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Anna [14]
3 years ago
12

A hydrogen atom has a radius of 2.5 x 10^11m

Chemistry
2 answers:
Delvig [45]3 years ago
7 0
Ndhsjsjd
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annandale
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g100num [7]3 years ago
4 0

Answer:

The radius of a Magnesium atom is (1.5).10^{-10}m

Explanation:

The question is wrong and a graph is missing.

It should be '' A hydrogen atom has a radius of (2.5).10^{-11}m ''

I also add the missing graph.

If we take a look at the graph, we can see that using that scale the atomic radius of the Hydrogen is 6mm and the atomic radius of Magnesium is 36mm

Therefore, the radius of the Magnesium atom

\frac{36mm}{6mm}=6

is six times bigger than the Hydrogen atom

If we multiply by 6 the radius of the Hydrogen atom ⇒

(6).(2.5).10^{-11}m=(1.5).10^{-10}m

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Verizon [17]
The answer is Ice spheres
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Which of the following would contain 48 electrons?
Helga [31]
The answer is cadmium its got 48 electrons which is y its number 48 on the period table
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A 25.0-mL sample containing Cu2+ gave an instrument signal of 25.2 units (corrected for a blank). When exactly 0.500 mL of 0.027
irga5000 [103]

Answer:

The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.

Explanation:

The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:

C = \frac{n}{V} \\0.0275 = \frac{n}{0.0005} \\

n = 1.375x10⁻⁵ mol

The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):

1.375x10⁻⁵ mol _________ 19.9 units

        x              _________  25.2 units

x = 1.741x10⁻⁵mol

Finally, we can calculate the Cu²⁺ concentration :

C = 1.741x10⁻⁵mol / 0.025 L

C = 6.964x10⁻⁴ M

7 0
3 years ago
What are oil and waters properties of matter?
yulyashka [42]

Answer:

Two other beneficial properties of the oil include the inhibition of corrosion of metal surfaces, such as steel, and the removal of dirt and detritus via detergency.

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4 0
3 years ago
How many grams of NaF should be added to 500 mL of a 0.100 M solution of HF to make a buffer with a pH of 3.2
Korolek [52]

Answer:

2.25g of NaF are needed to prepare the buffer of pH = 3.2

Explanation:

The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:

pH = pKa + log [A-] / [HA]

<em>Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF). </em>

The moles of HF are:

500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF

Replacing:

3.2 = 3.17 + log [A-] / [0.0500moles]

0.03 = log [A-] / [0.0500moles]

1.017152 = [A-] / [0.0500moles]

[A-] = 0.0500mol * 1.017152

[A-] = 0.0536 moles NaF

The mass could be obtained using the molar mass of NaF (41.99g/mol):

0.0536 moles NaF * (41.99g/mol) =

<h3>2.25g of NaF are needed to prepare the buffer of pH = 3.2</h3>
4 0
2 years ago
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