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Ugo [173]
2 years ago
6

How many molecules are in 0.25 grams of dinitrogen pentoxide?

Chemistry
2 answers:
ZanzabumX [31]2 years ago
7 0

Answer: 0.014\times 10^{23}molecules

Explanation: Moles can be calculated by using the formula:

\text{no of moles}=\frac{\text{given mass}}{\text{Molecular mass}}

\text{no of moles}=\frac{0.25g}{108g/mol}=0.002moles

1 mole of N_2O_5 contains =6.023\times 10^{23}molecules

0.002 mole of N_2O_5 contains =\frac{6.023\times 10^{23}}{1}\times 0.002=0.014\times 10^{23}molecules

Vesna [10]2 years ago
4 0
In order to get the answer, you have to know the formula of how to convert grams to mole and by that you can get the number of molecules.
The answer is 2 moles of N2O5. This is shown by the solution:
(0.25 g N2O5) (1 mol/ 108 g)=2.31 molecules
The answer is 2 molecules.
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A 1.0-gram sample of solid iodine is placed in a tube and the tube is sealed after all of the air is removed. The tube and the s
yKpoI14uk [10]

Answer:

27.0

Explanation:

Because Mass can neither be created nor be destroyed hence total mass of sample of iodine and tube remain equal as it is sealed.

7 0
2 years ago
6.0 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 13
vodomira [7]

<u>Answer:</u>

<em>The molecular formula of X is given as C_7 H_6 O_3</em>

<em></em>

<u>Explanation:</u>

Moles $C O_{2}=\frac{\text { mass }}{\text { molar mass }}=\frac{13.39 \mathrm{g}}{44.01 \mathrm{g} \text { per mole }}=0.304 \mathrm{mol}$\\\\moles $\mathrm{C}=$ moles $\mathrm{CO}_{2}=0.304 \mathrm{mol}$

mass $C=$ moles $\times$ molar mass $=0.304 \mathrm{mol} \times 12 \frac{g}{m o l}=3.65g$\\\\moles $\mathrm{H}_{2} \mathrm{O}=\frac{2.35 \mathrm{g}}{18.02 \mathrm{g} \text { permole }}=0.130 \mathrm{mol}$\\\\moles $\mathrm{H}=2 \times$ moles $\mathrm{H}_{2} \mathrm{O}=0.130 \times 2=0.260 \mathrm{mol}$\\\\Mass $\mathrm{H}=0.260 \mathrm{mol} \times 1.008 \frac{g}{\mathrm{mol}}=0.262 \mathrm{g}$

mass O = Total mass of the compound - (mass of C + mass of H)

=6.0 g - ( 3.65 + 0.262 ) g

=2.09 g

moles $O=\frac{2.09 g}{16 g \text { per mole }}=0.131 \mathrm{mol}$

Least moles is for O that is 0.131mol and dividing all by the least we get

$\begin{aligned} C &=\frac{0.304}{0.131}=2.3 \\\\ H &=\frac{0.260}{0.131}=2 \\\\ O &=\frac{0.131}{0.131}=1 \end{aligned}$

Since 2.3 is a fraction it has to be converted to a whole number so we multiply all the answers by 3

\\$C 2.3 \times 3=7$\\\\$H 2 \times 3=6$\\\\$O 1 \times 3=3$

So the empirical formula is C_7 H_6 O_3

Empirical formula mass

=(7 \times 12) +(6\times1.008)+(3\times16)=138.048g

$n=\frac{\text { molar mass }}{\text { empirical formula mass }}=\frac{138}{138.048}=1$

Molecular formula =n × empirical formula

=1 \times C_7 H_6 O_3

Compound X  = C_7 H_6 O_3  is the Answer

8 0
2 years ago
What is the pH of a 0.75 M HNO3 solution
Sauron [17]
Hello!

The dissociation reaction of HNO₃ is the following:

HNO₃ → H⁺ + NO₃⁻

This is a strong acid, so the concentration of HNO₃ would be the same as the concentration of H⁺. The formula for pH is the following:

pH=-log([H_3O^{+}])=-log(0,75M)=0,12

So, the pH would be 0,12

Have a nice day!
4 0
2 years ago
Read 2 more answers
Which solution will have a higher boiling point: A solution containing 15 grams of sucrose (C12H22O11) in 25 grams of water or a
Angelina_Jolie [31]
A solute rises the boiling point of a solution, in direct relation with the number of particles added to the solution. Sucrose remains a molecule, does not separate into anything. NaCl gives Na+ + Cl-. 

<span>Molar mass of sucrose is 12*12+22*1+11*16=144+22+176=342 </span>

<span>105g sucrose is 105/342=0.3moles ---> 0.3 moles of particles (molecules) </span>

<span>Molar mass of NaCl is 23+35.5=58.5 </span>

<span>35 grams of NaCl is 35/58.5=0.598 ----->0.598*2=1.1974 moles of particles (ions, Na+ and Cl-) </span>

<span>So, 35 grams of sodium chloride in 500 grams of water will have a higher boiling point</span>
5 0
2 years ago
A 25-ml sample of river water was titrated with 0.0010 m k2cr2o7 and required 8.3 ml to reach the end point. what is the chemica
arsen [322]

First let us calculate the moles of K2Cr2O7 that was supplied.

moles K2Cr2O7 = 0.0010 M * 0.0083 L = 8.3x10^-6 mol

 

From the chemical formula itself, we see that there are 7 O for every mole of K2Cr2O7 or 3.5 O2. Therefore:

moles O2 = 8.3x10^-6 mol K2Cr2O7 * (3.5 mol O2 / 1 mol K2Cr2O7)

moles O2 = 2.905x10^-5 mol O2

 

Calculating for the mass of O2 in mg:

mass O2 = 2.905x10^-5 mol O2 * (32 g / mol) * (1000 mg / g)

mass O2 = 0.9296 mg

 

Therefore the chemical oxygen demand (COD) is:

COD = 0.9296 mg / (0.025 L)

<span>COD = 37.184 mg/L</span>

7 0
2 years ago
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