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Trava [24]
3 years ago
6

How many moles of sulfur dioxide are in 2.26x10^33 sulfur dioxide molecules?

Chemistry
2 answers:
m_a_m_a [10]3 years ago
4 0

Answer:

It would be about 3 moles

Alexus [3.1K]3 years ago
3 0

Answer:moles = no. of molecules / Avogadro's number

          = 2.26 x 10^33 / 6.022 x 10^23

          = 3752906011

Round to significant figures which is 3 = 3.75 x 10^9 mol

Explanation:

The formula for finding how many moles of a substance when given the amount of molecules is: moles = number of molecules / Avogadro's number

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How many are molecules ( or formula) in each sample?
andre [41]

Answer:

  • 4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}
  • 16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

<u>Explanation</u>:

<u>Number of molecules for 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}</u>

\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:

Atomic mass of Na + H + C + 3(O)  = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }

55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)

=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}

<u>Number of molecules for for \left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}</u>

\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.

=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

8 0
3 years ago
0.785 moles of N2,fill a balloon at 1.5 atm and 301 K.<br> What is the volume of the balloon?
Tcecarenko [31]

Answer:

V = 12.93 L

Explanation:

Given data:

Number of moles = 0.785 mol

Pressure of balloon = 1.5 atm

Temperature = 301 K

Volume of balloon = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will put the values.

V = nRT/P

V = 0.785 mol × 0.0821 atm.L/ mol.K × 301 K / 1.5 atm

V = 19.4 L /1.5

V = 12.93 L

7 0
3 years ago
If 1.0 gram of hydrogen reacts with 19.0 grams of fluorine, then what is the percent by mass of fluorine in the compound that is
astraxan [27]
The reaction between hydrogen (H2) and fluorine (F2) is given below,
                                   H2 + F2 ---> 2HF
One mole of both hydrogen and fluorine yields to 2 moles of hydrogen fluoride. This can also be expressed as, 2 grams of hydrogen and 38 grams of fluorine will form 40 grams of hydrogen fluoride. From the given, only 20 grams of HF is formed with 19 g of it being fluorine. Thus, the percentage fluorine of the compound formed is 95%. 
8 0
3 years ago
Read 2 more answers
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mihalych1998 [28]
This isn’t anything related to chemistry dude
7 0
3 years ago
I MARK AS BRAINLIEST IF YOU ANSWER QUICK!
Sedaia [141]

Answer:

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Explanation:

hope this helps :)

5 0
3 years ago
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