em>:

The rate of diffusion (<em>r</em>) of a gas is inversely proportional to the square root of its molar mass (<em>M</em>).

$r = \frac{1 }{\sqrt{M}}$

If you have two gases, the ratio of their rates of diffusion is

$\frac{r_{2}}{r_{1}} = \sqrt{\frac{M_{1}}{M_{2}}}$

Squaring both sides, we get

$(\frac{r_{2}}{r_{1}})^{2} = \frac{M_{1}}{M_{2}}$

Solve for <em>M</em>₂:

$M_{2} = M_{1} \times (\frac{r_{1}}{r_{2}})^{2}$

$M_{2} = \text{39.95 g/mol} \times (\frac{\text{3.2 cm/s}}{\text{4.5 cm/s}})^{2}= \text{39.95 g/mol} \times (0.711 )^{2}$

$= \text{39.95 g/mol} \times 0.506 = \textbf{20 g/mol}$

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3 years ago

I need help with this question

KiRa [710]

A challenge they could face is that the plants are not used to the climate in the new area . (sorry if it’s wrong)

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3 years ago

Which of the following represents the velocity of a light wave?

Softa [21]

Answer:

Explanation:

c = fλ

where" c "is the speed of light ,

• "f" is the frequency of the electromagnetic waves,

• " λ" is its wavelength.

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3 years ago

C. If 62.9 g of lead (II) chloride is produced, how many grams of lead (II) nitrate were

melomori [17]

Answer:

Mass = 76.176 g

Explanation:

Given data:

Mass of lead(II) chloride produced = 62.9 g

Mass of lead(II) nitrate used = ?

Solution:

Chemical equation:

Pb(NO₃)₂ + 2HCl → PbCl₂ + 2HNO₃

Number of moles of lead(II) chloride:

Number of moles = mass/molar mass

Number of moles = 62.9 g/ 278.1 g/mol

Number of moles = 0.23 mol

Now we will compare the moles of lead(II) chloride with Pb(NO₃)₂ from balance chemical equation:

PbCl₂ : Pb(NO₃)₂

1 : 1

0.23 : 0.23

Mass of Pb(NO₃)₂:

Mass = number of moles × molar mass

Mass = 0.23 mol × 331.2 g/mol

Mass = 76.176 g

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4 years ago