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Ivenika [448]
3 years ago
6

If you stacked Avogadro's number of pennies one on top of the other on Earth's surface, how far would the stack extend (in kilom

eters)? [For comparison, the sun is about 150 million km from Earth and the nearest star (Proxima Centauri) is about 40 trillion km from Earth.] Express your answer using four significant figures.
Chemistry
1 answer:
attashe74 [19]3 years ago
5 0

Answer:

Avogadro number of pennies will extend to a distance of 6.02 * 10¹⁷ km

<em>Note: The question is missing some parts. The complete question is as follows;</em>

<em>A penny has a thickness of approximately 1.0 mm . If you stack ed Avogadro's number of pennies one on top of the other on Earth 's surface, how far would the stack extend (in km)? [For comparison, the sun is about 150 million km from Earth and the nearest star (Proxim a Centauri) is about 40 trillion km from Earth.]</em>

Explanation:

Avogadro number = 6.02 * 10²³

thickness of a penny = 1.0 mm

I mm = 0.001 m

Thickness of Avogadro number of pennies stacked one upon another will be:

6.02 * 10²³ * 0.001 m = 6.02 * 10²⁰ m

Distance in km;

1 m = 0.001 km

therefore, 6.02 * 10²⁰ m = 6.02 * 10²⁰ * 0.001 km = 6.02 * 10¹⁷ km

Avogadro number of pennies will extend to a distance of 6.02 * 10¹⁷ km

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Which of the following descriptions of the storage requirements for alkali metals is the most accurate? Select the correct answe
elena55 [62]

Answer:

A. They must be kept separate from moisture and oxygen

Explanation:

Alkali metals refers to a group of chemical elements in the periodic table. It is the common name given to Group 1 elements (excluding hydrogen). They are generally metallic in nature with physical properties like shining lustre, silvery in appearance etc. Alkali metals include lithium, Pottasium, Sodium, Rubidium etc.

One important chemical property of alkali metals is their high reactivity i.e. they tend to react quickly with other substances. For example, alkali metals react vigorously with water to form hydroxides. They also react with oxygen to form oxides. Due to this highly reactive properties, they are kept or stored away from moisture (water) and oxygen (present in air).

7 0
2 years ago
What is the concentration of an unknown Mg(OH)2 solution if it took an average of 15.4mL of
vova2212 [387]

Answer:

0.077M

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2HCl + Mg(OH)2 —> MgCl2 + 2H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 2

The mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question.

Concentration of base Cb =...?

Volume of base (Vb) = 10mL

Concentration of acid (Ca) = 0.1M

Volume of acid (Va) = 15.4mL

Step 3:

Determination of the concentration of the base, Mg(OH)2.

The concentration of the base can be obtained as follow:

CaVa/CbVb = nA/nB

0.1 x 15.4 /Cb x 10 = 2/1

Cross multiply to express in linear form

Cb x 10 x 2 = 0.1 x 15.4

Divide both side by 10 x 2

Cb = (0.1 x 15.4) /(10 x 2)

Cb = 0.077M

Therefore, the concentration of the base, Mg(OH)2 is 0.077M

7 0
3 years ago
A beaker was half filled with freshly distilled H2O and placed on a hot plate. As the temperature of the water reached 100°C, vi
jeka57 [31]

Answer:

If anything but H2O were found, it would be evidence of a chemical change. Therefore, 4 is the only viable evidence that it was a physical change of H2O liquid to H2O gas.

4 0
3 years ago
Read 2 more answers
A 72.0 mL aliquot of a 1.40 M solution is diluted to a total volume of 248 mL. A 124 mL portion of that solution is diluted by a
Aliun [14]

Answer: 0.20 M

Explanation:

According to the dilution law,

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 1.40 M

V_1 = volume of stock solution = 72.0 ml

M_2 = molarity of diluted solution = m

V_2 = volume of diluted solution = 248 ml

1.40\times 72.0=m\times 248

m=0.41M

Now 124 mL portion of this prepared solution is diluted by adding 133 mL of water.

According to the dilution law,

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 0.41 M

V_1 = volume of stock solution = 124 ml

M_2 = molarity of diluted solution = m

V_2 = volume of diluted solution = (124 +133) ml = 257 ml

0.41\times 124=m\times 257

m=0.20M

Thus the final concentration of the solution is 0.20 M.

8 0
3 years ago
In laboratory experiment, a NOVDEC Student was
Ede4ka [16]

Answer:

i. Molar mass of glucose = 180 g/mol

ii. Amount of glucose = 0.5 mole

Explanation:

<em>The volume of the glucose solution to be prepared</em> = 500 cm^3

<em>Molarity of the glucose solution to be prepared</em> = 1 M

i. Molar mass of glucose (C_1_2H_6O_6) = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol

ii.<em> mole = molarity x volume</em>. Hence;

amount (in moles) of the glucose solution to be prepared

                 = 1 x 500/1000 = 0.5 mole

3 0
2 years ago
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