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vredina [299]
3 years ago
9

How many grams is 4.2X10^24 atoms of sulfur?

Chemistry
1 answer:
ipn [44]3 years ago
7 0
First find the number of moles of sulfur using dimensional analysis with avogadro’s number as the conversion factor. 4.2*10^24 atoms * (1 mol/6.022*10^23 atoms) = 7.0 mol sulfur. The molar mass of sulfur is 32.06 g/mol, which is found on the periodic table as sulfur’s (S) atomic weight. Use dimensional analysis again with the molar mass of sulfur as the conversion factor. 7.0 mol * 32.06 g/mol = 224.42 g sulfur. Since the problems gives us two significant figures, round the mass of sulfur to 220 grams, or 2.2 * 10^2 g.
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4 years ago
C. A pure substance has "one set of universal properties". What does this mean?
ICE Princess25 [194]

A pure substance has "one set of universal properties". This means they have some of the universal properties in common.

<h3>The definition of universal property</h3>

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4 0
1 year ago
What is the function of this instrument?
chubhunter [2.5K]
To measure weight of a item
7 0
4 years ago
How many hydrogen atoms are in 89.5 g of<br> C6H6 ?<br> Answer in units of atoms.
Elodia [21]

Solution :

Molar mass of C_6H_6 is :

M = 6×12 + 6×1 g

M = 78 g

78 gram of C_6H_6 contains 6.022 \times 10^{23} molecules.

So, 89.5 gram of C_6H_6 contains :

n = 6.022 \times 10^{23} \times \dfrac{89.5}{78}\\\\n = 6.91 \times 10^{23}

Now, from the formula we can see that one molecule of C_6H_6 contains 2 hydrogen atom . So, number of hydrogen atom are :

h = 2\times 6.91 \times 10^{23}\\\\h = 1.38 \times 10^{22}\ atoms

Hence, this is the required solution.

8 0
3 years ago
Under standard-state conditions, which of the following half-reactions occurs at the cathode during the electrolysis of aqueous
Slav-nsk [51]

Answer:

The following reaction will occur at cathode:

Ni^{+2}(aq)+2e--->Ni(s)

Explanation:

The two half reaction during electrolysis of aqueous nickel sulfate will be

a) anode reaction :

Water will undergo oxidation and will evolve oxygen gas at anode as shown in the given reaction:

2H_{2}O(l) ----> O_{2}(g) + 4H^{+}(aq) + 4e

b) Cathode reaction: The reduction of Nickel ion will occur by gain of two electrons as shown in the given equation:

Ni^{+2}(aq)+2e--->Ni(s)

Thus the overall reaction will be:

2H_{2}O(l) + 2Ni^{+2}----> O_{2}(g) + 4H^{+}(aq) + 2Ni(s)

6 0
3 years ago
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