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Basile [38]
3 years ago
13

How many atoms are in 355 mol Na?

Chemistry
1 answer:
eimsori [14]3 years ago
5 0

Answer:

2.138 x 10^26 atoms Na

Explanation:

atoms Na = 355 mol Na x (6.022 x 10^23 atoms Na/1 mol Na) = 2137.81 x 10^23 atoms Na = 2.138 x 10^26 atoms Na

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How many milligrams of product can be produced from the complete Diels-Alder reaction of 180. mg of anthracene and 100. mg of ma
ZanzabumX [31]

Answer: It will be produced 276,3 mg of product

Explanation: The reaction of anthracene (C14H10) and maleic anhydride (C4H2O3) produce a compound named 9,10-dihydroanthracene-9,10-α,β-succinic anhydride (C18H12O3), as described below:

C14H10 + C4H2O3 → C18H12O3

The reaction is already balanced, which means to produce 1 mol of C18H12O3 is necessary 1 mol of anthracene and 1 mol of maleic anhydride.

1 mol of C14H10 equals 178,23 g. As it is used 180 mg of that reagent, we have 0,001 mol of anthracene. With it, the reaction produces 0,001 mol of C18H12O3.

As 1 mol of C18H12O3 equals 276,3 g, the mass produced is 276,3 mg.

3 0
3 years ago
an engineering team is conducting a trial launch of a few rocket . which part is the engineering process is the team in ?
brilliants [131]

Answer:

Evaluate the result (Apex)

Explanation:

7 0
3 years ago
6. What volume would be occupied by 7.75 g of this same substance?
Kaylis [27]
D = m/V 
<span>V = m/d
V= 7.75 g / 1.71 g/cm^3
V= 4.53 c</span>m^3
7 0
4 years ago
If 26.2 mL of AgNO3 is needed to precipitate all the Cl− ions in a 0.785-mg sample of KCl (forming AgCl), what is the molarity o
sweet-ann [11.9K]

<u>Answer:</u>

<em>The molarity of the AgNO_3 solution is 4.02 \times 10^4 M </em>

<em></em>

<u>Explanation:</u>

The Balanced chemical equation is

1AgNO_3 (aq) +1KCl (aq) > 1 AgCl (s)+1KNO_3 (aq)

Mole ratio of AgNO_3 : KCl is 1 : 1

So moles AgNO_3  = moles KCl

Moles KCl = \frac {mass}{molarmass}

= \frac {0.785 mg}{(39.1+35.5 g per mol)}

= \frac {0.000785 g}{74.6 g  per mol}

= 0. 0000105 mol KCl

= 0.0000105 mol AgNO_3

So  Molarity

= \frac {moles of solute}{(volume of solution in L)}

= \frac {0.0000105 mol}{26.2 mL}

=\frac {0.0000105 mol}{0.0262 L}

= 0.000402M or mol/L is the Answer

(Or) 4.02 \times 10^4 M is the Answer

6 0
3 years ago
Calculate these masses.
Alex
A) mass / volume = density
m/6.00cm3 = 13.5939g/cm3
m = 13.5939g/cm3 • 6.00cm3
m = 81.6g

B) mass / volume = density
m/25.0cm3 = 0.702g/cm3
m = 0.702cm3 • 25.0g/cm3
m = 17.6g

1mL = 1cm3
8 0
4 years ago
Read 2 more answers
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