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2 years ago

What is the volume of NH4 NO3(s) — N2O(g) + 2H2O(g)

Chemistry

1 answer:

tatiyna2 years ago

8 0

Answer:

Explanation:

NH4NO3(s) N2O(g) + 2H2O(g) What is the total volume of gases produced at 546 K and 1.00 atm pressure when 240 g of ammonium nitrate undergoes the above decomposition reaction? A) 9 × 22.4 L B) 3 × 22.4 L C) 12 × 22.4 L D) 18 × 22.4 L E) 6 × 22.4 L

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Where are the protons in an atom located?

densk [106]

Protons and neutrons are found in the nucleus. They group together in the center of the atom.

The answer is D

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3 years ago

Nitric acid is a key industrial chemical, largely used to make fertilizers and explosives. The first step in its synthesis is th

kolbaska11 [484]

Answer: When using 645 L /s of O2 in a temperature and pressure of 195°C, 0.88 atm respectively, we will get 0.355Kg /s NO

Explanation:

First we review the equation that represents the oxidation process of the NH3 to NO.

4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l)

Second we gather the information what we are going to use in our calculations.

O2 Volume Rate = 645 L /s

Pressure = 0.88 atm

Temperature = 195°C + 273 = 468K

NO molecular weight = 30.01 g/mol

Third, in order to calculate the amount of NO moles produced by 645L/ s of O2, we must find out, how many moles (n) are 645L O2 by using the general gas equation PV =n RT

Let´s keep in mind that using this equation our constant R is 0.08205Lxatm/Kxmol

PV =n RT

n= PV / RT

n= [ 0.88atm x 645L/s] / [ (0.08205 Lxatm/Kxmol) x 468K]

n= 14.781 moles /s of O2

Fourth, now by knowing the amount of moles of O2, we can use the equation to calculate how many moles of NO will be produced and then with the molecular weight, we will finally know the total mass per second .

14.781 moles /s of O2 x 4moles NO / 5 moles O2 x 30.01g NO / 1 mol NO x 1Kg NO /1000g NO = 0.355Kg /s NO

6 0

3 years ago

50.g of NaNO3 was dissolved in 1250 mL of water. what is the molality of the solution? [ Molar mass of NaNO3 = 85 g/mol

Viefleur [7K]

Answer:

Approximately $0.47\; \rm mol \cdot L^{-1}$ (note that $1\; \rm M = 1 \; \rm mol \cdot L^{-1}$ .)

Explanation:

The molarity of a solution gives the number of moles of solute in each unit volume of the solution. In this $\rm NaNO_3$ solution in water,

Let $n$ be the number of moles of the solute in the whole solution. Let $V$ represent the volume of that solution. The formula for the molarity $c$ of that solution is:

$\displaystyle c = \frac{n}{V}$ .

In this question, the volume of the solution is known to be $1250\; \rm mL$ . That's $1.250\; \rm L$ in standard units. What needs to be found is $n$ , the number of moles of $\rm NaNO_3$ in that solution.

The molar mass (formula mass) of a compound gives the mass of each mole of units of this compound. For example, the molar mass of $\rm NaNO_3$ is $85\; \rm g \cdot mol^{-1}$ means that the mass of one mole of

$\displaystyle n = \frac{m}{M}$ .

For this question,

$\begin{aligned}&n\left(\mathrm{NaNO_3}\right) \\ &= \frac{m\left(\mathrm{NaNO_3}\right)}{M\left(\mathrm{NaNO_3}\right)}\\&= \frac{50\; \rm g}{85\; \rm g \cdot mol^{-1}} \\& \approx 0.588235\; \rm mol\end{aligned}$ .