All categories

2 years ago

How many moles of nitrogen, N, are in 0.286 mil of ammonium nitrate?

Chemistry

1 answer:

Arisa [49]2 years ago

8 0

Answer:

0.572

Explanation:

The formula for ammonium nitrate is

NH4NO3

The trick here is to realize the 4 applies only to the hydrogen. The 3 applies only to the oxygen.

There are 2 nitrogens in the formula. So if the entire molecule represents 0.286 mols. The individual mols of each part of the formula is just multiplied by the number of moles of the molecule. In formula form

Number of moles of the atoms making up the molecule = number of atoms from the formula * mols of the molecule.

2*0.286 = 0.572 moles of Nitrogen

You might be interested in

Which compound is most likely to have a sweet smell? which compound is most likely to have a sweet smell? ch3−ch2−nh2 o∥ch3−c−o−

Nady [450]

The compound that will have a sweet smell would be the one, whereby the molecular formula closely resembles that of an ether

R-O-R.

I believe the third one

5 0

2 years ago

1. What is the wave speed of a wave that has a frequency of 100 Hz and a wavelength of 0.30 m?

Sav [38]

Wave speed = frequency x wavelength
= 100 x 0.3
= 30 m/s.

Hope this helps!

5 0

2 years ago

Read 2 more answers

If nitrogen and hydrogen combine in a combustion reaction, what would the product of the reaction be?

kow [346]

Co2

Explanation:

CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)

4 0

2 years ago

Read 2 more answers

How are genes and alleles related

Montano1993 [528]

Alleles are inside a gene and genes are made up by alleles. Also, a gene is DNA, so the allele is like piece of DNA inside a gene<span>.

Hope this helps:)
</span>

7 0

2 years ago

25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca

Kamila [148]

Answer:

Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

Explanation:

Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

7 0

2 years ago